2026 Gauss Contests
Solutions
(Grade 7 and 8)
May 11 to May 22, 2026
(in North America and South America)
May 11 to May 22, 2026
(outside of North American and South America)
May 11 to May 22, 2026
(in North America and South America)
May 11 to May 22, 2026
(outside of North American and South America)
©2026 University of Waterloo
In the diagram, there are \(3\)
columns containing triangles that each contain \(2\) triangles.
So, there are \(3\times2=6\)
triangles.
Answer: (C)
When \(3\) is subtracted from \(8\), the result is \(5\). The integer that must replace the square is \(8\).
Answer: (B)
On a number line, the number \(-10\) is \(10\) units left of \(0\). The number \(9\) is \(9\) units right of \(0\). Each of the remaining three choices is closer to \(0\) than \(-10\) or \(9\). Of the given choices, \(-10\) is farthest away from \(0\) on a number line.
Answer: (A)
Reading from the graph, Daia donated \(\$12\), Joe donated \(\$6\), Bel donated \(\$10\), Susie donated \(\$8\), and Zara donated \(\$2\).
The total amount of money that they donated was \(\$12+\$6+\$10+\$8+\$2=\$38\).
Answer: (D)
Beginning at the rightmost digit and moving left, \(5\) is in the ones (units) place, \(3\) is in the tens place, \(6\) is in the hundreds place, \(2\) is in the thousands place, and \(1\) is in the ten-thousands place. Thus, the digit in the thousands place is \(2\).
Answer: (B)
Since \(\angle PQR\) is a straight angle, its measure is \(180\degree\). Thus, the angles with measures \(40\degree\) and \(x\degree\) have a sum of \(180\degree\), or \(40+x=180\) and so \(x=180-40=140\).
Answer: (B)
Suppose that each side of the square has length \(n~\text{cm}\), where \(n\) is a whole number. Then the perimeter
of the square is \(4\times
n~\text{cm}\). This means that the perimeter of the square in
centimetres is a multiple of \(4\). Of
the choices given, only \(32~\text{cm}\) is a multiple of \(4\).
(We note that in this case, the square has side length \(n=8~\text{cm}\).)
Answer: (E)
On the second day, Mauricio walks \(11+3=14\) minutes. On the third day he walks \(17\) minutes. On the fourth day, he walks \(20\) minutes. On the fifth day, he walks \(23\) minutes. In total, Mauricio walks \(11+14+17+20+23=85\) minutes over the five days.
Answer: (A)
Following the reflection, the resulting shape is a mirror image of the original shape in the horizontal line. The resulting image has the same size and shape as the original. The following diagram has the original shape, the horizontal line, and the reflected image below it.
The correct answer is the figure shown in (D).
Answer: (D)
The sequence contains the integers from \(1\) to \(43\) inclusive with the even integers removed. The first even integer removed is \(2=2\times1\). The last even integer removed is \(42=2\times21\), and so the \(21\) even integers between \(1\) and \(43\) were removed. Thus, the number of terms in the sequence is \(43-21=22\).
Answer: (D)
The mean (average) of Lana's \(4\) jumps is \(\dfrac{1.60\text{ m}+1.65\text{ m}+1.85\text{ m}+1.90\text{ m}}{4}=\dfrac{7\text{ m}}{4}=1.75\text{ m}\).
(In place of the calculation above, we may have noticed that \(1.75\) is "in the middle" of \(1.65\) and \(1.85\) (thus is the mean of these two numbers), and \(1.75\) is also in the middle of \(1.60\) and \(1.90\), and thus is the mean of the four numbers.)
Answer: (E)
We seek two positive integers that differ by \(3\) and whose sum is \(27\). With a small amount of trial and error, we determine that \(15-12=3\) and \(15+12=27\), and so Elsa is \(12\).
Answer: (C)
The area of \(\triangle PQR\) is
\(\dfrac12\times8\text{ cm}\times6\text{
cm}=24\text{ cm}^2\).
The area of \(\triangle PST\) is \(\dfrac12\times4\text{ cm}\times3\text{ cm}=6\text{
cm}^2\).
The area of the shaded region is equal to the area of \(\triangle PQR\) minus the area of \(\triangle PST\), which is \(24\text{ cm}^2-6\text{ cm}^2=18\text{
cm}^2\).
Answer: (A)
There are exactly \(9\) such integers. These are: \(2026\), \(2062\), \(2206\), \(2260\), \(2602\), \(2620\), \(6022\), \(6202\), \(6220\).
Answer: (E)
The net shown in (C) is not the net of a standard
six-sided die.
On a standard six-sided die, the faces containing the numbers \(3\) and \(4\) must be opposite one another. In the
net shown in (C), the numbers \(3\) and
\(4\) are adjacent to one another, and
thus when this net is folded into a die, the face containing the \(3\) will be adjacent to the face containing
the \(4\), and so they will not be
opposite one another.
Answer: (C)
When two standard six-sided dice are rolled, the minimum sum that
can be rolled is \(1+1=2\), and so Riah
must move forward at least \(2\)
squares on each roll.
This means that Riah cannot land on the square numbered \(1\), she can land on at most one of the
squares numbered \(11\) and \(12\), and she can land on at most one of
the squares numbered \(16\) and \(17\).
To achieve the maximum score possible, the goal is to land on as many
shaded squares as possible, recognizing that a choice must be made
between squares \(11\) and \(12\), and also between squares \(16\) and \(17\).
To begin the game, Riah cannot roll a \(1\), but she can roll a \(3\), thus adding \(6\) points to her score. She can then roll
\(3\) again, adding \(4\) points to her score, followed by a roll
of \(2\) to add \(5\) more points to her score. Riah is now
on the square numbered \(8\), she has
\(6+4+5=15\) points, and she has
collected the maximum number of points possible to this point in the
game.
From the square numbered \(8\), Riah
can roll a \(3\) and collect \(7\) points or she can roll a \(4\) and collect \(8\) points. Since landing on square \(11\) or \(12\) does not affect her ability to land on
square \(16\) or \(17\), Riah maximizes her points by rolling
\(4\) and collecting \(8\) points. She is now on the square
numbered \(12\) and has \(15+8=23\) points.
From the square numbered \(12\), Riah
can roll a \(4\) and collect \(10\) points or she can roll a \(5\) and double her points. Since adding
\(10\) points gives Riah \(23+10=33\) and doubling her points gives
\(23\times2=46\), then rolling a \(5\) and landing on the square numbered
\(17\) maximizes Riah's points. There
are no more points that can be collected before finishing the
game.
