2025 Canadian Team Mathematics Contest
Solutions
April 2025
April 2025
©2025 University of Waterloo
Since \(m=4\), then \((m+1)(m+2) = (4+1)(4+2) = 5 \times 6 = 30\).
Answer: \(30\)
The sum of the digits of \(n\)
is \(7+A+6+6+5+B+2+A+7 = 33+ 2A + B\),
which we are told equals \(46\).
Therefore \(2A + B = 46-33 =
13\).
The sum of the rightmost two digits of \(n\) is \(A+7\), which we are told equals \(11\). Therefore \(A=4\).
Since \(2(4) + B = 13\), we also have
that \(B=5\), and therefore \(A+B = 4+5 = 9\).
Answer: \(9\)
Let \(C\) be the capacity of the pot in litres. The given information says that \(\dfrac{2}{3}C - 3 = \dfrac{1}{2} C\). Then \(3 = \dfrac{2}{3} C - \dfrac{1}{2} C = \dfrac{1}{6}C\). Therefore \(C=3(6)=18\) L.
Answer: \(18\) L
Solution 1:
We observe the following: \[\begin{align*}
1+2+3+4+5 &= 15 \\
2+3+4+5+6 &= 20 \\
3+4+5+6+7 &= 25 \\
4+5+6+7+8 &= 30\end{align*}\] Note that \(15\), \(20\), and \(25\) are not multiples of \(6\), and \(30\) is a multiple of \(6\).
Since we require that all \(5\)
consecutive integers be positive, we conclude that \(30\) is the smallest such multiple of \(6\).
Solution 2:
Call our \(5\) consecutive integers
\(a-2, a-1, a , a+1, a+2\), for some
integer \(a \geq 3\). Then the sum of
our \(5\) consecutive integers is \((a-2) + (a-1) + a + (a+1 ) + (a+2) =
5a\).
Since we want \(5a\) to be a positive
multiple of \(6\), this means that
\(5a = 6b\) for some positive integer
\(b\).
Since \(5\) and \(6\) have a greatest common divisor of \(1\), \(a\)
must be a multiple of \(6\). The
smallest positive multiple of \(6\) is
\(6\) itself, and we note that the only
solution with \(a=6\) is \(b=5\).
Therefore the smallest positive multiple of \(6\) that can be expressed as a sum of 5
consecutive positive integers is \(5a=5(6)=30\).
Answer: \(30\)
Since \(\dfrac{AB}{BC}=\dfrac{2}{3}\), we may write
\(AB=2x\) and \(BC=3x\) for some positive real number \(x\).
It follows that \(AC=AB+BC=2x+3x=5x\).
Since \(ABIH\) and \(DEFJ\) are congruent squares, we calculate
that \[AG=AH+HG = AH+DE = 2x+2x=4x.\]
Therefore the area of \(ACEG\) is \(2420=(4x)(5x)=20x^2\). Dividing by \(20\) tells us that \(x^2 = 121\). Since \(x>0\) we conclude that \(x=11\).
Finally, the perimeter of \(ACEG\) is
\(2(AC)+2(AG) = 2 (5x) + 2(4x) = 18x = 18(11)
= 198\).
Answer: \(198\)
Let \(N\) be a possible number
of stamps. The given conditions imply that there exist non-negative
integers \(a\), \(b\), and \(c\) such that \[\begin{align*}
N &= 4a + 1 \\
N &= 5b+ 2 \\
N &= 9c\end{align*}\] We will find the smallest positive
\(c\) for which this system has a
solution; since \(N=9c\) then this will
also determine the smallest positive integer \(N\).
When \(c=1\) we have that \(N=9\), and we want the equations \(9=4a+1\) and \(9=5b+2\) to both have integer solutions.
The first equation has a non-negative integer solution, but the second
equation has no integer solution.
When \(c=2\) we have that \(N=18\), so we want \(18=4a+1\) and \(18=5b+2\) to both have integer solutions
but neither of these have integer solutions.
