2025 Gauss Contests
Solutions
(Grade 7 and 8)
Wednesday, May 14, 2025
(in North America and South America)
Thursday, May 15, 2025
(outside of North American and South America)
Wednesday, May 14, 2025
(in North America and South America)
Thursday, May 15, 2025
(outside of North American and South America)
©2025 University of Waterloo
If each of \(12\) friends gives \(\$5\), in total they give \(12\times \$5=\$60\) to the charity.
Answer: (C)
The regular hexagon is divided into \(6\) triangles having equal area. Since \(1\) of these \(6\) triangles is shaded, then \(\frac16\) of the area of the hexagon is shaded.
Answer: (E)
Reading from the graph, Dae ate \(6\) apples, Joe ate \(3\), Etta ate \(5\), Susie ate \(4\), and Vinh ate \(1\). Thus, Dae ate the greatest number of
apples.
(We may have simply noted that the person who ate the greatest number of
apples is the person with the ’highest bar’.)
Answer: (A)
The equal-arm scale shows that \(2\) squares have the same mass as \(4\) circles.
If we split the \(2\) squares and the
\(4\) circles each into two equal
groups, then \(1\) square has the same
mass as \(2\) circles.
Answer: (B)
A square with side length 8 has area \(8\times8=64\). Since \(8\times2=16\), \(2\times(8+8)=2\times16=32\), \(4\times8=32\), and \(8\times8\times8\times8=4096\), then \(8\times8\) is the only expression given that is equal to the area of the square.
Answer: (C)
Since \(\angle PQR\) is a straight angle, its measure is \(180\degree\). Thus, the angles with measures \(130\degree\) and \(x\degree\) add to \(180\degree\), or \(130+x=180\) and so \(x=180-130=50\).
Answer: (C)
In a list of numbers having exactly one mode, the mode is the
number that appears in the list the greatest number of times. Since the
mode of the given list is \(8\), and
the list has exactly one mode, then \(8\) must appear more times than each of the
other numbers in the list.
If \(n=8\), then \(8\) appears in the list three times, which
is more than any other number in the list, and so the list has exactly
one mode, which is \(8\), as
required.
If the value of \(n\) is \(15\) or \(3\), then the list has three numbers that
each appear twice, and thus the list does not have exactly one
mode.
If \(n=9\), then \(9\) appears more times than any other
number and so the mode is \(9\).
If \(n=10\), then both \(8\) and \(9\) each appear twice (which is more than
any other number in the list), and thus the list does not have exactly
one mode.
Answer: (D)
Solution 1:
To measure \(1\) cup of flour, Sam
fills his \(\frac12\) cup container
\(2\) times (since \(\frac12\times2=1\)).
To measure \(2\) cups of flour, Sam
fills his \(\frac12\) cup container
\(2\times2=4\) times.
To measure \(2\frac12\) cups of flour,
Sam fills his \(\frac12\) cup container
\(4+1=5\) times.
Solution 2:
Since \(2\frac12\) cups is equal to \(\frac52\) cups, and \(5\times\frac12=\frac52\), then Sam fills his \(\frac12\) cup container 5 times.
Answer: (E)
There are \(7\) days in a week. If June \(1\) is a Tuesday, then by moving successively \(7\) days later in the month, we get June \(8\), \(15\), \(22\), and \(29\) are also Tuesdays. Thus, June \(30\) is a Wednesday.
Answer: (C)
When viewed from the opposite side of the window, text appears
reflected horizontally.
That is, P U G  F O R  S A L E appears
as . The letters that look the same when viewed from both sides of the
window are A, O and U and so there are \(3\) such letters.
Answer: (A)
The \(x\)-coordinate of \((2,6)\) is \(5\) less than the \(x\)-coordinate of \((7,0)\), and so \(R\) must be translated right \(5\).
The \(y\)-coordinate of \((2,6)\) is \(6\) more than the \(y\)-coordinate of \((7,0)\), and so \(R\) must be translated down \(6\).
Thus, the translation right \(5\), down
\(6\) will move \(R(2,6)\) to \((7,0)\).
Answer: (E)
A train stops at Waterloo Station at 6:25 a.m. and every \(3\) minutes thereafter, and so a train
stops only at times which differ from 6:25 a.m. by a multiple of \(3\). Of the given answers, a train stops at
the station at 6:28 a.m. (difference is \(3\) minutes), at 6:40 a.m. (difference is
\(15\) minutes), and at 6:55
a.m. (difference is \(30\)
minutes).
A bus stops at Waterloo Station at 6:25 a.m. and every \(5\) minutes thereafter, and so a bus stops
only at times which differ from 6:25 a.m. by a multiple of \(5\). Of the given answers, a bus stops at
the station at 6:30 a.m. (difference is \(5\) minutes), at 6:40 a.m. (difference is
\(15\) minutes), and at 6:55 a.m.
(difference is \(30\) minutes).
Thus, the next time that both a train and a bus stop at Waterloo Station
at the same time is 6:40 a.m.
Note: The lowest common multiple of both \(3\) and \(5\) is \(15\). This tells us that a train and a bus
will stop at the station every \(15\)
minutes after 6:25 a.m. This confirms that 6:40 a.m. is the next time
that both a train and a bus will stop at the station at the same
time.
Answer: (D)
The pattern repeats in blocks of the \(4\) numbers \(2\), \(0\), \(2\), \(5\). Since \(50=12\times4+2\), then the \(50\) numbers in the pattern contain \(12\) complete blocks of \(2\), \(0\), \(2\), \(5\)
followed by the first \(2\) numbers in
the block, \(2\), \(0\).
The number \(5\) appears once in each
of the \(12\) blocks and does not
appear as the 49th or 50th number. Thus, the number \(5\) appears \(12\) times.
Answer: (C)
Since \(\dfrac{28}{32}+\dfrac{1}{\Box}=1\) and \(\dfrac{28}{32}+\dfrac{4}{32}=\dfrac{32}{32}=1\), then \(\dfrac{1}{\square}=\dfrac{4}{32}\).
Reducing \(\dfrac{4}{32}\) to lowest terms, we get \(\dfrac{4}{32}=\dfrac{1}{8}\), and so the number that goes in the box is \(8\).
Answer: (E)
In the grid shown, the first column and first row show the possible numbers that may appear on the top faces of the two dice. The inside of the grid shows the sum of those corresponding two numbers.
| \(+\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) |
|---|---|---|---|---|---|---|
| \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) |
| \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) | \(8\) |
| \(3\) | \(4\) | \(5\) | \(6\) | \(7\) | \(8\) | \(9\) |
| \(4\) | \(5\) | \(6\) | \(7\) | \(8\) | \(9\) | \(10\) |
| \(5\) | \(6\) | \(7\) | \(8\) | \(9\) | \(10\) | \(11\) |
| \(6\) | \(7\) | \(8\) | \(9\) | \(10\) | \(11\) | \(12\) |
Of the \(36\) possible combinations of rolls, a sum of \(7\) can happen \(6\) different ways, a sum of \(8\) can happen \(5\) different ways, a sum of \(9\) can happen \(4\) different ways, a sum of \(10\) can happen \(3\) different ways, and a sum of \(11\) can happen \(2\) different ways. Thus, of the given sums, \(11\) is the least likely to occur.
Answer: (E)
For the greatest possible result, we want the sum of the first
two 2-digit numbers to be as large as possible, and the third 2-digit
number to be as small as possible.
Since the tens digit contributes more to the value of a number than the
ones digit, we choose the two largest digits, \(7\) and \(8\), as the tens digits of the first two
2-digit numbers, and the smallest of the given digits, \(1\), as the tens digit of the third 2-digit
number.
Of the remaining digits, \(3\), \(6\), and \(2\), we choose the two largest digits,
\(3\) and \(6\), as the ones digits of the first two
2-digit numbers, and \(2\) as the ones
digit of the third 2-digit number, which becomes \(12\).
Since \(73+86=76+83\), it does not
matter how we pair the tens digits with the ones digits for the first
two 2-digit numbers.
Thus, the greatest possible result is \(73+86-12=147\).
Answer: (C)
Solution 1:
We begin by recognizing that written as a fraction in lowest terms,
\(60\%\) is equal to \(\dfrac35\).
Thus, if Savanah tossed \(3\) tails out
of \(5\) tosses, then \(60\%\) of the tosses that she made were
tails.
Since the result of her last toss was tails, then following Savanah’s
second last toss she had tossed \(2\)
tails out of \(4\) tosses, meaning that
\(50\%\) of the tosses that she made to
that point were tails, as required.
Therefore, Savanah made \(5\) tosses in
total.
Solution 2:
We proceed to work backward from the answers given.
If Savanah made \(3\) tosses, and 60%
of those tosses were tails, then she tossed \(3\times\dfrac{60}{100}=\dfrac{180}{100}=1.8\)
tails. She must toss a whole number of tails, and so this is not
possible.