Thus, the greatest number of points that Riah can finish the game with
is \(46\).
Answer: (B)
Asha could make one cut, cutting the ribbon into \(2\) pieces, each with length \(\dfrac{36\text{ m}}{2}=18~\text{m}\).
She could make two cuts, cutting the ribbon into \(3\) pieces, each with length \(\dfrac{36\text{ m}}{3}=12~\text{m}\).
Since the length of each of the equal pieces must be a whole number of
metres, then the number of pieces must be a positive divisor of \(36\).
The positive divisors of \(36\) that
are greater than \(1\) (since there
must be at least one cut and so at least \(2\) pieces) are \(2\), \(3\), \(4\), \(6\), \(9\), \(12\), \(18\), and \(36\), and so there are \(8\) possibilities for the length of the
smaller ribbons.
The possible lengths of the smaller pieces are \(18~\text{m}\), \(12~\text{m}\), \(\dfrac{36\text{ m}}{4}=9~\text{m}\), \(\dfrac{36\text{ m}}{6}=6~\text{m}\), \(\dfrac{36\text{ m}}{9}=4~\text{m}\), \(\dfrac{36\text{ m}}{12}=3~\text{m}\), \(\dfrac{36\text{ m}}{18}=2~\text{m}\), and
\(\dfrac{36\text{
m}}{36}=1~\text{m}\).
Answer: (E)
We begin by determining the values of the digits \(B\) and \(C\) when \(A<9\). At the end of the solution we
will show that the values of \(B\) and
\(C\) are the same when \(A=9\).
Suppose that when \(A<9\), the value
of \(0.ABC\) rounded to the nearest
tenth is \(0.T00\).
Since \(0.T00\) is \(0.024\) greater than \(0.ABC\), then \(0.ABC+0.024=0.T00\).
The thousandths digit of \(0.T00\) is
0, and so the sum of the thousandths digits, \(C\) and \(4\), ends in \(0\). Since \(C\) is at most \(9\), then \(C+4=10\), and so \(C=6\).
In this case, the 'carry' from the thousandths column to the hundredths
column is \(1\).
The hundredths digit of \(0.T00\) is
\(0\), and so the sum of the hundredths
digits, \(B\) and \(2\), added to the carry of \(1\), ends in \(0\). Since \(B\) is at most \(9\), then \(B+2+1=10\), and so \(B=7\).
In this case, the carry from the hundredths column to the tenths column
is \(1\), and so \(T=A+1\).
For any choice of the non-negative digit \(A\), where \(A<9\), \(0.A76+0.024=0.T00\) where \(T=A+1\).
As examples, \(0.876+0.024=0.900\) and
\(0.076+0.024=0.100\).
Finally, we show that when \(A=9\), the
values of \(B\) and \(C\) remain the same.
The value of \(0.9BC\) when rounded to
the nearest tenth is greater than \(0.9BC\), and therefore is equal to \(1\).
In this case, \(0.9BC+0.024=1\) or
\(0.9BC=1-0.024=0.976\), and so \(B=7\) and \(C=6\).
Therefore, for all possible values of the non-negative digit \(A\), we get \(B+C=7+6=13\).
Answer: (C)
Between \(100\) and \(199\) inclusive, the numbers that are
divisible by \(4\) are: \[4\times25=100,
4\times26=104, 4\times27=108, 4\times28=112, \dots,
4\times49=196\] Thus, there are \(49-25+1=25\) such numbers.
Of these, \(100\), \(112\) and \(120\) are the only numbers whose digit sum
is less than \(5\).
Therefore, between \(100\) and \(199\) inclusive, there are \(25-3=22\) numbers that are divisible by
\(4\) and for which the sum the digits
of each number is \(5\) or more.
Answer: (C)
Of all \(85\) cones sold, \(\dfrac{11}{17}\times85=55\) cones contained some chocolate. Of all \(85\) cones sold, \(\dfrac{10}{17}\times85=50\) cones contained some vanilla. The total \(55+50=105\) is more than the number of cones \((85)\) because it counts each twist cone exactly twice. Thus, \(105-85=20\) must equal the number of twist cones. That is, \(20\) cones were twist, \(55-20=35\) cones were chocolate, and \(50-20=30\) cones were vanilla, as shown in the Venn diagram.
Answer: (E)
\(PQRS\) is a square with \(PS=RS=30\text{ cm}\).
Since \(PU+UV+VS=PS=30\text{ cm}\) and
\(PU=UV=VS\), then \(UV=\dfrac{30\text{ cm}}{3}=10\text{
cm}\).
\(TUVW\) is a square with \(UT=TW=VW=UV=10\text{ cm}\).
\(TXRW\) is a parallelogram, and so
\(XR=TW=10\text{ cm}\).
The height of \(TXRW\) is the vertical
distance between its two parallel and horizontal sides \(TW\) and \(XR\). This vertical height is equal to
\(RS+VW=30\text{ cm}+10\text{ cm}=40\text{
cm}\).
The area of parallelogram \(TXRW\) is
the product of its base and its height, which is equal to \(10\text{ cm}\times40\text{ cm}=400\text{
cm}^2\).
Answer: (B)
Beverly writes down the value of \(\text{sum}\), which is \(0\), as the first term of the
sequence.
She then sets \(\text{sum} = \text{sum} + \text{a} = 0 + 1 = 1\).
Beverly then writes down the value of \(\text{sum}\), which is \(1\), as the second term of the
sequence.
Since \(\text{a}\neq 9\), she
sets \(\text{a} = \text{a} + 2 = 1 + 2 = 3\) and returns to Step 1.
In the table below, we keep track of the values of \(\text{sum}\) and \(\text{a}\), as well as the terms of the sequence as Beverly progresses through the steps.
| Values of \(\text{sum}\) | \(0\) | \(0+1=1\) | \(1+3=4\) | \(4+5=9\) | \(9+7=16\) | \(16+9=25\) |
|---|---|---|---|---|---|---|
| Sequence terms | \(0\) | \(1\) | \(4\) | \(9\) | \(16\) | \(25\) |
| Values of \(\text{a}\) | \(1\) | \(1+2=3\) | \(3+2=5\) | \(5+2=7\) | \(7+2=9\) | stop since \(\text{a}=9\) |
Therefore, the last term in Beverly's sequence is \(25\).