We continue this search in the table below. Since we are starting with
\(c=1\) and increasing \(c\) by \(1\) each time, our search is guaranteed to
find the smallest possible value of \(N\).
| \(c\) | \(N\) | \(N=4a+1\,?\) | \(N=5b+2\,?\) |
|---|---|---|---|
| \(1\) | \(9\) | yes | no |
| \(2\) | \(18\) | no | no |
| \(3\) | \(27\) | no | yes |
| \(4\) | \(36\) | no | no |
| \(5\) | \(45\) | yes | no |
| \(6\) | \(54\) | no | no |
| \(7\) | \(63\) | no | no |
| \(8\) | \(72\) | no | yes |
| \(9\) | \(81\) | yes | no |
| \(10\) | \(90\) | no | no |
| \(11\) | \(99\) | no | no |
| \(12\) | \(108\) | no | no |
| \(13\) | \(117\) | yes | yes |
Therefore, the smallest possible number of stamps is \(117\).
Editorial Comment: Imposing that \(N=4a+1\) for some non-negative integer \(a\) is equivalent to imposing that \(N\) has remainder \(1\) when divided by \(4\). The Chinese Remainder Theorem tells us there is nothing special about the given remainders (\(0\) when divided by \(9\), \(1\) when divided by \(4\), and \(2\) when divided by \(5\)) since no two of \(4,5,9\) have a common factor other than \(\pm 1\). That is, \(0\), \(1\), and \(2\) can be replaced with any possible remainders, and the system will still have a solution!
Answer: \(117\)
Since our three integers are between \(0\) and \(10\), and are odd, they will each belong to the set \(\{ 1,3,5,7,9 \}\). Denote this set by \(S\). Consider an ordered triple \((a,b,c)\), where \(a\), \(b\), and \(c\) are integers from \(S\), possibly not distinct. We wish to count how many such triples satisfy \(a^2 b^2 c^2 = 2025\). Since \(a\), \(b\), and \(c\) are positive, this is equivalent to counting how many such triples satisfy \(abc=45 = 5 \times 9\).
Since \(5\) is the only multiple of \(5\) in \(S\), exactly one of \(a,b,c\) must be \(5\). Therefore the product of the other two numbers is \(9\), and the remaining two numbers are both \(3\), or one of them is \(1\) and the other of them is \(9\).
In summary, \(a,b,c\) are \(5,1,9\) in some order, or are \(5,3,3\) in some order. There are \(3!=6\) ways that \(a,b,c\) can be \(1,5,9\) in some order, and there are \(3\) ways that \(a,b,c\) can be \(3,3,5\) in some order. Therefore there are \(6+3=9\) triples in which we can have \(abc=45\).
Since there are \(5\) choices for each of \(a,b,c\), there are \(5^3\) different triples \((a,b,c)\). Therefore, the probability that the product of the squares of the three numbers equals \(2025\) is \(\dfrac{9}{5^3} = \dfrac{9}{125}\).
Answer: \(\dfrac{9}{125}\)
Note that \(xy \neq 0\), \(yz \neq 0\), and \(xz \neq 0\). Thus we may take the reciprocals of each equation to get the system \[\dfrac{4x+y}{xy} = \dfrac{10}{3} \dfrac{y+z}{yz}=\dfrac{10}{3} \dfrac{4x+z}{xz} = 14\] Also, \(x \neq 0\), \(y \neq 0\), and \(z \neq 0\) since \(xy \neq 0\) and \(xz \neq 0\). Let \(a =\dfrac{1}{x}\), \(b=\dfrac{1}{y}\), and \(c=\dfrac{1}{z}\). Note that \(\dfrac{4x+y}{xy}= \dfrac{4}{y} + \dfrac{1}{x}=4b+ a\). In terms of these new variables, the system therefore transforms to \[4b + a = \dfrac{10}{3} c + b = \dfrac{10}{3} 4c+a = 14\] This is a system of \(3\) linear equations in \(3\) unknowns.
The first equation, when solved for \(b\), gives \(b = \frac{1}{4} ( \frac{10}{3}-a)\). The third equation, when solved for \(c\), gives \(c = \frac{1}{4} ( 14-a)\). Substituting these expressions for \(b\) and \(c\) into the second equation yields \(\frac{1}{4} ( 14-a) + \frac{1}{4} ( \frac{10}{3}-a) = \frac{10}{3}\). Solving for \(a\) yields that \(a=2\). Then the first equation gives \(b=\frac{1}{3}\) and the third equation gives \(c=3\).
Therefore, \(x=\frac{1}{2}\), \(y=3\), and \(z=\frac{1}{3}\).