If Savanah made \(9\) tosses, then she
tossed \(9\times\dfrac{60}{100}=\dfrac{540}{100}=5.4\)
tails. Again, this is not possible.
If Savanah made \(8\) tosses, then her
second last toss was her 7th toss. However, it is not possible for
Savanah to toss \(50\%\) tails in \(7\) tosses (since one half of \(7\) is not a whole number). (We could have
instead shown that \(60\%\) of \(8\) is also not an integer as we did in the
first two cases.)
Similarly, Savanah could not have made \(10\) tosses since one half of \(9\) tosses is not an integer.
By process of elimination, the answer must be \(5\) tosses. We can confirm that if she
tosses \(2\) tails in her first \(4\) tosses, then \(50\%\) of her tosses are tails, and if her
final toss is a tail, then she has tossed \(3\) tails out of \(5\) tosses, which is \(60\%\) tails, as required.
Answer: (D)
The sum of the four angle measurements in a quadrilateral is
\(360\degree\).
The sum of the five angle measurements given is \(62\degree+85\degree+99\degree+108\degree+114\degree=468\degree\).
The difference, \(468\degree-360\degree=108\degree\), is the
measure of the angle that is not in the quadrilateral.
(We may confirm that \(62\degree+85\degree+99\degree+114\degree=360\degree\).)
Answer: (D)
Each integer is either an odd number or it is an even
number.
Thus, each of the \(10\) students will
check off exactly one box from the first pair of boxes.
Similarly, every integer greater than \(1\) is either a prime number or it is a
composite number, and so each of the \(10\) students will check off exactly one
box from the second pair of boxes.
Thus, every student checks off exactly \(2\) of the first \(4\) boxes.
This means that every student checks off exactly \(2\) of the \(5\) boxes if their card is not
numbered with a perfect square, and they check off exactly \(3\) of the \(5\) boxes if their card is
numbered with a perfect square.
The card numbered \(16\) is the only
card numbered with a perfect square, and so each of the remaining \(10-1=9\) students have a card that is not
numbered with a perfect square.
Thus, there are \(9\) students that
check off exactly two boxes.
Answer: (B)
We begin by extending \(PT\) to point \(U\) on \(QR\), as shown.
Since \(PT\) is parallel to \(SR\), then \(TU\) is parallel to \(SR\).
Similarly, since \(TS\) is parallel to
\(QR\), then \(TS\) is parallel to \(UR\).
Since \(TU\) is parallel to \(SR\) and \(TS\) is parallel to \(UR\), then \(TURS\) is a parallelogram, and so \(TU=SR\) and \(UR=TS=6\) cm.
In \(\triangle PQU\), the sum of the
measures of the three angles is \(180\degree\), and so \(\angle
PUQ=180\degree-60\degree-60\degree=60\degree\).
Thus, \(\triangle PQU\) is an
equilateral triangle, and so \(QU=PU=PQ=10\) cm.
The perimeter of \(PQRST\) is \[\begin{align*}
PQ+QR+SR+TS+PT&=PQ+QU+UR+SR+TS+PT \ \ \text{(since }QR=QU+UR)\\
&=PQ+QU+UR+TU+TS+PT \ \ \text{(since }SR=TU)\\
&=PQ+QU+UR+PT+TU+TS \ \ \text{(reordering some sides)}\\
&=PQ+QU+UR+PU+TS \ \ \text{(since }PT+TU=PU)\\
&=10\text{ cm}+10\text{ cm}+6\text{ cm}+10\text{ cm}+6\text{ cm}\\
&=42\text{ cm}\end{align*}\]
Answer: (A)
A circle with radius \(1\) cm
has area \(\pi\times(1 \text{ cm})^2=\pi \text{ cm}^{2}\).
A circle with radius \(5\) cm has area
\(\pi\times(5 \text{ cm})^2=25\pi\text{ cm}^{2}\).
A circle with radius \(x\) cm has area
\(\pi\times (x \text{ cm})^2=x^2\pi\text{ cm}^{2}\).
Since the mean area of the three circles is \(30\pi\text{ cm}^{2}\), then the sum of the areas of the
three circles is \(3\times30\pi \text{
cm}^2=90\pi\text{ cm}^{2}\).
Since \(\pi+25\pi+x^2\pi=26\pi+x^2\pi\), we get
\(26\pi+x^2\pi=90\pi\) or \(x^2\pi=64\pi\) or \(x^2=64\), and so \(x=8\) (since \(x>0\)).
Answer: (D)
Solution 1:
The probability that at least one colour is not used is equal to 1
minus the probability that all three colours are used.
Each of the three doors can be painted one of three different colours,
and so there are \(3\times3\times3=27\)
different ways to paint the doors with no restrictions.
If all \(3\) colours are used, then the
first door can be painted \(3\)
different colours, the second door can be painted \(2\) different colours, and the third door
must be painted the \(1\) remaining
colour.
The probability that all three colours are used is thus \(\dfrac{3\times2\times1}{27}=\dfrac{6}{27}\),
and so the probability that at least one colour is not used is \(1-\dfrac{6}{27}=\dfrac{27}{27}-\dfrac{6}{27}=\dfrac{21}{27}\),
which is equal to \(\dfrac{7}{9}\).
Solution 2:
The probability that at least one colour is not used is equivalent to
the probability that at most two different colours are used.
Thus, the probability that at least one colour is not used is equal to
the sum of the probability that exactly one colour is used and the
probability that exactly two different colours are used.
We begin by determining the probability that exactly one colour is
used.
Each of the three doors can be painted one of three different colours,
and so there are \(3\times3\times3=27\)
different ways to paint the doors with no restrictions.
If exactly one colour is used, there are 3 choices for the colour and
each door must be painted with that colour, and so the probability that
exactly one colour is used is \(\dfrac{3}{27}\).
If exactly two colours are used, then two doors are painted the same
colour and the remaining door is painted a different colour.
There are \(3\) choices for the first
colour and then \(2\) choices for the
second colour, and thus \(3\times2=6\)
ways to choose the two colours. Once the two colours are chosen, the
doors can be painted with these colours in \(3\) different ways. For example, if the
colour chosen for two of the doors is black (B) and the colour for the
remaining door is white (W), the doors can be painted BBW, BWB or
WBB.
Therefore, the number of ways to paint the doors with exactly two
different colours is \(6\times3=18\),
and so the probability that exactly two different colours are used is
\(\dfrac{18}{27}\).
Finally, the probability that at least one colour is not used is \(\dfrac{3}{27}+\dfrac{18}{27}=\dfrac{21}{27}=\dfrac{7}{9}\).
Answer: (A)
Solution 1:
We are being asked to find the number of multiples of \(18\) between \(111\,000\) and \(111\,999\) inclusive.
Since \(18\times 6166=110\,988\) and
\(18\times6167=111\,006\), then the
smallest multiple of \(18\) greater
than or equal to \(111\,000\) is \(111\,006\). (We can find the numbers \(6166\) and \(6167\) by dividing \(111\,000\) by \(18\) and rounding the result both down and
up to the nearest integer.)
Since \(18\times 6222=111\,996\) and
\(18\times6223=112\,014\), then the
largest multiple of \(18\) less than or
equal to \(111\,999\) is \(111\,996\). (We can similarly find the
numbers \(6222\) and \(6223\) by dividing \(111\,999\) by \(18\) and rounding the result both down and
up to the nearest integer.)
All multiples of \(18\) from \(18\times6167\) to \(18\times6222\) inclusive satisfy the given
conditions, and so there are \(6222-6167+1=56\) different possibilities
for \(N\).
Solution 2:
Since \(N=111\,abc\) is divisible by
\(18\), then \(N\) is divisible by both \(2\) and \(9\) since \(2\) and \(9\) have no factors in common and \(2\times9=18\).
Since \(N\) is divisible by \(2\), then \(N\) is even and so its units digit, \(c\), must equal \(0\), \(2\), \(4\), \(6\), or \(8\).
An integer is divisible by \(9\)
exactly when the sum of its digits is divisible by \(9\).
Thus, the sum of the digits of \(N\),
which is equal to \(1+1+1+a+b+c=a+b+c+3\), must be a multiple
of \(9\).
The smallest possible value of \(a+b+c+3\) is \(3\) (when \(a=b=c=0\)) and its largest possible value
is \(30\) (when \(a=b=c=9\)).
The multiples of \(9\) in this range
are \(9\), \(18\) and \(27\), and so \(a+b+c=6\) or \(a+b+c=15\) or \(a+b+c=24\).
To summarize to this point, \(N=111\,abc\) is divisible by \(18\) exactly when \(c\) is equal to \(0\), \(2\), \(4\), \(6\), or \(8\), and \(a+b+c=6\) or \(a+b+c=15\) or \(a+b+c=24\).
We proceed by setting \(c\) equal to
each of the possible even digits and determining the number of possible
values for \(a\) and \(b\).
When \(c=0\), we get \(a+b=6\) or \(a+b=15\) or \(a+b=24\).