Answer: (E)
In the second bowl, the ratio of the number of blueberries to the
number of raspberries is \(2:5\), and
so the number of blueberries is a positive integer multiple of \(2\), and the number of raspberries is the
same positive integer multiple of \(5\).
For example, there could be \(2\times1=2\) blueberries and \(5\times1=5\) raspberries, or \(2\times2=4\) blueberries and \(5\times2=10\) raspberries, and so on.
Thus, the number of raspberries in the second bowl has units digit \(5\) or \(0\).
There are a total of \(89\)
raspberries, and so the number of raspberries in the first bowl must
have units digit \(4\) (when the number
of raspberries in the second bowl has units digit \(5\)), or it must have units digit \(9\) (when the number of raspberries in the
second bowl has units digit \(0\)).
In the first bowl, the ratio of the number of blueberries to the number
of raspberries is \(3:7\), and so the
number of raspberries is a positive integer multiple of \(7\).
The positive integer multiples of \(7\)
less than \(89\) that have units digit
\(4\) or \(9\) are \(14\), \(49\) and \(84\).
If the number of raspberries in the first bowl is \(14\), the number of raspberries in the
second bowl is \(89-14=75\).
If there are \(14\) raspberries in the
first bowl, there are \(3\times\dfrac{14}{7}=6\) blueberries in the
first bowl (\(6:14=3:7\)).
If there are \(75\) raspberries in the
second bowl, there are \(2\times\dfrac{75}{5}=30\) blueberries in
the second bowl (\(30:75=2:5\)).
In this case, the total number of blueberries is \(6+30=36\).
If the number of raspberries in the first bowl is \(49\), the number of raspberries in the
second bowl is \(89-49=40\).
If there are \(49\) raspberries in the
first bowl, there are \(3\times\dfrac{49}{7}=21\) blueberries in
the first bowl (\(21:49=3:7\)).
If there are \(40\) raspberries in the
second bowl, there are \(2\times\dfrac{40}{5}=16\) blueberries in
the second bowl (\(16:40=2:5\)).
In this case, the total number of blueberries is \(21+16=37\).
If the number of raspberries in the first bowl is \(84\), the number of raspberries in the
second bowl is \(89-84=5\).
If there are \(84\) raspberries in the
first bowl, there are \(3\times\dfrac{84}{7}=36\) blueberries in
the first bowl (\(36:84=3:7\)).
If there are \(5\) raspberries in the
second bowl, there are \(2\)
blueberries in the second bowl.
In this case, the total number of blueberries is \(36+2=38\).
Thus, the smallest possible number of blueberries is \(36\).
Answer: (A)
The integer \(abc\) is divisible
by \(5\), and so \(c=5\) (since \(c=0\) is not possible).
The integer \(cde=5de\) is divisible by
\(11\). The integers \(5de\) having different non-zero digits that
are divisible by \(11\) are \(517\), \(528\), \(539\), \(561\), \(572\), \(583\) and \(594\).
Next, we determine the smallest five-digit integer for which \(5de\) is equal to one of the seven
possibilities above. The smallest five-digit integers \(abcde\) with different digits have \(a=1\) and \(b=2\).
Since the digits must be different, then \(12\,517\), \(12\,528\), \(12\,561\), and \(12\,572\) are not possible. Thus, the
smallest possibility is \(12\,539\).
The final condition for a Sorrol number is that \(bcd=b5d\) is divisible by \(3\), and so the sum of the digits \(b+5+d\) is a multiple of \(3\).
Since \(2+5+3=10\) is not a multiple of
\(3\), then \(12\,539\) is not a Sorrol number.
The next smallest possibility is \(12\,583\).
In this case, \(2+5+8=15\) is a
multiple of \(3\), and so \(12\,583\) is the smallest Sorrol
number.
Next, we determine the largest Sorrol number.
The largest five-digit integers \(abcde\) with different digits have \(a=9\) and \(b=8\).
Since the digits must be different, then \(98\,594\), \(98\,583\), \(98\,539\), and \(98\,528\) are not possible. Thus, the
largest possibility is \(98\,572\).
However, \(8+5+7=20\) is not divisible
by \(3\).
Similarly, \(98\,561\) and \(98\,517\) do not satisfy the divisibility
by \(3\) condition.
The next largest five-digit integers \(abcde\) have \(a=9\) and \(b=7\).
In this case, the largest possibility is \(97\,583\) (since \(97\,594\) does not have distinct digits),
but this does not satisfy the divisibility by \(3\) condition.
Checking the next largest possibility \(97\,561\) (since \(97\,572\) does not have distinct digits),
we get \(7+5+6=18\) which is divisible
by \(3\), and so \(97\,561\) is the largest Sorrol
number.
The positive difference between the largest Sorrol number and the
smallest Sorrol number is \(97\,561-12\,583=84\,978\).
Answer: (A)
There are \(9\) non-lettered
squares that do not share an edge with a lettered square. In the grid
shown, each of these squares has been labelled with a \(0\).
Also, there are \(8\) non-lettered
squares that share an edge with exactly one lettered square. In the grid
shown, each of these squares has been labelled with a \(1\).
Finally, there are \(4\) non-lettered
squares that share an edge with exactly two lettered squares. In the
grid shown, each of these squares has been labelled with a \(2\).
| \(0\) | \(1\) | \(0\) | \(1\) | \(0\) |
| \(1\) | \(A\) | \(2\) | \(B\) | \(1\) |
| \(0\) | \(2\) | \(0\) | \(2\) | \(0\) |
| \(1\) | \(C\) | \(2\) | \(D\) | \(1\) |
| \(0\) | \(1\) | \(0\) | \(1\) | \(0\) |
Four or fewer of the numbered squares are to be shaded in. There is
no single numbered square that shares a side with all four lettered
squares, so at least two numbered squares must be shaded.
We proceed to count the number of ways to shade the grid by considering
the following three cases: exactly two squares are shaded, exactly three
squares are shaded, and exactly four squares are shaded.
Case 1: Exactly two of the numbered squares are shaded.
Shading the two \(2\)s in the middle row is one way to shade the grid, as shown.
Shading the two \(2\)s in the middle
column is a second way to shade the grid. These are the only possible
ways to shade the grid with exactly two shaded squares.
In Case 1, there are \(2\) ways to
shade the grid.
Case 2: Exactly three of the numbered squares are shaded.
We break this case into four subcases by considering the number of squares labelled \(2\) that are shaded.