Answer: \((x,y,z) = \big(\frac{1}{2},3, \frac{1}{3}\big)\)
The circle with radius \(5\) has a circumference of \(10 \pi\). The \(72\degree\) angle given in the question tells us that the top circular face of the frustum will have circumference equal to \(\dfrac{360-72}{360}=\dfrac{4}{5}\) times the circumference of the circle with circumference \(10 \pi\), which equals \(8 \pi\). Therefore the diameter of the top circular face is \(8\).
Similarly, the bottom circular face of the frustum has diameter \(\dfrac{4}{5}\times(20 \pi)=16\pi\) and diameter of \(16\). Also, since \(OA=5\) and \(OB=10\), we conclude that \(AB=10-5=5\).
The diagram below is a cross section obtained by cutting the frustum with the vertical plane that contains points \(A\) and \(B\), where we call the center of the top circular face \(E\).
We divide the diagram in half using a vertical line through point \(E\). Focusing on the left half after this division, we label point \(F\) as shown, and extend \(AB\) and \(EF\) to meet at point \(G\):
Since \(\angle BFG = 90\degree\), by the Pythagorean Theorem, we have that \(EF = \sqrt{5^2 - (8-4)^2} = 3\). Since \(\triangle BFG\) and \(\triangle AEG\) are similar, and since \(AE = \frac{1}{2} BF\), we have that \(EG=\frac{1}{2}FG\), so \[FG = 2(EG) = 2(FG-EF)\] which can be rearranged to get \(FG=2(EF)=6\). Using this and \(EF=3\) gives \[EG=FG-EF=6-3=3.\]
The volume of the frustum is equal to the volume of the cone with radius \(BF\) and height \(FG\), minus the volume of the cone with radius \(AE\) and height \(EG\). Computing this quantity, we have \[\begin{align*} \frac{\pi}{3}(BF)^2(FG) - \frac{\pi}{3}(AE)^2(EG) &= \frac{\pi}{3}\left((BF)^2(FG) - (AE)^2(EG)\right) \\ &= \frac{\pi}{3}\left(8^2(6) - 4^2(3)\right) \\ &= \frac{\pi}{3}(3)(4^2)(2^2(2)-1) \\ &= 16\pi(7) \\ &= 112\pi\end{align*}\]
Answer: \(112\pi\)
We will name the positions in an arrangement of \(20\) students as position \(1\) through position \(20\), from left to right. For some \(1 \leq i \leq 19\), we say that an arrangement has a transition from position \(i\) to position \(i+1\) if the students at these two positions have different coloured hats.
How many arrangements have a transition from position \(1\) to position \(2\)? There are \(8\) options for the red-hatted students,
\(12\) options for the green-hatted
students, and \(2\) options for which
goes in position \(1\) and which goes
in position \(2\). There are then \(18!\) ways to arrange the remaining \(18\) students (remember, we are assuming
that the students are all different). Therefore, the number of sequences
with a transition from position \(1\)
to position \(2\) is \[2 \times 8 \times 12 \times 18!\]
Similarly, this number is also the number of transitions that occur from
spot \(i\) to spot \(i+1\) for all \(1
\leq i \leq 19\).
Therefore, the total number of transitions is \(19 \times 2 \times 8 \times 12 \times
18!\). The average is computed to be \[\dfrac{19 \times 2 \times 8 \times 12 \times
18!}{20!} = \dfrac{12 \times 8 \times 2}{20} =
\dfrac{48}{5}\]
Answer: \(\dfrac{48}{5}\)
\(\sqrt{25}=5\) and \(\sqrt{49}=7\), so \(\big(\sqrt{25}+\sqrt{49}\big)^2=(5+7)^2=12^2=144\).
Answer: \(144\)
Converting to decimals, we want to count the odd integers between
\(\dfrac{13}{2}=6.5\) and \(\dfrac{37}{2}=18.5\).
These integers are \(7\), \(9\), \(11\), \(13\), \(15\), and \(17\), so there are \(6\).
Answer: \(6\)
Substituting \(3w=x\) into \(2x=y\) gives \(2(3w)=y\) or \(6w=y\).
Substituting \(6w=y\) into \(5y=z\) gives \(5(6w)=z\) or \(30w=z\).
Since \(w\neq 0\), we can divide by
\(w\) to get \(30=\dfrac{z}{w}\).
Answer: \(30\)
Using the definition of \(5\lozenge
x\), we get \(5x+5+x=41\) which
is equivalent to \(6x=36\).
Dividing both sides by \(6\) gives
\(x=6\).
Answer: \(6\)
Suppose the shaded area is \(x\). Then the area of each square is \(3x\).