When \(a+b=6\), the possible values for
the digits \(a\) and \(b\) (written as ordered pairs \((a,b)\)) are \((0,6)\), \((1,5)\), \((2,4)\), \((3,3)\), \((4,2)\), \((5,1)\), and \((6,0)\). (Alternately, when \(a+b=6\), \(a\) can equal any integer from \(0\) to \(6\) inclusive and then \(b=6-a\).)
Thus, there are \(7\) ordered pairs of
digits \((a,b)\) when \(c=0\) and \(a+b=6\), and so there are \(7\) possible values of \(N\).
When \(a+b=15\), the possible pairs of
digits \((a,b)\) are \((6,9)\), \((7,8)\), \((8,7)\), and \((9,6)\), and so there are \(4\) possible values of \(N\).
When \(a+b=24\), there are no possible
pairs of digits \((a,b)\) since \(a+b\) is at most \(9+9=18\).
When \(c=2\), we get \(a+b=6-2=4\) or \(a+b=15-2=13\) or \(a+b=24-2=22\).
We continue to count the number of pairs of digits \((a,b)\) given each of the possible values
of \(c\) and summarize those results in
the tables below.
| \(c=0\) | \(a+b=6\) | \(a+b=15\) | \(a+b=24\) | Number of values of \(N\) |
| Number of pairs \((a,b)\) | \(7\) | \(4\) | \(0\) | \(11\) |
| \(c=2\) | \(a+b=4\) | \(a+b=13\) | \(a+b=22\) | Number of values of \(N\) |
| Number of pairs \((a,b)\) | \(5\) | \(6\) | \(0\) | \(11\) |
| \(c=4\) | \(a+b=2\) | \(a+b=11\) | \(a+b=20\) | Number of values of \(N\) |
| Number of pairs \((a,b)\) | \(3\) | \(8\) | \(0\) | \(11\) |
| \(c=6\) | \(a+b=0\) | \(a+b=9\) | \(a+b=18\) | Number of values of \(N\) |
| Number of pairs \((a,b)\) | \(1\) | \(10\) | \(1\) | \(12\) |
| \(c=8\) | \(a+b=-2\) | \(a+b=7\) | \(a+b=16\) | Number of values of \(N\) |
| Number of pairs \((a,b)\) | \(0\) | \(8\) | \(3\) | \(11\) |
Therefore, the number of possibilities for \(N\) is \(11+11+11+12+11=56\).
Answer: (C)
We begin by naming shapes shown by the thick lines 1, 2, 3, 4, and 5, as shown in the diagram.
We also adopt the notation [row number, column number] to refer to
the contents of specific squares in the diagram.
For example, we are given that \([2,3]\) is \(D\) (this is in shape 3), and \([2,4]\) is \(A\) (this is in shape 2).
It is important to note that the solution that follows is one of many different ways to arrive at the final answer.
Since \([3,2]\) is \(C\), then \([2,1]\) cannot be \(C\) (shape 1 already has \(C\)), and \([2,2]\) cannot be \(C\) (column 2 already has \(C\)), and thus the \(C\) in row 2 must be \([2,5]\).
Since row 5 already has an \(E\), then
the \(E\) in shape 5 cannot be in row
5, and thus \([4,2]\) is \(E\).
Row 2 is missing \(B\) and \(E\) and since column 2 has an \(E\), then \([2,1]\) is \(E\) and \([2,2]\) is \(B\).
The letters determined to this point are included in the diagram
below.
Since shape 2 already contains \(A\)
and row 1 is missing \(A\), then \([1,1]\) is \(A\).
Also, column 4 already contains \(A\)
and so \([5,4]\) is not \(A\) which means that \([5,2]\) is \(A\) (since shape 5 is missing \(A\)).
The letters determined to this point are included in the diagram
below.
Shape 1 must contain \(D\) somewhere
in column 1 and so \([5,1]\) is not
\(D\). Since shape 5 is missing \(D\), then \([5,4]\) is \(D\).
Shape 4 is missing \(A\), \(B\) and \(D\). Since column 4 already contains \(A\) and \(D\), then \([3,4]\) cannot be \(A\) or \(D\), and thus \([3,4]\) is \(B\).
Finally, shape 1 is missing \(B\) and
\(D\). Since row 3 contains \(B\), then \([3,1]\) cannot be \(B\), and so the letter in the shaded
square, \([4,1]\), is \(B\) as shown.
The completed diagram is included below. Given the initial 5 letters and their locations, there is exactly one way to complete this diagram.
Answer: (B)
We adopt the notation [row number, column number] to be equal to
the contents of a specific square in the grid.
Since \([1,1]\) is \(1\) and adjacent numbers increase by a
fixed integer \(a>0\) moving left to
right within each row, then \([1,2]\)
is \(1+a\), \([1,3]\) is \(1+2a\), and so on to the end of the row
where \([1,8]\) is \(1+7a\).
Similarly, adjacent numbers increase by a fixed integer \(b>0\) moving top to bottom within each
column, and so the contents of each square in row 2 is \(b\) greater than the adjacent square in row
1.
That is, \([2,1]\) is \(1+b\), \([2,2]\) is \(1+a+b\), \([2,3]\) is \(1+2a+b\), and so on to the end of the row
where \([2,8]\) is \(1+7a+b\).
Continuing in this way in row 3, we get \([3,1]\) is \(1+2b\), \([3,2]\) is \(1+a+2b\), \([3,3]\) is \(1+2a+2b\), and so on to \([3,8]\) which is equal to \(1+7a+2b\).
We continue this pattern in the table below, and include each of the
entries in column 5 since the focus of the question is on this
column.
| 1 | \(1+a\) | \(1+2a\) | \(1+3a\) | \(1+4a\) | \(1+5a\) | \(1+6a\) | \(1+7a\) |
| \(1+b\) | \(1+a+b\) | \(1+2a+b\) | \(1+3a+b\) | \(1+4a+b\) | \(1+5a+b\) | \(1+6a+b\) | \(1+7a+b\) |
| \(1+2b\) | \(1+a+2b\) | \(1+2a+2b\) | \(1+3a+2b\) | \(1+4a+2b\) | \(1+5a+2b\) | \(1+6a+2b\) | \(1+7a+2b\) |
| \(1+3b\) | \(1+a+3b\) | \(\vdots\) | \(\vdots\) | \(1+4a+3b\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
| \(1+4b\) | \(1+a+4b\) | \(1+4a+4b\) | |||||
| \(1+5b\) | \(1+a+5b\) | \(1+4a+5b\) | |||||
| \(1+6b\) | \(1+a+6b\) | \(1+4a+6b\) | |||||
| \(1+7b\) | \(1+a+7b\) | \(1+4a+7b\) | \(1+7a+7b\) |
We are given that the number in the bottom right corner of the grid
is less than \(75\), and so \(1+7a+7b<75\) or \(7a+7b<74\).
Dividing each term of this inequality by \(7\), we get \(\dfrac{7a}{7}+\dfrac{7b}{7}<\dfrac{74}{7}\)
or \(a+b<\dfrac{74}{7}\) and since
\(a\) and \(b\) are positive integers, then \(a+b\leq10\).
Since \(a+b\leq10\), then by
multiplying each term by \(4\), we get
\(4a+4b\leq40\).
In the table above, \([5,5]\) is \(1+4a+4b\). Since \(4a+4b\leq40\), then \(1+4a+4b\leq41\) and so \([5,5]\) cannot equal \(45\).
Further, since \(b\) is a positive
integer, then each of the entries in column \(5\) above row 5 is less than \(1+4a+4b\) and thus cannot equal \(45\).
Thus if \(45\) appears in column 5 of
this grid, then it can only appear in rows 6, 7 and 8.
We begin by noting that \(a\) and
\(b\) are positive integers, and since
\(a+b\leq10\), then \(a\leq9\) and \(b\leq9\).
If \([6,5]\) is 45, then \(1+4a+5b=45\) or \(4a+5b=44\).
If \(a=1\), then \(4\times1+5b=44\) or \(5b=40\), and so \(b=8\).
We confirm that when \(a=1\) and \(b=8\), then \(a+b\leq10\) and so the number appearing in
the bottom right corner of the grid is less than \(75\).
Thus, in the arithmetic grid with \(a=1\) and \(b=8\), \([6,5]\) is 45.
If \(a=2\), then \(4\times2+5b=44\) or \(5b=36\) or \(b=\dfrac{36}{5}\) (which is not an
integer), and so there is no such arithmetic grid in which \([6,5]\) is \(45\) when \(a=2\).
We could continue substituting the remaining values of \(a\) from \(3\) to \(9\) to determine for which values of \(a\), \(4a+5b=44\) and \(a+b\leq10\) and \(b\) is a positive integer.
Alternately, we might notice that \(4a\) is a multiple of \(4\), as is \(44\).
Thus, if \(4a+5b=44\) then \(5b\) must also be a multiple of \(4\) and so \(b\) must be a multiple of \(4\).