Subcase 2a: No square labelled \(2\) is shaded.
If no square labelled \(2\) is
shaded, then each of the three shaded squares must be labelled \(0\) or \(1\).
It is not possible to shade three of the squares labelled with \(0\)s or \(1\)s while satisfying the condition that
each of the four lettered squares shares a side with exactly one shaded
square.
(You should confirm this for yourself before reading on.)
In Subcase 2a, there are \(0\) ways to
shade the grid.
Subcase 2b: Exactly one square labelled \(2\) is shaded.
There are \(4\) choices for which
square labelled \(2\) to shade.
This square shares a side with exactly two lettered squares, and so the
remaining two shaded squares must each be labelled \(1\).
For each choice of the square labelled \(2\), how many choices exist for the two
squares labelled \(1\)?
To answer this question, consider the following example. Suppose that the square labelled \(2\) between the \(A\) and \(B\) is shaded, as shown.
The two squares labelled \(1\) that share an edge with \(A\) cannot be shaded (since \(A\) already shares an edge with a shaded square). Similarly, the two squares labelled \(1\) that share an edge with \(B\) cannot be shaded.
Square \(C\) does not share an edge
with a shaded square, and so one of the two squares labelled \(1\) that share an edge with \(C\) must be shaded. There are \(2\) choices here.
Similarly, one of the two squares labelled \(1\) that share an edge with \(D\) must be shaded. There are \(2\) choices here. For each of the possible
choices, no shaded square shares an edge with another shaded square.
Thus for each of the \(4\) choices
of the square labelled \(2\), there are
\(2\times2\) choices for the two
squares labelled \(1\), by the symmetry
of the lettered and numbered squares within the grid.
Thus, in Subcase 2b, there are \(4\times2\times2=16\) ways to shade the
grid. One possible way to shade the grid is shown here.
Subcase 2c: Exactly two squares labelled \(2\) are shaded.
As was shown in Case 1, there are \(2\) possible ways that two squares labelled
\(2\) can be shaded.
For each of these possibilities, the remaining shaded square must be
labelled \(0\).
Of the \(9\) squares labelled \(0\), \(3\)
of these share an edge with at least one of the shaded squares labelled
\(2\) (these are the \(0\)s that are in the same row or column as
the \(2\)s).
Thus, for each of the \(2\) choices for
the two squares labelled \(2\), there
are \(9-3=6\) choices for the remaining
square labelled \(0\).
For each of the possible choices, no shaded square shares an edge
with another shaded square.
Thus in Subcase 2c, there are \(2\times6=12\) ways to shade the grid.
One possible way to shade the grid is shown here.
Subcase 2d: Exactly three squares labelled \(2\) are shaded.
All possible choices of shading three squares labelled \(2\) result in two lettered squares sharing
a side with two shaded squares.
Thus in Subcase 2d, there are \(0\)
ways to shade the grid.
In Case 2, there are a total of \(16+12=28\) ways to shade the grid.
Case 3: Exactly four of the numbered squares are shaded.
Similar to Case 2, we break this case into subcases by considering the number of squares labelled \(2\) that are shaded. As in Subcase 2d, it is not possible to shade three or more squares labelled \(2\).
Subcase 3a: No square labelled 2 is shaded.
If no square labelled \(2\) is
shaded, then each of the four shaded squares must be labelled \(1\).
Each of the \(2\) squares labelled
\(1\) that share an edge with the \(A\) can be shaded.
Similarly, there are \(2\) squares
labelled \(1\) that share an edge with
each of \(B\), \(C\) and \(D\) that can be shaded.
For each of the possible choices, no shaded square shares an edge with
another shaded square.
Thus in Subcase 3a, there are \(2\times2\times2\times2=16\) ways to shade
the grid. One possible way to shade the grid is shown here.
Subcase 3b: Exactly one square labelled \(2\) is shaded.
With exactly one shaded square labelled \(2\), two squares labelled \(1\) must be shaded, and one square labelled
\(0\) must be shaded.
As was shown in Subcase 2b, for each of the \(4\) choices of the square labelled \(2\), there are \(2\times2=4\) ways for the two squares
labelled \(1\) to be shaded.
The number of choices of the square labelled \(0\) is dependent on how the two squares
labelled \(1\) are shaded.
For example, suppose that the square labelled \(2\) between the \(A\) and the \(B\) is shaded.
Then, the \(4\) ways to shade the
squares labelled \(1\) are shown in the
four figures below.
Next, we consider the number of ways to choose the final shaded
square labelled \(0\).
In each of the four figures, the \(2\)
squares labelled \(0\) that share an
edge with the shaded square labelled \(2\) cannot be shaded.
This leaves \(9-2=7\) choices for the
square labelled \(0\) in each
figure.
In Figure 1, there are an additional \(4\) squares labelled \(0\) that share an edge with a shaded square
labelled \(1\), leaving \(7-4=3\) choices for the square labelled
\(0\) in this figure.
Thus in Figure 1, there are \(3\) ways
to choose the remaining squares. Similarly, there are \(3\) ways to choose the remaining squares in
Figure 2 and Figure 3, and so there are \(3+3+3=9\) ways to choose the squares.
In Figure 4, there are only \(3\)
squares labelled \(0\) that share an
edge with a shaded square labelled \(1\), leaving \(7-3=4\) choices for the square labelled
\(0\) in this figure.
Thus, for this choice of the square labelled \(2\), there are \(4\) ways to choose the remaining squares in
Figure 4.
Thus, for this choice of the square labelled \(2\), there are \(9+4=13\) ways to choose the remaining
squares.
Again by symmetry, the counting shown for this example is the same for
each of the \(4\) possible choices of
the square labelled \(2\), and so in
Subcase 3b, there are \(4\times13=52\)
ways to shade the grid.
Subcase 3c: Exactly two squares labelled \(2\) are shaded.
With exactly two shaded squares labelled \(2\), the remaining two shaded squares must
be labelled \(0\).
There are \(2\) ways to choose the two
shaded squares labelled \(2\).
For each of these choices, there are \(3\) squares labelled \(0\) that share an edge with a shaded
square, and so there are \(9-3=6\)
squares labelled \(0\) from which to
choose two.
For each of the possible choices, no shaded square shares an edge with
another shaded square.
There are \(6\) ways to choose the
first square labelled \(0\), followed
by \(5\) ways to choose the second
square labelled \(0\). However, this
double counts the number of possibilities since choosing, for example,
the \(0\) in the top left corner
followed by the \(0\) in the top right
corner, is the same pair of squares that we get when choosing them in
the reverse order.