The area of the entire figure is the sum of the areas of the two
squares, but with the overlap subtracted since it should only be
included once in total rather than once for each square.
Thus, the area of the figure is \(3x+3x-x=5x\), so the fraction of the figure
that is shaded is \(\dfrac{x}{5x}=\dfrac{1}{5}\).
Answer: \(\dfrac{1}{5}\)
The first \(180\,\text{km}\)
took a total of \(\dfrac{180\,\text{km}}{45\,\text{km/hr}}=4\,\text{hr}\).
The second \(180\,\text{km}\) took a
total of \(\dfrac{180\,\text{km}}{90\,\text{km/hr}}=2\,\text{hr}\).
The trip took \(6\) hours, so the
average speed for the trip was \(\dfrac{360\,\text{km}}{6\,\text{hr}}=60\,\text{km/hr}\).
Answer: \(60\,\text{km/hr}\)
The given condition is that \((10 \times A+B)(10 \times C+D) = 100 \times B + 10 \times E +D\), which is equivalent to \(100 \times A \times C + 10 \times (A \times D+B \times C) + B \times D = 100 \times B + 10 \times E + D\).
We see that \(B \times D\) and \(D\) must have the same remainder when divided by \(10\). By considering all the possibilities for \(B\) and \(D\) (there are \(5(4)=20\) such possibilities), we see that either \(B=1\), or \(B=3\) and \(D=5\).
If \(B=1\), the identity becomes \[100 \times A \times C + 10 \times (A \times D+C) + D = 100 + 10 \times E + D\] Note that the right hand side is less than \(200\), while the left hand side is greater than \(200\), since \(A \times C \geq 6\), since \(A,C\) cannot be \(1\).
If \(B=3\) and \(D=5\), the identity becomes \[100 \times A \times C + 10 \times (5 \times A+3 \times C) + 15 = 300 + 10 \times E + 5\] Note that the right hand side is less than \(400\). We see that the only way the left hand side can be less than \(400\) is if \(A \times C=2\), since the only possible values of \(A \times C\) are \(2\), \(4\), and \(8\) (since \(3\) and \(5\) have already been used). Therefore, \(A\) and \(C\) are \(1\) and \(2\) in some order, and \(E\) is \(4\). The identity becomes
\[200 + 10 \times(5 \times A+3 \times C) + 15 = 300 + 40 + 5\] which means that \(5 \times A+3 \times C=13\). Since \(A=1\) and \(C=2\) is not a solution to this, and \(A=2\) and \(C=1\) is a solution, we conclude that \(A=2\), \(B=3\), \(C=1\), \(D=5\), \(E=4\). Confirming, we indeed have that \(23 \times 15=345\).
Answer: \(A=2\), \(B=3\), \(C=1\), \(D=5\), \(E=4\)
By symmetry, the final answer is four times the length of a
single chord. We will only work out the length of the chord determined
by \(x=3\), and then multiply our final
answer by \(4\).
Let \(O\) be the centre of the circle,
let \(A\) be the intersection of the
chord with the \(x\)-axis, and let
\(B\) be the upper intersection of the
chord with the circle.
Consider right-angled \(\triangle
OAB\): we have that \(OB=5\) and
\(OA=3\).
By the Pythagorean Theorem, \(AB = \sqrt{5^2 -
3^2} = 4\).
Therefore the total length of this chord is \(2\times 4=8\), and so the sum of all four
length is \(4\times 8 = 32\).
Answer: \(32\)
The prime numbers less than \(20\), and greater than \(1\), are \(2\), \(3\), \(5\), \(7\), \(11\), \(13\), \(17\), \(19\). Since there are exactly \(8\) such prime numbers on that list, this means that the \(8\) chips must be labelled precisely with these numbers.
The number of ways to select \(2\)
of the \(8\) prime numbers, where we
don’t care about the order of selection, is \({8 \choose 2} = \frac{8 \times 7}{2}= 28\).
In what follows, we treat pairs as being un-ordered.
We count the number of ways that two of these chips have a units digit
of 9 when the labels are multiplied.
If one prime number has units digit of \(1\), the other prime number must have units
digit of \(9\) - only the pair \(11,19\) satisfies these conditions.
If one prime number has units digit \(3\), the other prime number must also have
units digit \(3\) - only the pair \(3,13\) satisfies these conditions.