The only remaining positive integer \(b\) for which \(b\leq9\) and \(b\) is a multiple of 4 is \(b=4\).
When \(b=4\), we get \(4a+5\times4=44\) or \(4a=24\) and so \(a=6\).
We again confirm that when \(a=6\) and
\(b=4\), then \(a+b\leq10\) and so the number appearing in
the bottom right corner of the grid is less than \(75\).
In the arithmetic grid with \(a=6\) and
\(b=4\), \([6,5]\) is \(45\).
Thus, there are exactly \(2\) such
grids in which \(45\) appears in row 6,
column 5.
If \([7,5]\) is \(45\), then \(1+4a+6b=45\) or \(4a+6b=44\) or \(2a+3b=22\).
In this case, we similarly notice that \(2a\) is a multiple of \(2\), as is \(22\).
Thus, if \(2a+3b=22\) then \(3b\) must also be a multiple of \(2\) and so \(b\) must be a multiple of \(2\).
We proceed by checking values of \(b\)
which are equal to positive even integers.
When \(b=2\), we get \(2a+3\times2=22\) or \(2a=16\) and so \(a=8\).
In this case, \(a+b\leq10\) and so the
number appearing in the bottom right corner is less than \(75\).
In the arithmetic grid with \(a=8\) and
\(b=2\), \([7,5]\) is \(45\).
When \(b=4\), we get \(2a+3\times4=22\) or \(2a=10\) and so \(a=5\).
In this case, \(a+b\leq10\).
In the arithmetic grid with \(a=5\) and
\(b=4\), \([7,5]\) is \(45\).
When \(b=6\), we get \(2a+3\times6=22\) or \(2a=4\) and so \(a=2\).
In this case, \(a+b\leq10\).
In the arithmetic grid with \(a=2\) and
\(b=6\), \([7,5]\) is \(45\).
When \(b=8\), we get \(2a+3\times8=22\) which is not possible
since \(a>0\).
Thus, there are exactly \(3\) such
grids in which \(45\) appears in row 7,
column 5.
If \([8,5]\) is \(45\), then \(1+4a+7b=45\) or \(4a+7b=44\).
We notice that \(4a\) is a multiple of
\(4\), as is \(44\).
Thus, if \(4a+7b=44\) then \(7b\) must also be a multiple of \(4\) and so \(b\) must be a multiple of \(4\).
We proceed by checking the two possible values of \(b\), namely \(b=4\) and \(b=8\).
When \(b=4\), we get \(4a+7\times4=44\) or \(4a=16\) and so \(a=4\).
In this case, \(a+b\leq10\) and so the
number appearing in the bottom right corner is less than \(75\).
In the arithmetic grid with \(a=4\) and
\(b=4\), \([8,5]\) is \(45\).
When \(b=8\), we get \(4a+7\times8=44\) which is not possible
since \(a>0\).
Thus, there is exactly \(1\) such grid
in which \(45\) appears in row 8,
column 5, and so there are \(2+3+1=6\)
grids in total.
Answer: (A)
In each row, exactly \(3\) of the \(6\) or one-half of the circles are shaded. Thus, one-half of the \(24\), that is, \(12\) circles are shaded. Alternately, we could just count the \(12\) shaded circles.
Answer: (B)
If each of \(4\) children shared \(36\) pieces equally, then each child received \(\dfrac{36}{4}=9\) pieces.
Answer: (D)
There are \(7\) days in each
week, and so there are \(2\times7=14\)
days in two weeks.
Thus, the date two weeks from May 12 is May 26 (since \(12+14=26\)).
Answer: (D)
Solution 1:
The time 2 hours after 8:45 a.m. is 10:45 a.m., and the time 45 minutes after 10:45 a.m. is 11:30 a.m.
Solution 2:
Two hours and 45 minutes is 15 minutes less than 3 hours.
The time 3 hours after 8:45 a.m. is 11:45 a.m., and the time 15 minutes
before 11:45 a.m. is 11:30 a.m.
Answer: (C)
If \(7x-3=60\), then \(7x=60+3\), so \(7x=63\) and \(x=\dfrac{63}{7}=9\).
Answer: (A)
There are 2 ways to colour the triangle and \(3\) ways to colour the square, and so there
are \(2\times3=6\) ways to colour the
figure.
We can write the \(6\) colourings of
the figure as ordered pairs in which the first colour listed in the pair
is the colour of the triangle and the second is the colour of the
square. These are: (red, blue), (red, purple), (red, green), (yellow,
blue), (yellow, purple), and (yellow, green).
Answer: (D)
When the point \((5,7)\) is
reflected in the \(x\)-axis, the
resulting point is \((5,-7)\).
When a point is reflected in the \(x\)-axis, its \(x\)-coordinate does not change and its
\(y\)-coordinate changes sign (provided
that the original point is not on the \(x\)-axis).
Answer: (A)
Reading from the graph, \(5\)
students voted for Spring, \(15\) voted
for Summer, \(5\) voted for Fall, and
\(10\) voted for Winter.
Therefore, (A) Fall and Spring received the same number of votes, (B)
Winter received more votes than Spring, (C) \(35\) students participated in the survey,
and (E) \(5\) students voted for Fall,
are all true statements.
Since \(15\) students voted for Summer
and \(15\) is less than half of \(35\), then (D) is the statement that is
false.
Answer: (D)
Solution 1:
The sum of the original five digits is \(5+2+8+7+9=31\).
The largest multiple of \(4\) that is
less than \(31\) is \(28\), and \(28\) is \(3\) less than \(31\).
However, \(3\) is not among the list of
five digits, and thus Ruhab cannot erase a \(3\).
The next largest multiple of \(4\) that
is less than \(31\) is \(24\).
Since \(31-24=7\), and \(7\) does appear in the original list of
five digits, then if Ruhab erases the \(7\), the sum of the remaining four digits
is a multiple of \(4\).
(We may confirm that \(5+2+8+9=24\).)
Solution 2:
We may erase each of the five digits one at a time and in each case,
determine the sum of the remaining four digits.
Doing so, we get \[5+2+8+7=22; \ 5+2+8+9=24;
\ 5+2+7+9=23;\] \[5+8+7+9=29; \
2+8+7+9=26\] Of these sums, only \(24\) is a multiple of \(4\) and so Ruhab erased the digit \(7\).
Answer: (D)
The perfect squares are \(1^2=1\), \(2^2=4\), \(3^2=9\), and so on.
There are \(18\) integers between \(3\) and \(20\) inclusive.
Of these \(18\) integers, only \(2^2=4\), \(3^2=9\) and \(4^2=16\) are perfect squares.
(We note that \(1^2=1\) and \(5^2=25\) are not among the integers
given.)
The probability of randomly selecting a perfect square from the integers
given is \(\dfrac{3}{18}=\dfrac{1}{6}\).
Answer: (C)
Since \(\dfrac{28}{32}+\dfrac{1}{\Box}=1\) and
\(\dfrac{28}{32}+\dfrac{4}{32}=\dfrac{32}{32}=1\),
then \(\dfrac{1}{\square}=\dfrac{4}{32}\).
Reducing \(\dfrac{4}{32}\) to lowest
terms, we get \(\dfrac{4}{32}=\dfrac{1}{8}\), and so the
number that goes in the box is \(8\).
Answer: (E)
There are \(60 \text{ minutes}\)
in one hour.
Since \(3\times20\text{ minutes}=60 \text{
minutes}\), and \(3\times1.5 \text{
km}= 4.5 \text{ km}\), then Leticia can walk \(4.5 \text{ km}\) in one hour.
Walking at this same rate, Leticia can walk \(4\times4.5\text{ km}=18 \text{ km}\) in
\(4 \text{ hours}\).
Answer: (A)
Ordered from smallest to largest, the list is \[1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,6,6,6,6,6,6\]
This list contains \(1+2+3+4+5+6=21\)
integers.
The median is the middle number in an ascending (or descending) list of
numbers.
The 11th integer in the list of \(21\)
integers above has \(10\) integers
before it and \(10\) integers following
it, and thus is the median of the list.
The 11th integer and the median of the given list is \(5\).
Answer: (D)
Opposite angles are equal in measure and so \(t=50\degree\) since it is opposite the
angle whose measure is given as \(50\degree\).
By a property of parallel lines, the angles whose measures are \(t\) and \(x\) form a "C pattern", and are
supplementary.
Thus \(t+x=180\degree\) or \(x=180\degree-50\degree=130\degree\).
Since the angle whose measure is \(y\)
is opposite the angle whose measure is \(x\), then \(y=x=130\degree\).
Since \(t+y=50\degree+130\degree=180\degree\), then
from the given answers, \(t\) and \(y\) are the pair of angles whose measures
sum to \(180\degree\).
Of the answers given, we confirm that no other pair of angles have
measures whose sum is \(180\degree\) by
determining the measures of each of the missing angles, as shown.