Thus, the number of ways to choose two squares labelled \(0\) is \(\dfrac{6\times5}{2}=15\).
In Subcase 3c, there are \(2\times15=30\) ways to shade the
grid.
One possible way to shade the grid is shown here.
In Case 3, there are a total of \(16+52+30=98\) ways to shade the grid.
Having considered all possible cases, there are a total of \(2+28+98=128\) ways to shade the grid.
Answer: (E)
In a list of numbers, the mode is the number that occurs most
frequently.
Thus, the mode in the given list of numbers is \(2\).
Answer: (A)
After Liliana gives Abigail \(\$4\), she still owes her \(\$11-\$4=\$7\).
Answer: (D)
The perimeter is \(12 \text{ cm}\). Thus, \(3 \text{ cm} + 5 \text{ cm} + x \text{ cm}=12 \text{ cm}\), and so \(x=12-8=4\).
Answer: (C)
There are \(60\) seconds in \(1\) minute. The faucet drips at a rate of \(1\) drop every \(10\) seconds and so, \(\dfrac{60}{10}=6\) drops will fall in \(1\) minute.
Answer: (E)
The distance between \(0\) and
\(4\) along the number line is \(4\).
The tick marks divide this distance into \(16\) equal lengths, and so the distance
between adjacent tick marks is \(\dfrac{4}{16}=0.25\).
The number \(T\) is \(3\) tick marks to the right of \(0\), and thus the value of \(T\) is \(3\times0.25=0.75\).
Answer: (C)
The measure of each of the three angles in an equilateral
triangle is \(\dfrac{180\degree}{3}=60\degree\).
Thus, \(\angle RSU=\angle
TSU=60\degree\).
Since \(\angle RST=\angle RSU+\angle
TSU\), then the measure of \(\angle
RST=60\degree+60\degree=120\degree\).
Answer: (D)
The total number of students surveyed was \(10+5+9+6=30\).
Of the \(30\) students surveyed, \(6\) students arrive by car.
Thus, the percentage of students that get to school by car is \(\dfrac{6}{30}\times100\%=\dfrac{1}{5}\times100\%=20\%\).
Answer: (C)
Each digit of Livy's code is a different integer from \(0\) to \(9\) inclusive, meaning that there are \(10\) possible digits.
Since each digit of the code is different, Livy knows that the last
digit of the code is not equal to one of the first \(3\) digits. This leaves \(10-3=7\) possible digits from which Livy
will choose the last digit. The probability that Livy will choose the
correct last digit from these \(7\) on
her first try is \(\dfrac{1}{7}\).
Answer: (B)
The net shown in (B) could not be the net of the cube.
Using the net in (B), we begin by labelling three squares "bottom", "O"
and "P", as shown.
We then fold the net so that the face labelled bottom is the bottom face of the cube, and the two faces labelled O and P, folded upward, become two adjacent vertical faces of the cube.
The bottom of the cube, along with the two folded vertical faces are shown in the second diagram.
We ignore the other three faces of the cube, since the faces labelled O and P are sufficient for demonstrating that the net in (B) cannot be folded to give the required band around the cube.
Answer: (B)
The first three even integers in the list are \(2=2\times1\), \(4=2\times 2\), and \(6=2\times3\).
Each even integer in the list is of the form \(2\times n\) where \(n\) is a positive integer.
Thus, the 15th even integer is \(2\times
15=30\).
The first three odd integers in the list are each one less than the
first three even integers in the list, respectively. These are \(1=2\times1-1\), \(3=2\times 2-1\), and \(5=2\times3-1\).
Each odd integer in the list is of the form \(2\times n-1\) where \(n\) is a positive integer.
Thus, the 25th odd integer is \(2\times
25-1=49\).
The result of subtracting the 15th even integer from the 25th odd
integer is \(49-30=19\).
Answer: (C)
Since \(p+q+r=p+r+q=19\) and \(p+r=8\), then \(8+q=19\) and so \(q=19-8=11\).
Answer: (B)
Brock is seated in the chair between Abel and Duan, as shown in
Figure 1.
Edith is not beside Duan, and thus must be beside Abel, as shown in
Figure 2.
Callie must be seated in the final seat. Therefore, Abel and Callie are
seated to Edith's immediate left and right, as shown in Figure 3.
Confirm for yourself that switching the positions of Abel and Duan in
each figure results in Edith and Callie switching positions in Figure 3,
but does not change the final result.
Answer: (A)
The mass of a serving tray is \(4\) times the mass of a plate, and so the
combined mass of a tray and 6 plates is equivalent to the mass of \(4+6=10\) plates.
If the mass of \(10\) plates is \(3000 \text{ g}\), then the mass of each
plate is \(\dfrac{3000\text{ g}}{10}=300
\text{ g}\).
Answer: (E)
Solution 1:
Each hour, Anna jogs \(6\text{ km}\)
and Yao jogs \(8\text{ km}\), and so
combined they jog \(6\text{ km}+8\text{
km}=14\text{ km}\).
After \(2\) hours, they combine to jog
\(2\times14\text{ km}=28\text{ km}\),
and after \(3\) hours, they combine to
jog \(3\times14\text{ km}=42\text{
km}\).
Since they started \(42\text{ km}\)
apart, then after \(3\) hours they
meet. After \(3\) hours, Anna has
jogged \(3\text{ h}\times 6\text{
km/h}=18\text{ km}\), Yao has jogged \(3\text{ h}\times8\text{ km/h}=24\text{
km}\), and so Yao has travelled \(24\text{ km}-18\text{ km}=6\text{ km}\)
farther than Anna.
Solution 2:
Anna jogs at \(6\text{ km/h}\) and
Yao jogs at \(8\text{ km/h}\), and so
they are moving toward one another at a constant rate of \(6\text{ km/h}+8\text{ km/h}=14\text{
km/h}\).
Thus combined Anna and Yao will travel \(42\text{ km}\), and meet after \(\dfrac{42\text{ km}}{14\text{ km/h}}=3\)
hours.
Since Yao jogs \(8\text{ km/h}-6\text{
km/h}=2\text{ km/h}\) faster than Anna jogs, then Yao travels
\(2\text{ km/h}\times 3\text{ h}=6\text{
km}\) farther than Anna.
Answer: (A)
Half of the stack, or \(25\text{
cm}\), is removed from the stack and so \(25\text{ cm}\) remains in the stack.