If one prime number has units digit \(7\), the other prime number must also have
units digit \(7\) - only the pair \(7,17\) satisfies these conditions.
The prime number \(2\) cannot be used
in any product, since multiples of \(2\) don’t end in \(9\). Similarly, the prime number \(5\) cannot be used in any product.
Therefore there are \(3\) different
ways to choose \(2\) of the \(8\) prime number and get a units digit of
\(9\).
Therefore the probability that the units digit of the product is \(9\) is \(\dfrac{3}{28}\).
Answer: \(\dfrac{3}{28}\)
Denote the \(5\) digits as \(A,B,C,D,E\). Then the condition we impose
is that \[A \times B \times C \times D \times
E -1 = A+B+C+D+E.\] Let \(L=A \times B
\times C \times D \times E-1\) and let \(R=A+B+C+D+E\). Note that none of the digits
can be \(0\), since then we would have
\(L=-1\) and \(R \geq 0\), hence \(L \neq R\).
We will find the smallest such number by using the most possible number
of \(1\)s, followed by the most
possible number of \(2\)s, etc., and
then arranging these from left to right.
All \(5\) of the digits cannot be \(1\), since then we would have \(L=0\) and \(R=5\).
Could \(4\) of the digits be \(1\)? Assume \(A=B=C=D=1\), and consider \(E\) as a variable digit. Then \(L=E-1\) and \(R=4+E\). The condition \(L=R\) is equivalent to the equation \(E-1 = 4+E\), which has no digit solution. Therefore \(4\) of the digits cannot be \(1\).
Could \(3\) of the digits be \(1\)? Assume \(A=B=C=1\), and consider \(D\), \(E\) as variable digits. Then the constraint \(L=R\) is equivalent to \(D \times E-1 = 3 + D + E\). This is equivalent to \((D-1)\times (E-1)= 5\). The only digit solution to this equation is for one digit to be \(2\) and for the other digit to be \(6\).
The above shows that the smallest number will use three \(1\)s, one \(2\), and one \(6\). Therefore, the smallest possible integer is \(11126\).
Answer: \(11126\)
Since \(64-54=10\), for every \(60\) minutes that have elapsed on Clock A, the difference between Clock S and Clock F increases by \(10\) minutes. Initially, this difference is \(0\). When Clock S reads 7:20 p.m. and Clock F reads 8:10 p.m. there is a difference of \(50\) minutes between them. This means that \(5(60)=300\) minutes, or \(5\) hours, have elapsed on Clock A since all three clocks were set to the correct time.
Since \(64-60=4\), after \(5\) hours have elapsed on Clock A, we have that \(4(5)=20\) additional minutes have elapsed on Clock F. Therefore, when Clock F reads 8:10 p.m., Clock A will read 7:50 p.m. Therefore, Joni set all three clocks to the correct time \(5\) hours before 7:50 p.m., giving us 2:50 p.m.
Answer: 2:50 p.m.
Suppose the first two numbers are \(a\) and \(b\) in that order. Then the sequence is
\[a, b, a+b, a+2b, 2a+3b, 3a+5b,
5a+8b\] and the sum of the numbers in the sequence is \[a + b + (a+b) + (a+2b) + (2a+3b) + (3a+5b) +
(5a+8b) = 13a + 20b\] The seventh number in the sequence is given
to be \(97\), so we get the equation
\(5a+8b=97\).
The sum of the seven numbers in the sequence is given to be \(245\), so we get the equation \(13a+20b=245\).
Multiplying \(5a+8b=97\) by \(5\) gives \(25a+40b=485\), and multiplying \(13a+20b=245\) by \(2\) gives \(26a+40b=490\).
Subtracting these two equations gives \(a=5\). Substituting into \(5a+8b=97\) gives \(25+8b=97\), so \(8b=72\), or \(b=9\).
Hence, \(a=5\) and \(b=9\), so the third number in the sequence
is \(a+b=5+9=14\).
Answer: \(14\)
Substituting \(x=n\) gives \(-16 = n^2+9n-n\), which is equivalent to
\(n^2+8n+16=0\).
This equation factors as \((n+4)^2=0\),
so we conclude that \(n=-4\).
Hence, \(f(x)=x^2+9x-(-4)=x^2+9x+4\),
so \(f(-2) =
(-2)^2+9(-2)+4=4-18+4=-10\).
Answer: \(-10\)
We will refer to the four dice as die \(A\), \(B\), \(C\), and \(D\) as shown in the diagram below.