Answer: (D)
Solution 1:
The mean age of the first three students is \(13\), and so the sum of their ages is \(13\times3=39\).
The mean age of the four students is \(14\), and so the sum of their ages is \(14\times4=56\).
The age of the fourth student is the difference between these two sums,
which is \(56-39=17\) years old.
Solution 2:
In an ordered list of \(3\)
consecutive integers, the average of the list is always the middle
integer. Can you see why?
Therefore, three students whose ages are consecutive integers and whose
mean age is \(13\) are \(12\), \(13\) and \(14\) years old.
Suppose the fourth student is \(x\)
years of age.
Since the mean age of the four students is \(14\), then \(\dfrac{12+13+14+x}{4}=14\) or \(39+x=14\times4\), and so \(x=56-39=17\).
The fourth student is \(17\) years
old.
Answer: (E)
There is \(1\) dog for every
bowl of food, and so if there were \(6\) dogs, then there would be \(6\) bowls of food.
There are \(2\) dogs for every bowl of
water, and so these same \(6\) dogs
would need \(\frac{6}{2}=3\) bowls of
water.
There are \(3\) dogs for every bowl of
treats, and so the \(6\) dogs would
need \(\frac{6}{3}=2\) bowls of
treats.
Thus, \(6\) dogs require \(6\) bowls of food, \(3\) bowls of water, and \(2\) bowls of treats, or \(6+3+2=11\) bowls in total.
There are \(11\) bowls for every \(6\) dogs, and since there are \(77=11\times7\) bowls, then there are \(7\) groups of \(6\) dogs, or \(6\times7=42\) dogs.
Answer: (C)
In \(\triangle CBD\), \(CB=BD\) and so \(\angle BCD\) and \(\angle BDC\) have equal measures. The sum
of the angles in a triangle is 180\(\degree\), and since \(\angle CBD=90\degree\), then \(\angle BCD=\angle BDC=45\degree\).
(We confirm that \(90\degree+45\degree+45\degree=180\degree\).)
Since \(\angle CDE\) is a straight
angle, then \(\angle BDE+\angle
BDC=180\degree\) or \(\angle
BDE=180\degree-45\degree=135\degree\).
In \(\triangle BDE\), \(BD=DE\) and so \(\angle DBE\) and \(\angle DEB\) have equal measures. The sum
of the angles in a triangle is 180\(\degree\), and since \(\angle BDE=135\degree\), then \(\angle DBE=\dfrac{180\degree-135\degree}{2}=
22.5\degree\).
Finally, since \(\angle ABC\) is a
straight angle, then \(\angle ABE+\angle
DBE+\angle DBC=180\degree\) or \(\angle
ABE=180\degree-22.5\degree-90\degree=67.5\degree\).
Answer: (B)
In the grid shown, the first column and first row
show the possible numbers that may appear on the top faces of the two
dice.
The inside of the grid shows the product of the corresponding two
numbers.
| \(\times\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) |
|---|---|---|---|---|---|---|
| \(1\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) |
| \(2\) | \(2\) | \(4\) | \(6\) | \(8\) | \(10\) | \(12\) |
| \(3\) | \(3\) | \(6\) | \(9\) | \(12\) | \(15\) | \(18\) |
| \(4\) | \(4\) | \(8\) | \(12\) | \(16\) | \(20\) | \(24\) |
| \(5\) | \(5\) | \(10\) | \(15\) | \(20\) | \(25\) | \(30\) |
| \(6\) | \(6\) | \(12\) | \(18\) | \(24\) | \(30\) | \(36\) |
Of the results, a product of \(4\) appears \(3\) times, a product of \(6\) appears \(4\) times, a product of \(9\) appears \(1\) time, a product of \(15\) appears \(2\) times, and a product of \(8\) appears \(2\) times. Thus, of the possible products given, \(6\) is the most likely to occur.
Answer: (B)
We begin by determining the prime factorization of \(2025\). \[\begin{align*} 2025&= 25\times81\\ &=5\times5\times9\times9\\ &=5\times5\times3\times3\times3\times3\\ &=3^4\times5^2\end{align*}\] There are exactly \(15\) positive factors of \(2025\). These are:
\(1\), \(3\), \(3^2\), \(3^3\), \(3^4\), \(5\), \(5^2\), \(3\times5\), \(3^2\times5\), \(3^3\times5\), \(3^4\times 5\), \(3\times5^2\), \(3^2\times5^2\), \(3^3\times5^2\), and \(3^4\times 5^2\)
We are asked to express \(2025\) as the product of two positive integers \(n\) and \(m^2\), where \(m^2\) is a perfect square. Of the \(15\) positive factors, the following are perfect squares:
\(1\), \(3^2\), \(3^4\), \(5^2\), \(3^2\times5^2\), and \(3^4\times 5^2\)
These are all the possible values of \(m^2\), and so the values of \(m\) are: \(1\), \(3\), \(5\), \(3^2\), \(3\times5\), and \(3^2\times5\).
Thus, the ordered pairs of positive integers \((m,n)\) for which \(m^2\times n=2025\) are:
\((1,3^4\times5^2)\), \((3,3^2\times5^2)\), \((5,3^4)\), \((3^2,5^2)\), \((3\times5,3^2)\), \((3^2\times5,1)\)
Therefore, there are \(6\) such
ordered pairs.
It is interesting to note that each of the \(6\) values of \(n\) is also a perfect square. Can you see
why this occurs?
Answer: (E)
In the first diagram shown, we label the vertices of the polygon and the length \(ST=c\), since \(ST=QR\).
Next, we extend \(UT\) by a length equal to \(SR\), and we extend \(QR\) by a length equal to \(ST\), as shown in the second diagram.
Each of the angles in the polygon is a right angle, and so these two
extended line segments are perpendicular to each other and will meet at
a point that we label \(V\).
That is, \(STVR\) is a rectangle with
\(TV=SR=b\) and \(RV=ST=c\).
Each of the following expressions is equal to the perimeter of the
original polygon \[\begin{align*}
&PQ+QR+SR+ST+TU+PU\\
=\ & PQ+QR+ST+SR+TU+PU\ (\text{reordering the lengths})\\
=\ & PQ+QR+RV+TV+TU+PU\ (\text{since }RV=ST \text{ and } TV=SR)\\
=\ & PQ+QV+UV+PU\ (\text{since }QR+RV=QV \text{ and }
TV+TU=UV)\end{align*}\] which is the perimeter of \(PQVU\).
Each of the angles in \(PQVU\) is a
right angle, and \(PQ=PU\), and thus
\(PQVU\) is a square.
Since \(PQ=UV=UT+TV=a+b\), and \(PU=QV=QR+RV=c+c=2c\), then \(a+b=2c\).
Summarizing, the perimeter of the original polygon is equal to the
perimeter of square \(PQVU\), and each
side length of square \(PQVU\) can be
expressed as \(a+b\) or as \(2c\) since \(a+b=2c\).
If each of the \(4\) side lengths is
expressed as \(a+b\), then the
perimeter of \(PQVU\) (and thus the
perimeter of the original polygon), is equal to \((a+b)+(a+b)+(a+b)+(a+b)=4a+4b\).
If \(3\) side lengths are expressed as
\(a+b\) and \(1\) side length is expressed as \(2c\), then the perimeter is \((a+b)+(a+b)+(a+b)+(2c)=3a+3b+2c\).
If \(2\) side lengths are expressed as
\(a+b\) and \(2\) side lengths are expressed as \(2c\), then the perimeter is \((a+b)+(a+b)+(2c)+(2c)=2a+2b+4c\).
If \(1\) side length is expressed as
\(a+b\) and \(3\) side lengths are expressed as \(2c\), then the perimeter is \((a+b)+(2c)+(2c)+(2c)=a+b+6c\).
Finally, if all \(4\) sides lengths are
expressed as \(2c\), the perimeter is
\((2c)+(2c)+(2c)+(2c)=8c\).
Of the expressions given, \(a+b+7c\)
remains, and since \(a+b+7c=2c+7c=9c\)
is not equal to the perimeter, then the correct answer is (B).
Answer: (B)
\(H\) is a perfect square and is
\(1\) more than \(D\).
The perfect squares from \(1\) to \(8\) inclusive are \(1\) and \(4\). Since \(H\) is one more than \(D\), \(H\)
cannot be equal to \(1\) (since \(D\neq0\)), and so \(H=4\) and \(D=3\).
\(B\) is the largest prime number in
the set, and so \(B=7\).
\(C\) is a multiple of both \(G\) and \(D\). Since \(D=3\), then \(C=6\) and \(G\) is equal to either \(1\) or \(2\) (since \(6\) is a multiple of each of these).
The value of \(B+G\) is even, and since
\(B=7\), then \(G=1\).
The letters which have not been assigned values are \(A\), \(E\)
and \(F\), and the integers which have
not been assigned are \(2\), \(5\) and \(8\).