Of the \(25\text{ cm}\) removed, \(\dfrac15\times 25\text{ cm}=5\text{ cm}\)
are put back onto the stack.
The final height of the stack is \(25\text{
cm}+5\text{ cm}=30\text{ cm}\).
Answer: (C)
Using the Pythagorean Theorem, we get \(x^2+x^2=\sqrt{8}^2\) or \(2x^2=8\). So, \(x^2=4\) or \(x=2\) (since \(x>0\)).
Thus, the right-angled isosceles triangle has area \(\dfrac12\times 2\text{ cm}\times2\text{
cm}=2\text{ cm}^2\).
A square with side length \(8\text{
cm}\) has area \(8\text{
cm}\times8\text{ cm}=64\text{ cm}^2\).
Thus, the smallest number of these triangles needed to completely cover
the square is \(\dfrac{64\text{ cm}^2}{2\text{
cm}^2}=32\).
Answer: (D)
Since \(P4R\) is a three-digit
integer, then \(P\neq0\). \[\begin{array}{ccccc}
&&\!\!\!P\!\!\!&\!\!4\!\!&\!\!\!R\!\!\! \\ +
&&\!\!\!7\!\!\!&\!\!\!Q\!\!\!&\!\!\!S\!\!\! \\ \hline
&&\!\!\!T\!\!\!&\!\!U\!\!\!&\!\!\!1\!\!\!
\end{array}\] If \(P>2\),
then \(P+7>9\) and so \(T>9\). In this case, \(TU1\) would be a four-digit integer which
is not possible, and so \(P=2\).
In the hundreds column \(P+7=2+7=9\).
So, there can be no carry from the tens column to the hundreds column
for the same reason as just described, and thus \(T=9\). The remaining digits are \(0\), \(3\), \(5\), \(8\).
The ones (units) digit of the sum is \(1\), and so the only possibilities for
\(R\) and \(S\) are \(3\) and \(8\), in some order.
Since \(3+8=11\), there is a carry of
\(1\) from the ones column to the tens
column.
Thus, the sum in the tens column is \(1+4+Q=5+Q\).
Since there is no carry from the tens column to the hundreds column,
then \(5+Q=U\), which gives \(Q=0\) and \(U=5\).
The two possible sums are shown below.
Answer: (D)
Of all \(85\) cones sold, \(\dfrac{11}{17}\times85=55\) cones contained
some chocolate.
Of all \(85\) cones sold, \(\dfrac{10}{17}\times85=50\) cones contained
some vanilla.
The total \(55+50=105\) is more than
the number of cones \((85)\) because it
counts each twist cone exactly twice. Thus, \(105-85=20\) must equal the number of twist
cones. That is, \(20\) cones were
twist, \(55-20=35\) cones were
chocolate, and \(50-20=30\) cones were
vanilla, as shown in the Venn diagram.
Answer: (E)
The distance between two parallel lines is constant. Thus, the
height of parallelogram \(ABFE\) is
equal to the height of \(\triangle
CDG\).
Suppose that height is \(h\). In this
case, the ratio of the area of \(ABFE\)
to the area of \(\triangle CDG\) is
\(AB\times h:\dfrac12\times CD\times
h\).
Substituting and simplifying, this ratio is equal to \(8\times h:\dfrac12\times12\times
h=8:6=4:3\).
Answer: (E)
Written as a mixed fraction, \(\dfrac{90}{11}=8\dfrac{2}{11}\), and so
\(a+\dfrac{1}{b+\frac1c}=8+\dfrac{2}{11}\).
If \(a=8\), then \(\dfrac{1}{b+\frac1c}=\dfrac{2}{11}\).
Since \(\dfrac{2}{11}\) has numerator
\(2\), we rewrite the previous equation
as \(\dfrac{2\times1}{2\times\left(b+\frac1c\right)}=\dfrac{2}{11}\).
Both numerators are now equal to \(2\),
and so \(2\times\left(b+\frac1c\right)=11\) or \(b+\dfrac1c=\dfrac{11}{2}\).
Written as a mixed fraction, \(\dfrac{11}{2}=5\dfrac{1}{2}\), and so \(b+\dfrac1c=5+\dfrac{1}{2}\).
If \(b=5\), then \(\dfrac{1}{c}=\dfrac{1}{2}\), and so \(c=2\).
Therefore, \(a+\dfrac{1}{b+\frac1c}=8+\dfrac{1}{5+\frac12}\)
and \(a+b+c=8+5+2=15\).
Answer: (D)
Solution 1:
We assign each of \(r\), \(s\) and \(t\) a different number from the list \(3\), \(4\), \(5\), and then determine the value of \(r\times s+t\) for each possibility.
If \(r=3\), \(s=4\) and \(t=5\), then \(r\times s+t=3\times 4+5=17\), which is
odd.
If \(r=3\), \(s=5\) and \(t=4\), then \(r\times s+t=3\times 5+4=19\), which is
odd.
If \(r=4\), \(s=3\) and \(t=5\), then \(r\times s+t=4\times 3+5=17\), which is
odd.
If \(r=4\), \(s=5\) and \(t=3\), then \(r\times s+t=4\times 5+3=23\), which is
odd.
If \(r=5\), \(s=3\) and \(t=4\), then \(r\times s+t=5\times 3+4=19\), which is
odd.
If \(r=5\), \(s=4\) and \(t=3\), then \(r\times s+t=5\times 4+3=23\), which is
odd.
We have considered all possible ways to assign the values \(3\), \(4\)
and \(5\), and so there are \(6\) such ways for which \(r\times s+t\) is odd.
Solution 2:
First, recognize that both \(3\) and
\(5\) are odd, and \(4\) is even.
When each of \(r\), \(s\) and \(t\) is assigned a different number from the
list \(3\), \(4\), \(5\), then either \(r\) and \(s\) are both odd, or exactly one of \(r\) and \(s\) is odd.
If \(r\) and \(s\) are both odd, then \(r\times s\) is odd and \(t=4\), and so \(r\times s+t\) is odd.
If exactly one of \(r\) and \(s\) is odd, then \(r\times s\) is even and \(t\) is odd, and so \(r\times s+t\) is odd.
Thus, the value of \(r\times s+t\) is
always odd.
When assigning the values \(3\), \(4\), \(5\), there are \(3\) choices for the value of \(r\), followed by \(2\) choices for the value of \(s\), and then \(1\) choice for the value of \(t\).