Consider the face on die \(A\) that
is touching a face on die \(B\). Call
this Face \(x\) for the moment.
This face does not have \(5\) or \(3\) dots on it, since other faces on die
\(A\) have these numbers of dots.
Face \(x\) is also not opposite the
faces with \(5\) and \(3\) dots on them, so Face \(x\) does not have \(7-5=2\) or \(7-3=4\) dots on it, either.
Therefore, Face \(x\) has either \(1\) dot or \(6\) dots on it.
The total number of dots on Face \(x\)
and the face on die \(B\) that it
touches is \(8\). Since there are at
most \(6\) dots on any face of any die
and \(1+6=7<8\), Face \(x\) cannot have \(1\) dot on it.
Therefore, Face \(x\) has \(6\) dots on it.
Let Face \(y\) be the face on die
\(B\) that is touching Face \(x\). By the given condition, Face \(y\) has \(8-6=2\) dots on it.
The face on die \(B\) that is opposite
Face \(y\) has \(7-2=5\) dots on it. Therefore, the face on
die \(C\) that touches die \(B\) has \(8-5=3\) dots on it.
Now consider die \(C\). The face on top
has \(6\) dots, the face touching die
\(B\) has \(3\) dots, and so the face opposite the face
touching die \(B\) has \(7-3=4\) dots.
The face on the bottom (opposite the top face) has \(7-6=1\) dot on it, so the face touching die
\(D\) must have either \(2\) or \(5\) dots on it.
If it had \(5\) dots on it, then the
face on die \(D\) that touches die
\(C\) would have \(8-5=3\) dots on it. However, the face with
\(3\) dots on die \(D\) must be the bottom face, since it has
to be opposite the side with \(4\) dots
on it.
We conclude that the face on die \(C\)
that touches die \(D\) must have \(2\) dots on it.
Therefore, the face on die \(D\) that
touches die \(C\) has \(8-2=6\) dots on it. This face is opposite
the face marked with \(X\), so the face
marked with \(X\) has \(7-6=1\) dot on it.
Answer: \(1\)
We will use that an integer is divisible by \(9\) exactly when the sum of its digits is
divisible by \(9\).
Consider the Anderson number of \(n=999\). To compute the sum of its digits,
we need to compute the sum of the digits of the integers \(1\), \(2\), \(3\), and so on up to \(999\).
It will be useful to think of all integers from \(0\) through \(999\) as three-digit numbers, with leading
digits of \(0\) allowed. For example,
we can think of \(0\) as \(000\), \(5\) as \(005\), and \(45\) as \(045\).
Since digits of \(0\) do not contribute
to the sum of the digits, we can find the sum of the digits of the
Anderson number of \(999\) by finding
the sum of the digits of \(000\), \(001\), \(002\), \(003\), and so on to \(999\).
By allowing digits of \(0\), we get
that each of the \(10\) digits, \(0\) through \(9\), occurs the same number of times among
the three-digit numbers \(000\) through
\(999\). There are a total of \(1000\) integers, each with \(3\) digits, so the total number of digits
is \(3000\). There are \(10\) digits, so each digit occurs \(300\) times.
Therefore, the sum of the digits of the Anderson number of \(999\) is \[300(0+1+2+3+4+5+6+7+8+9)=300\times
45=13500\] Therefore, the sum of the digits of the Anderson
number of \(1000\) is \(13500+1+0+0+0=13501\). Similarly, the sum
of the digits of the Anderson number of \(1001\) is \(13501+1+0+0+1=13503\).
Continuing in this way, we can compute the sum of the digits of the
Anderson numbers of the next few four-digit integers. The results are
summarized in the following table.
| \(\boldsymbol{n}\) | Sum of digits of Anderson number of \(\boldsymbol{n}\) |
|---|---|
| \(1000\) | \(13501\) |
| \(1001\) | \(13503\) |
| \(1002\) | \(13506\) |
| \(1003\) | \(13510\) |
| \(1004\) | \(13515\) |
| \(1005\) | \(13521\) |
| \(1006\) | \(13528\) |
| \(1007\) | \(13536\) |
One can check that \(n=1007\) is the smallest four-digit integer \(n\) for which the Anderson number of \(n\) is divisible by \(9\).
Answer: \(1007\)
Suppose \(n\) has \(k\) digits and leading digit \(d\). Then \(n=10^{k-1}\times d + m\) where \(m\) is the integer obtained by removing the
leading digit from \(n\).