Since \(5\) and \(8\) are in the same row and \(A\) and \(E\) are the two remaining unassigned
letters that are in the same row, then \(A\) and \(E\) are \(5\) and \(8\) in some order, which leaves \(F=2\).
Answer: (A)
Solution 1:
The opposite sides of \(ABCD\) are
parallel, and so \(ABCD\) is a
parallelogram. The opposite sides of a parallelogram are equal in
length, and so \(AB=CD\) and \(AD=BC\).
Since the value of \(r+s+t\) is equal
to a constant, we may choose any location for \(B(0,r)\) provided that it satisfies the
given condition \(r<0\). We choose
\(r=-2\), so that the \(y\)-coordinate of \(B(0,r)\) is equal to the \(y\)-coordinate of \(A(-3,-2)\), and so \(AB\) is a horizontal line segment as
shown.
In this case, the length of \(AB\) is equal to the positive difference between the \(x\)-coordinates of \(A\) and \(B\), which is \(0-(-3)=3\).
\(CD\) is parallel to \(AB\), and so \(CD\) must also be a horizontal line
segment. Thus, points \(C(6,10)\) and
\(D(s,t)\) have equal \(y\)-coordinates, and so \(t=10\). Further, \(CD\) has the same length as \(AB\), and so \(6-s=3\) or \(s=3\).
Therefore, the value of \(r+s+t=-2+3+10=11\).
Solution 2:
The opposite sides of \(ABCD\) are
parallel, and so \(ABCD\) is a
parallelogram. The opposite sides of a parallelogram are equal in
length, and so \(AB=CD\) and \(AD=BC\).
Since \(AB\) and \(CD\) are parallel and equal in length, then
the vertical distance between \(A\) and
\(B\) must equal the vertical distance
between \(C\) and \(D\), and the horizontal distance between
\(A\) and \(B\) must equal the horizontal distance
between \(C\) and \(D\).
The vertical distance between two points is equal to the non-negative
difference between their \(y\)-coordinates, and so \(-2-r=t-10\) (assuming \(r\leq-2\) and \(t\geq10\) as in the diagram).
Simplifying, we get \(-2+10=t+r\) and
so \(t+r=8\).
The horizontal distance between two points is equal to the non-negative
difference between their \(x\)-coordinates, and so \(0-(-3)=6-s\) (assuming \(s<6\) as in the diagram).
Simplifying, we get \(0+3=6-s\) or
\(s=6-3\) and so \(s=3\).
Therefore, the value of \(r+s+t=(r+t)+s=8+3=11\). From Solution 1, we
note that \(r=-2\), \(s=3\), \(t=10\) are values satisfying the given
conditions and for which \(r+s+t=11\).
Answer: (B)
Solution 1:
Let the last three digits of \(n\)
be \(abc\). That is, \(n\) has units digit \(c\), tens digit \(b\) and hundreds digit \(a\).
(By the end of this solution, we will have demonstrated why considering
only the last three digits of \(n\) was
sufficient.)
The units digit of the product \(2013\times
n\) is equal to the units digit of \(3\times c\). \[\begin{array}[t]{cccccc}
&&\!\!\!2\!\!\!&\!\!\!0\!\!\!&\!\!\!1\!\!\!&\!\!\!3\!\!\!
\\ \times
&&&\!\!\!a\!\!\!&\!\!\!b\!\!\!&\!\!\!c\!\!\! \\
\hline
&\dots&\!\!\!2\!\!\!&\!\!\!0\!\!\!&\!\!\!2\!\!\!&\!\!\!5\!\!\!
\end{array}\]
Since the units digit of the product \(2013\times n\) is 5, then the units digit
of \(3\times c\) is \(5\), and so \(c=5\).
(You should confirm for yourself that this is the only possible value of
\(c\).) \[\begin{array}[t]{cccccc}
&&\!\!2\!\!\!&\!\!\!0\!\!\!\!\!&\!1\!\!\!&\!\!\!3\!\!\!
\!\!\\ \times &&&&&\!\!\!\!5\!\!\! \!\!\!\\ \hline
&\!\!\!1\!\!\!\!\!\!\!&\!\!\!0\!\!\!\!&\!0\!\!\!\!\!&\!6\!\!\!&\!5\!\!\!
\end{array}\]
Continuing the long multiplication, the tens digit of \(n\) is \(b\), and so the tens digit of \(2013\times n\) is equal to the units digit
of \(6+3b\), as shown.
Since the units digit of \(6+3b\) is
\(2\), then the units digit of \(3b\) is \(6\), and so \(b=2\).
(You should confirm for yourself that this is the only possible value of
\(b\).) \[\begin{array}[t]{cccccc}
&&\!\!\!2\!\!\!\!\!\!\!\!\!\!\!\!\!&\!\!\!0\!\!\!\!\!\!\!\!\!\!&\!\!\!1\!\!\!&\!\!\!3\!\!\!
\\ \times &&&&\!\!\!b\!\!\!&\!\!\!5\!\!\! \\ \hline
&\!\!\!1\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!&\!\!\!0\!\!\!\!\!\!\!\!\!\!\!\!\!&\!\!\!0\!\!\!\!\!\!\!\!\!&\!\!\!6\!\!\!&\!\!\!5\!\!\!
\\
&&&\dots&\!\!\!3b\!\!\!&
\\ \hline &&&\dots&\!\!\!2\!\!\!&\!\!\!5\!\!\!
\end{array}\]
The multiplication completed to this point is shown to the
right.
We have determined that the last two digits of the product \(2013\times n\) are \(25\) exactly when the last two digits of
\(n\) are \(25\). \[\begin{array}[t]{cccccc}
&&\!\!\!2\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!&\!\!\!0\!\!\!\!\!\!\!\!\!\!&\!\!\!1\!\!\!&\!\!\!3\!\!\!
\\ \times &&&&\!\!\!2\!\!\!&\!\!\!5\!\!\! \\ \hline
&\!\!\!1\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!&\!\!\!0\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!&\!\!\!0\!\!\!\!\!\!\!\!\!\!&\!\!\!6\!\!\!&\!\!\!5\!\!\!
\\
&\!\!\!4\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!&\!\!\!0\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!&\!\!\!2\!\!\!\!\!\!\!\!\!&\!\!\!6\!\!\!&
\\ \hline &&&\dots&\!\!\!2\!\!\!&\!\!\!5\!\!\!
\end{array}\]
Continuing the long multiplication, the hundreds digit of \(n\) is \(a\), and so the hundreds digit of \(2013\times n\) is equal to the units digit
of \(1+0+2+3a\) (the \(1\) is the "carry" from the tens
column).
Since the units digit of \(3+3a\) is
\(0\), then the units digit of \(3a\) is \(7\), and so \(a=9\).
(You should confirm that this is the only possible value of \(a\).) \[\begin{array}[t]{cccccc}
&&\!\!\!2\!\!\!\!\!\!\!\!\!\!\!\!&\!\!\!0\!\!\!\!\!&\!\!\!1\!\!\!&\!\!\!3\!\!\!
\\ \times
&&&\!\!\!a\!\!\!\!&\!\!\!2\!\!\!&\!\!\!5\!\!\! \\
\hline
&\!\!\!1\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!&\!\!\!0\!\!\!\!\!\!\!\!\!\!\!&\!\!\!0\!\!\!\!&\!\!\!6\!\!\!&\!\!\!5\!\!\!
\\
&\!\!\!4\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!&\!\!\!0\!\!\!\!\!\!\!\!\!\!\!&\!\!\!2\!\!\!&\!\!\!6\!\!\!&
\\
&&\dots&\!\!\!3a\!\!\!&&
\\ \hline
&&\dots&\!\!\!0\!\!\!&\!\!\!2\!\!\!&\!\!\!5\!\!\!
\end{array}\]
The last three digits of \(2013\times
n\) are \(025\) exactly when the
last three digits of \(n\) are \(925\) (that is, \(a=9\), \(b=2\), \(c=5\) are the only possibilities for \(a\), \(b\), \(c\)).
The multiplication completed to this point is shown below. \[\begin{array}[t]{cccccccc}
&&&&\!\!\!2\!\!\!&\!\!\!0\!\!\!&\!\!\!1\!\!\!&\!\!\!3\!\!\!
\\
\times
&&&&&\!\!\!9\!\!\!&\!\!\!2\!\!\!&\!\!\!5\!\!\!
\\ \hline
&&&\!\!\!1\!\!\!&\!\!\!0\!&\!\!\!0\!\!&\!\!\!6\!\!\!&\!\!\!5\!\!\!
\\
&&&\!\!\!4\!\!&\!\!\!0\!&\!\!\!2\!\!&\!\!\!6\!\!\!&
\\
&\!\!\!1\!\!&\!\!\!8\!\!
&\!\!\!1\!\!&\!\!\!1\!&\!\!\!7\!\!&&
\\ \hline
&\!\!\!\!1\!\!\!&\!\!\!\!8\!\!\!&\!\!\!\!6\!\!\!&\!\!\!\!\!2\!\!\!&\!\!\!\!0\!\!\!&\!\!\!2\!\!\!&\!\!\!5\!\!\!