Since \(r\times s+t\) is odd for all
such choices, then there are \(3\times2\times1=6\) such ways to assign the
numbers.
Answer: (E)
The area of \(DMBN\) is equal to
the sum of the areas of triangles \(DMB\) and \(DNB\), and so we will first find these two
areas.
Each of \(DA\), \(DB\) and \(DC\) is a radius of the circle, and so
\(DA=DB=DC=5\).
Since \(\triangle DAB\) is an isosceles
triangle and \(M\) is the midpoint of
\(AB\), then \(DM\) is perpendicular to \(AB\) (\(DM\) is the height of \(\triangle DAB\)).
Using the Pythagorean Theorem in \(\triangle
DMB\), we get \(DB^2=DM^2+MB^2\)
and since \(MB=\dfrac{AB}{2}=2\text{
cm}\), then \(5^2=DM^2+2^2\).
Solving for \(DM\), we get \(DM^2=25-4=21\), and so \(DM=\sqrt{21}\text{ cm}\) (since \(DM>0\)).
Therefore, the area of \(\triangle
DMB\) is \(\dfrac12\times MB\times
DM=\dfrac12\times 2\text{ cm}\times \sqrt{21}\text{ cm}=\sqrt{21}\text{
cm}^2\).
We can similarly determine the area of \(\triangle DNB\).
Since \(NB=\dfrac12\times BC=3\text{
cm}\), then \(5^2=DN^2+3^2\).
Solving for \(DN\), we get \(DN^2=25-9=16\), and so \(DN=\sqrt{16}=4\text{ cm}\) (since \(DN>0\)).
Therefore, the area of \(\triangle
DNB\) is \(\dfrac12\times NB\times
DN=\dfrac12\times 3\text{ cm}\times 4\text{ cm}=6 \text{
cm}^2\).
Adding the two areas together, the area of \(DMBN\) is \(\sqrt{21}\text{ cm}^2+6\text{ cm}^2\),
which is \(10.6\text{ cm}^2\) when
rounded to one decimal place.
Answer: (B)
The probability that the second roll is a \(2\) is equal to the probability that the
first roll is odd and the second roll is a \(2\), added to the probability that the
first roll is even and the second roll is a \(2\).
The original die has \(2\) odd numbers
and \(4\) even numbers on its faces.
Thus the probability that the first roll is odd is \(\dfrac26\), and the probability that the
first roll is even is \(\dfrac46\).
If the first roll is odd, the numbers on the faces are changed to
\(2\), \(2\), \(2\), \(4\), \(14\), \(8\).
In this case, the probability that the second roll is a \(2\) is \(\dfrac36\), and so the probability that the
first roll is odd and the second roll is \(2\) is \(\dfrac26\times\dfrac36=\dfrac{6}{36}\).
If the first roll is even, the numbers on the faces are changed to
\(1\), \(1\), \(1\), \(2\), \(7\), \(4\).
In this case, the probability that the second roll is a \(2\) is \(\dfrac16\), and so the probability that the
first roll is even and the second roll is \(2\) is \(\dfrac46\times\dfrac16=\dfrac{4}{36}\).
Thus, the probability that the second roll is a \(2\) is equal to \(\dfrac{6}{36}+\dfrac{4}{36}=\dfrac{10}{36}=\dfrac{5}{18}\).
Answer: (D)
Suppose that when Serafine and Jamie stop playing catch, they
have a total of \(c\) catches and \(m\) misses.
Each catch increases the distance between them by \(2\times 60\text{ cm}=120\text{ cm}\), and
so \(c\) catches increases the distance
between them by \(120c\text{
cm}\).
Each miss decreases the distance between them by \(20\text{ cm}\), and so \(m\) misses decreases the distance between
them by \(20m\text{ cm}\).
They begin \(1500\text{ cm}\) apart and
when they stop they are \(1720\text{
cm}\) apart, and so \(1500+120c-20m=1720\).
Simplifying this equation, we get \(120c-20m=1720-1500\) or \(120c-20m=220\). Dividing each term by \(20\), we get \(6c-m=11\).
Since the question asks which number of throws is not possible, then we
let the number of throws be \(n\), and
so \(n=m+c\).
With \(6c-m=11\) and \(n=m+c\), we get the following equivalent
equations \[\begin{align*}
6c-m&=11\\
6c&=11+m\\
6c+c&=11+m+c\\
7c&=11+n\end{align*}\] and so \(n=7c-11\).
If \(n=21\), then \(7c-11=21\) or \(7c=32\), and so \(c=\dfrac{32}{7}\). However, \(\dfrac{32}{7}\) is not an integer and so
\(21\) throws is not possible.
We can confirm that when \(c\) is equal
to \(4\), \(12\), \(8\), and \(3\), then the number of throws, \(n=7c-11\), is equal to \(17\), \(73\), \(45\), and \(10\), respectively.
In each case, \(m=6c-11\) and so \(m\) is also a positive integer.
Of the given possibilities, \(21\)
throws is not possible.
Answer: (B)
Suppose the five distinct points are placed on a number line with
the leftmost point at \(0\).
The largest distance between a pair of these five points is \(16\) (since \(n<16\)), and so the rightmost point
appears on the number line at \(16\).
Next, we place the remaining three points on the number line at the
values \(p\), \(q\) and \(r\), where \(0<p<q<r<16\), as shown.
Since the ten distances between pairs of points are integers, then
each of \(p\), \(q\) and \(r\) is an integer.
The ten distances between pairs of points are:
Of the given distances, \(2\), \(3\), \(4\), \(5\), \(7\), \(9\), \(11\), \(13\), \(16\), \(n\), we have accounted for the distance
\(16\) by fixing endpoints at \(0\) and \(16\), and so we may ignore this distance
going forward.
The problem remains to determine all possible values of \(p\), \(q\), \(r\) for which the distances \(p\), \(q\), \(r\), \(q-p\), \(r-p\), \(r-q\), \(16-p\), \(16-q\), \(16-r\) (in some order) match the distances
\(2\), \(3\), \(4\), \(5\), \(7\), \(9\), \(11\), \(13\), \(n\) for some integer \(n\), where \(1\leq n\leq15\).