\end{array}\]
This shows that when \(n=925\), the
last four digits of the product \(2013\times
n\) are \(2025\), as
required.
Adding additional digits to \(n\) will
increase the value of \(n\), and since
we are asked for the smallest possible value of \(n\), we stop here.
Thus, the smallest possible value of \(n\) for which \(2013\times n\) has last four digits \(2025\),
is \(n=925\), and so the sum of the
digits of \(n\) is \(9+2+5=16\).
Solution 2:
We begin by showing that every positive integer having last two
digits \(25\) is a multiple of \(25\).
(It is worth noting that it is not true that every multiple of
\(25\) has last two digits \(25\).)
All positive integers whose last two digits are \(25\), are \(25\) more than some non-negative multiple
of \(100\).
That is, all positive integers whose last two digits are \(25\) can be expressed as \(100k+25\) for some integer \(k\geq0\).
Since \(100k\) is divisible by \(25\), and \(25\) is divisible by \(25\), then \(100k+25\) is divisible by \(25\).
Thus, every positive integer whose last two digits are \(25\) is a multiple of \(25\), and so \(2013\times n\) is a multiple of \(25\).
Since \(2013=3\times11\times61\) does
not have a prime factor of \(5\), then
\(2013\times n\) is a multiple of \(25\) exactly when \(n\) is a multiple of \(25\).
The last two digits of \(2013\times n\)
are equal to the two-digit number formed by the last two digits of the
product of \(13\) and the last two
digits of \(n\).
What are the last two digits of \(n\)?
Since \(n\) is a multiple of \(25\), then the last two digits of \(n\) could be \(25\), \(50\), \(75\), or \(00\).
(You should confirm that these are the only possibilities.)
Thus, the last two digits of \(2013 \times
n\) are equal to the last two digits of \(13\times25\) or \(13\times50\) or \(13\times75\) or \(13\times 00\), which are \(25\), \(50\), \(75\), and \(00\) respectively.
Since we require the last two digits of \(2013\times n\) to be \(25\), then the last two digits of \(n\) are \(25\).
We have reduced the problem to finding the smallest value of the
positive integer \(n\) with last two
digits \(25\) so that the last four
digits of \(2013\times n\) are \(2025\).
We substitute \(n=25, 125, 225, 325,425,
\dots\), and so on, in turn, into the product \(2013 \times n\).
Evaluating these products, we determine that \(2013\times 925 = 1\,862\,025\) is the first
time that the last four digits of \(2013\times
n\) are \(2025\).
Thus, the smallest possible value of \(n\) for which \(2013\times n\) has last four digits \(2025\) is \(n=925\), and the sum of the digits of \(n\) is \(9+2+5=16\).
Answer: (E)
Suppose the integer in the centre circle is \(a\).
Then each integer in a circle connected to the centre circle is either
\(d\) more than \(a\), which is \(a+d\), or it is \(d\) less than \(a\), which is \(a-d\).
The integers in the two circles connected to the centre could both be
\(a+d\), as in Figure 1 below, or they
could both be \(a-d\), as in Figure 2,
or one could be \(a-d\) and one could
be \(a+d\), as in Figure 3.
We note that in the last case (Figure 3), swapping locations of the
\(a-d\) and the \(a+d\) does not change the integers in the
final two empty circles (since they still depend on \(a+d\) and \(a-d\)), and thus does not change the sum of
the five integers.
Next, we explain why it is possible to place integers into the
remaining two circles (in each of the three cases above) so that the
positive difference between each pair of integers in connected circles
is \(d\).
For the case that began in Figure 1, the integers in the two empty
circles are either \(d\) more than
\(a+d\), which is \(a+2d\), or they are \(d\) less than \(a+d\), which is \(a\).
These integers could both be \(a+2d\),
as in Figure 1a below, or they could both be \(a\), as in Figure 1b, or one could be \(a+2d\) and one could be \(a\), as in Figure 1c.
We note that in the last case (Figure 1c), swapping locations of the
final two integers, \(a+2d\) and \(a\), does not change the sum of the five
integers.
For the case that began in Figure 2, the integers in the two empty
circles are either \(d\) more than
\(a-d\), which is \(a\), or they are \(d\) less than \(a-d\), which is \(a-2d\).
These integers could both be \(a\), as
in Figure 2a below, or they could both be \(a-2d\), as in Figure 2b, or one could be
\(a\) and one could be \(a-2d\), as in Figure 2c.
We again note that in the last case (Figure 2c), swapping locations of
the final two integers, \(a\) and \(a-2d\), does not change the sum of the five
integers.
Finally, for the case that began in Figure 3, each integer in an
empty circle must have a positive difference of \(d\) with both \(a-d\) and \(a+d\).
The integer \(d\) more than \(a-d\) is \(a\), and the integer \(d\) less than \(a-d\) is \(a-2d\).
The integer \(d\) more than \(a+d\) is \(a+2d\), and the integer \(d\) less than \(a+d\) is \(a\).
Thus, \(a\) is the only integer that
has a positive difference of \(d\) with
both \(a-d\) and \(a+d\), and so the integers in the two empty
circles must each be equal to \(a\), as
shown in Figure 3a.
Suppose that the sum of the five integers in the circles is \(S\).
For Case 1a (which corresponds to Figure 1a), adding the five integers
in the figure, we get\(S=a+(a+d)+(a+2d)+(a+d)+(a+2d)=5a+6d\).
In the table below, we determine the value of \(S\) for each of the \(7\) cases.
| Case 1a | Case 1b | Case 1c | Case 2a | Case 2b | Case 2c | Case 3a |
| \(S=5a+6d\) | \(S=5a+2d\) | \(S=5a+4d\) | \(S=5a-2d\) | \(S=5a-6d\) | \(S=5a-4d\) | \(S=5a\) |
We must determine the number of different integers \(d\), between \(1\) and \(20\) inclusive, for which at least one of
the seven expressions for \(S\) is
equal to \(54\) and \(a\) is an integer.
Consider Case 1a, from which we get \(5a+6d=54\).
Since both \(a\) and \(d\) are integers, and \(d\) is between \(1\) and \(20\) inclusive, we can systematically
substitute values of \(d\) into this
equation, and then solve for \(a\) to
determine if \(a\) is an integer.
For example if \(d=1\), we get \(5a+6\times1=54\) or \(5a=48\).
However, there is no integer \(a\) for
which \(5a=48\) and so \(d=1\) is not a possible value of \(d\) in Case 1a. Substituting \(d=2\) and \(d=3\) similarly give non-integer values of
\(a\).
When \(d=4\), we get \(5a+6\times4=54\) and so \(5a=30\) or \(a=6\).
In this case, the pair of integers \(d=4\) and \(a=6\) satisfy the equation \(5a+6d=54\).
Substituting \(d=4\) and \(a=6\) into Figure 1a, we get the following
diagram.
We can confirm that the positive difference between each pair of
integers in connected circles is \(4\)
(an integer between \(1\) and \(20\) inclusive), and the sum of the five
integers in the circles is \(54\), as
required. Thus \(d=4\) is a possible
value satisfying the given conditions.
We can systematically continue to substitute \(d=5,6,7,\dots, 20\) into \(5a+6d=54\) and solve the equation to
determine which values of \(d\) give
integer values of \(a\).
The next smallest value of \(d\) for
which \(a\) is an integer is \(d=9\). In this case, we get \(5a+6\times9=54\) and so \(5a=0\) or \(a=0\).
We could continue in this systematic way, however since there are \(20\) possible values of \(d\) and \(7\) cases to check, this would take a while
to complete. Instead, we might recognize that \(d=4\), \(a=6\) and \(d=9\), \(a=0\) are both solutions to \(5a+6d=54\).
Notice that from the first solution to the second, the value of \(d\) increases by \(5\), and the value of \(a\) decreases by \(6\).
Can you see why increasing \(d\) by
\(5\) and decreasing \(a\) by \(6\) gives the next possible pair of
integers for which \(5a+6d=54\)? (Hint:
Take a close look at the left side of the equation.)
If we increase \(d\) by \(5\) again, and decrease \(a\) by \(6\), we get \(d=9+5=14\) and \(a=0-6=-6\), and since \(5a+6d=5\times(-6)+6\times14=-30+84=54\),
then \(d=14\) and \(a=-6\) is a solution to the equation (and
in fact, this is the next smallest value of \(d\) that works).
The final integer value of \(d\)
between \(1\) and \(20\) inclusive for which \(5a+6d=54\) is \(d=14+5=19\), and in this case \(a=-6-6=-12\) or \((d,a)=(19, -12)\)
Therefore, Case 1a gives \(d=4,9,14\),
and \(19\), or \(4\) values of \(d\) which satisfy the given conditions.
We continue in this way for each of the first four cases, and summarize all possible integer solutions for those cases in the table below.
| Case 1a | \(5a+6d=54\) | \((d,a)=(4,6), (9,0), (14, -6), (19, -12)\) | \(d=4,9,14,19\) |
| Case 1b | \(5a+2d=54\) | \((d,a)=(2,10), (7,8), (12,6), (17,4)\) | \(d=2,7,12,17\) |
| Case 1c | \(5a+4d=54\) | \((d,a)=(1,10), (6,6), (11, 2), (16, -2)\) | \(d=1, 6, 11, 16\) |
| Case 2a | \(5a-2d=54\) | \((d,a)=(3,12), (8,14), (13, 16), (18,18)\) | \(d=3, 8, 13, 18\) |
Notice that after the first four cases shown above, all possible
values of \(d\) from \(1\) to \(20\) inclusive satisfy the given conditions
with the exception of \(d=5,10,15\),
and \(20\).
In Case 3a we get, \(5a=54\) and so
\(a\) is not an integer. Next, consider
Case 2b, \(5a-6d=54\).
Each value of \(d\) left to check
(\(d=5,10,15, 20\)) is a multiple of
\(5\), and so \(6d\) is a multiple of \(5\) for each of these possible values of
\(d\).
Since \(5a\) is also a multiple of
\(5\) for all possible integers \(a\), then \(5a-6d\) is the difference between two
multiples of \(5\), and thus is a
multiple of \(5\).
However, the right side of the equation \(5a-6d=54\) is not a multiple of \(5\) and so \(d\) cannot be equal to a multiple of \(5\).
In the final case, \(5a-4d=54\), it is
similarly not possible for \(d\) to be
equal to a multiple of \(5\).
Thus \(d\) can be equal to each of the
first \(20\) positive integers with the
exception of \(5,10,15\), and \(20\), and so there are \(20-4=16\) different possible values of
\(d\).
It is worth noting that there are many different ways to find the integer solutions to each of the \(7\) equations (cases) above. For example, the value of each of the terms \(6d\), \(2d\), \(4d\), \(-2d\), \(-6d\), \(-4d\), \(0d\) is even for all integers \(d\), and the right side of each equation, \(54\), is also even. This means that in each equation, the value of \(5a\) must be even, and so \(a\) is even. Further, when \(a\) is even, the units digit of \(5a\) is \(0\). Since the units digit of \(54\) is \(4\), what do we now know about the units digit of each term containing a \(d\), and in each case, what does that tell us about the possible values of \(d\)?
Answer: (E)
We begin by recognizing that in the given list, each of the
digits \(1\) through \(7\) occurs at least once as a units digit,
and at least once as a tens digit.
For example, the digit \(1\) occurs
twice as a units digit (\(11\) and
\(31\)), and three times as a tens
digit (\(11\), \(12\) and \(14\)).
Counting the number of times each of the digits \(1\) through \(7\) occurs as a units digit and as a tens
digit, we get:
| Digit | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) |
|---|---|---|---|---|---|---|---|
| Number of times occurring as a units digit | \(2\) | \(1\) | \(1\) | \(4\) | \(1\) | \(2\) | \(1\) |
| Number of times occurring as a tens digit | \(3\) | \(1\) | \(1\) | \(3\) | \(1\) | \(2\) | \(1\) |
The units digit of each number in the list matches the tens digit of
the number that follows it.
This tells us that if we ignore the tens digit of the first number in
the list and the units digit of the last number in the list, then the
number of times that each digit occurs as a units digit must be equal to
the number of times that it occurs as a tens digit.
Looking back to the table above, we see that this is true for all digits
except 1 and 4.
Since the digit 1 occurs twice as a units digit and three times as a
tens digit, then the tens digit of the first number in the list must be
equal to 1.
Similarly, the digit 4 occurs four times as a units digit and three
times as a tens digit, and so the units digit of the last number in the
list must be equal to 4.
Ignoring the number 14 for a moment, we separate the 11 remaining
numbers into two distinct lists, which we call \(A\) and \(B\). \[A:
11,12,23,31 \ \ \ \ \ \ B: 44,45,46,56,64,67,74\] Each digit in
\(A\) is less than or equal to \(3\), and each digit in \(B\) is greater than or equal to \(4\).
Since \(14\) is the only number given
that does not appear in \(A\) or \(B\), and \(14\) has a digit that appears in \(A\) and a digit that appears in \(B\), then \(14\) is the only number that can ’connect’
the numbers in \(A\) to those in \(B\).
Further, this tells us that the numbers in \(A\) must be arranged and then placed before
an arrangement of the numbers in \(B\),
with \(14\) appearing between the two
arrangements.
Also, the arrangement of the numbers in \(A\) must begin and end with a \(1\), and the arrangement of the numbers in
\(B\) must begin and end with a \(4\) (since \(14\) occurs between the two lists).
Next, we count the number of different ways to arrange the numbers in
\(A\), starting and ending with \(1\).
We begin by recognizing that each of the digits \(2\) and \(3\) occurs exactly once as a units digit
and once as a tens digit, and so \(12\), \(23\), \(31\) must appear together in this order
(the two \(2\)s must occur together and
the two \(3\)s must occur
together).
The list must begin and end with a \(1\), and so there are \(2\) possible locations for the \(11\) and thus \(2\) possible arrangements of the numbers in
\(A\): \(11,
12, 23, 31\), and \(12, 23, 31,
11\).
Next, we count the number of different ways to arrange the numbers in
\(B\), starting and ending with \(4\).
We begin by recognizing that each of the digits \(5\) and \(7\) occurs exactly once as a units digit
and once as a tens digit, and so \(45\), \(56\) must appear together in this order
(the two \(5\)s must occur together),
and \(67\), \(74\) must appear together in this order
(the two \(7\)s must occur
together).
The arrangement ends with a \(4\), and
thus cannot end with \(45\), \(56\), and so at least one more number must
immediately follow \(45\), \(56\).
There are two such possibilities: \(45, 56,
64\), and \(45, 56, 67, 74\)
(recall that \(67, 74\) must remain
together), which leads to exactly two distinct cases to consider.
Case 1: \(\boldsymbol{45, 56, 64}\) occur together in this order.
In this case, the remaining numbers are \(44\), \(46\), \(67\), and \(74\).
Since \(44\) has two equal digits, its
location in the arrangement of the \(B\) list cannot change the first digit in
the list (which must be \(4\)), and
cannot change the last digit in the list (which must also be \(4\)), and thus we ignore \(44\) for the moment.
The remaining numbers, \(46\), \(67\), \(74\) must occur together in this order. Can
you see why?
Since the blocks \(45, 56, 64\) and
\(46, 67, 74\) must each occur together
in their respective orders, this gives two possible arrangements of the
\(B\) list (ignoring the \(44\)).
These are: \(45, 56, 64, 46, 67, 74\),
and \(46, 67, 74, 45, 56, 64\).
Next, we determine the number of different ways to place \(44\) into each of these arrangements.
In the \(45, 56, 64, 46, 67, 74\)
arrangement, the \(44\) may appear at
the start, at the end, or between the \(64\) and \(46\), which gives \(3\) different arrangements of the \(B\) list. (These are: \(44, 45, 56, 64, 46, 67, 74\), and \(45, 56, 64, 46, 67, 74, 44\), and \(45, 56, 64, 44, 46, 67, 74\).)
In the \(46, 67, 74, 45, 56, 64\)
arrangement, the \(44\) may appear at
the start, at the end, or between the \(74\) and \(45\), which gives \(3\) more arrangements of the \(B\) list, or \(6\) in total for Case 1.
Case 2: \(\boldsymbol{45, 56, 67, 74}\) occur together in this order.
In this case, the remaining numbers are \(44\), \(46\), and \(64\).
We again begin by ignoring \(44\) for
the moment.
The remaining numbers, \(46, 64\) must
occur together in this order.
Since the blocks \(45, 56, 67, 74\) and
\(46, 64\) must each occur together in
their respective orders, this gives two possible arrangements of the
\(B\) list (ignoring the \(44\)).
These are: \(45, 56, 67, 74, 46, 64\)
and \(46, 64, 45, 56, 67, 74\).
In the \(45, 56, 67, 74, 46, 64\)
arrangement, the \(44\) may appear at
the start, at the end, or between the \(74\) and \(46\), which gives \(3\) more different arrangements of the
\(B\) list.
In the \(46, 64, 45, 56, 67, 74\)
arrangement, the \(44\) may appear at
the start, at the end, or between the \(64\) and \(45\), which gives \(3\) more arrangements of the \(B\) list, or \(6\) in total for Case 2.
Thus, there are a total of \(6+6=12\) different ways to arrange the \(B\) list.
There are \(2\) different ways to arrange the numbers in list \(A\), \(12\) different ways to arrange the numbers in list \(B\), and exactly \(1\) way to place the number \(14\) between arrangements of each of the two lists. Thus, the total number of arrangements of the given list is \(2\times12\times1=24\).
Answer: (B)