It is not possible for each of the values \(p\), \(q\) and \(r\) to equal \(1\), \(6\), \(8\), \(10\), or \(15\). Why? If for example \(p=1\), then \(16-p=15\). In this case, \(1\) and \(15\) are two distances not in the list
\(2\), \(3\), \(4\), \(5\), \(7\), \(9\), \(11\), \(13\). Since there is one unknown distance,
\(n\), then there can be at most one
distance not in the list \(2\), \(3\), \(4\), \(5\), \(7\), \(9\), \(11\), \(13\). This tells us that \(1\) is not a possible value of \(p\), \(q\) or \(r\).
Using this same argument, we can show that \(6\), \(8\), \(10\), and \(15\) are also not possible values of \(p\), \(q\)
and \(r\).
Next, we show that exactly one of \(p\), \(q\)
and \(r\) is equal to \(3\) or \(13\).
To get a distance of \(13\), we need
points at both \(0\) and \(13\), or at both \(1\) and \(14\), or at both \(2\) and \(15\), or at both \(3\) and \(16\).
We have shown that \(1\) and \(15\) are not possible values of \(p\), \(q\)
and \(r\), and so the middle two
possibilities are not permitted.
Thus, at least one of \(p\), \(q\) and \(r\) is equal to \(3\) or \(13\).
If two of \(p\), \(q\) and \(r\) are equal to \(3\) and \(13\), then the distances \(3\) and \(13\) would each appear twice in the list
(since \(16-3=13\) and \(16-13=3\)), which is not possible.
It follows that exactly one of \(p\),
\(q\) and \(r\) is equal to \(3\) or \(13\).
Let us assume that exactly one of \(p\), \(q\)
and \(r\) is equal to \(13\). We will deal with the possibility
that one of \(p\), \(q\) and \(r\) is equal to \(3\) at the end of the solution.
Suppose that \(q=13\). Then \(r=14\) since we have shown that \(r\) cannot equal \(15\).
However, if \(r=14\), then \(r-q=14-13=1\), but both \(1\) and \(14\) are not in the list of distances, so
this is not possible.
Similarly, we can show that \(p\neq
13\), and so \(r=13\), as
shown.
With \(r=13\), we get \(16-r=3\), and so the remaining distances
are \(2\), \(4\), \(5\), \(7\), \(9\), \(11\), \(n\), and the work above establishes that
all remaining distances are less than \(13\), and thus \(n\leq 12\).
With \(r=13\), we have shown that \(p\) and \(q\) cannot be equal to \(1\), \(3\), \(6\), \(8\), \(10\), \(13\), \(14\) and \(15\). The possible values of \(p\) and \(q\) are thus, \(2\), \(4\), \(5\), \(7\), \(9\), \(11\), and \(12\).
If \(p=2\), then \(16-p=14\) and since \(14\) is not in the list
of distances and \(n\neq 14\), then
\(p\neq 2\).
If \(q=12\), then \(q-0=12\) and \(r-q=13-12=1\) are two values not in the
list of distances, and so \(q\neq
12\).
Since \(q\leq 11\) and \(p<q\), then \(p\neq 11\). The remaining possible values
of \(p\) are \(4\), \(5\), \(7\), \(9\).
Since \(p\geq 4\) and \(p<q\), then \(q\neq 4\). The remaining possible values of
\(q\) are \(5\), \(7\), \(9\), \(11\).
This gives \(10\) possible pairs of
values \((p,q)\). These are: \((4,5)\), \((4,7)\), \((4,9)\), \((4,11)\), \((5,7)\), \((5,9)\), \((5,11)\), \((7,9)\), \((7,11)\), and \((9,11)\).
In each case, \(r=13\).
Recall that the remaining distances are \(2\), \(4\), \(5\), \(7\), \(9\), \(11\), and \(n\).
If \((p,q)=(4,5)\), then \(q-p=5-4=1\) and \(16-q=16-5=8\) are two distances not listed
and so \((p,q)\neq(4,5)\). We continue
our analysis of the remaining \(9\)
pairs in the table that follows.
| \(\boldsymbol{(p,q)}\) | Distance(s) not included in \(\boldsymbol{2}\), \(\boldsymbol{4}\), \(\boldsymbol{5}\), \(\boldsymbol{7}\), \(\boldsymbol{9}\), \(\boldsymbol{11}\) | Value of \(\boldsymbol{n}\)? |
|---|---|---|
| \((4,5)\) | \(q-p=1\) and \(13-q=8\) | not possible |
| \((4,7)\) | \(q-p=3\) and \(13-q=6\) | not possible |
| \((4,9)\) | \(16-p=12\) and \(p-0=13-q=4\) (gives two \(4\)s) | not possible |
| \((4,11)\) | \(16-p=12\) | \(n=12\) |
| \((5,7)\) | \(13-p=8\) and \(13-q=6\) | not possible |
| \((5,9)\) | \(13-p=8\) and \(q-p=13-q=4\) (gives two \(4\)s) | not possible |
| \((5,11)\) | \(13-p=8\) and \(q-p=6\) | not possible |
| \((7,9)\) | \(13-p=6\) and \(q-0=16-p=9\) (gives two \(9\)s) | not possible |
| \((7,11)\) | \(13-p=6\) | \(n=6\) |
| \((9,11)\) | \(13-q=q-p=2\) (gives two \(2\)s) | \(n=2\) |
Thus, the possible values of \(n\) are \(2\), \(6\), \(12\), and so there are \(3\) possible values of \(n\).
Note: Earlier in the solution we made the assumption that one of
\(p\), \(q\) and \(r\) was equal to \(13\), and then showed that \(r=13\). We could have similarly assumed
that one of \(p\), \(q\) and \(r\) was equal to \(3\), and then showed that \(p=3\).
Notice that having \(r=13\) is
symmetrical to having \(p=3\).
That is, when \(r=13\) we get the two
distances \(r-0=13\) and \(16-r=3\). When \(p=3\), we get the same two distances \(p-0=3\) and \(16-p=13\). The solution was shortened by
taking advantage of this symmetrical property of the values \(p\), \(q\)
and \(r\). Specifically, the reflection
of the values \(p\), \(q\), \(r\), that is, changing each of these to
\(16-p\), \(16-q\), \(16-r\) gives the same set of distances, and
thus the same value of \(n\). For
example, the reflection of \((0,p,q,r,16)=(0,4,11,13,16)\) is \((16-0, 16-p, 16-q, 16-r, 16-16)\) which is
equal to \((16,12,5, 3, 0)\) or \((0,3,5,12,16)\) which gives the same set of
distances as \((0,4,11,13,16)\), and
thus also gives \(n=12\).
Answer: (C)