Then we get the equation \(10^{k-1}d+m =
57m\) or \(10^{k-1}d=56m\).
Since \(56\) is divisible by \(7\), the integer \(10^{k-1}d\) must also be divisible by \(7\).
A power of \(10\) cannot be divisible
by \(7\), so this forces \(d\) to be a multiple of \(7\). Since \(d\) is a digit, \(d=7\).
Using this and dividing both sides of \(10^{k-1}d = 56m\) by \(7\) gives \(10^{k-1} = 8m\).
Regardless of the integer \(m\), \(8m\) has at least three prime factors of
\(2\). A power of \(10\) gets exactly one prime factor of \(2\) from each factor of \(10\), so this means \(k-1\geq 3\), or \(k\geq 4\).
If we take \(k=4\), then we have \(10^3=8m\), and so \(m=125\).
The integer \(n=7125\) has the desired
properties, and the argument above shows that it is the smallest.
Answer: \(7125\)
The table below summarizes the number of segments used for each digit.
| digit | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) | \(8\) | \(9\) |
|---|---|---|---|---|---|---|---|---|---|---|
| # segments | \(6\) | \(2\) | \(5\) | \(5\) | \(4\) | \(5\) | \(6\) | \(3\) | \(7\) | \(6\) |
The digit that uses the most segments is \(8\). If an integer has two digits equal to
\(8\), then these two \(8\)s take a total of \(7+7=14\) segments.
The only digit that uses \(17-14=3\)
segments is \(7\). Thus the three
integers \(788\), \(878\), and \(887\) are the only three integers that use
exactly \(17\) segments and have two
digits equal to \(8\).
If exactly one of the digits is \(8\), then the other two digits must use
\(17-7=10\) segments in total.
The only way to do this is for the two digits to use \(4\) and \(6\) segments or for them both to use \(5\) segments.
We first count the integers that have exactly one \(8\) and two digits that use \(5\) segments.
There are three choices for where to place the \(8\), and after the \(8\) is placed, there are three choices for
the digit that uses \(5\) segments
(either \(2\), \(3\), or \(5\)) in each of the other two positions.
This gives a total of \(3\times 3\times
3=27\) integers in this case.
We next count the integers that have \(8\) as a digit, a digit that uses \(4\) segments, and a digit that uses \(6\) segments.
There are three digits that use \(6\)
segments and \(4\) is the only digit
that uses \(4\) segments.
There are three places that the \(8\)
can go, then \(2\) places that the
\(4\) can go, and three choices of
which of \(0\), \(6\), and \(9\) to use (the position is determined
after \(8\) and \(4\) are placed).
This gives a total of \(3\times 2\times
3=18\) integers. However, this total counts the integers \(084\) and \(048\), which are not at least \(100\).
Therefore, there are \(18-2=16\)
integers in this case, and total of \(27+16=43\) integers that use exactly \(17\) segments and have exactly one digit
equal to \(8\).
The only way to express \(17\) as
the sum of three nonnegative integers where each integer is less than
\(7\) is \(6+6+5\).
Therefore, if we do not use the digit \(8\), the integer must have two digits that
use \(6\) segments and one digit that
uses \(5\) segments.
There are three choices of where to place the digit that uses \(5\) segments and three choices of which
digit to use. In all, there are \(3\times
3=9\) ways to select and place the digit that uses \(5\) segments.
Once this digit is placed, there are \(3\) ways to choose each digit that uses
\(6\) segments, for a total of \(9\times 3\times 3=81\) integers.
As in a previous case, we have counted some integers with leading digits
equal to \(0\).
If \(0\) is the leading digit, then
there are two choices of where to place the digit that uses \(5\) segments and three choices for that
digit, for a total of \(2\times 3=6\)
ways to place the digit that uses \(5\)
segments.
There are three choices of which digit to place in the remaining
position since it must be one of the three digits that use \(6\) segments. This gives a total of \(6\times 3=18\) integers that have a leading
digit of \(0\).
Therefore, there are \(81-18=63\)
integers between \(100\) and \(999\) inclusive that use exactly \(17\) segments and do not have \(8\) as a digit.
Adding the totals together, there are \(3+43+63=109\) integers.
Answer: \(109\)
Since \(OPQ\) is a quarter
circle, this means \(OB\) is a radius
of the same circle and so \(OB=1\).
As well, \(OB\) is a diagonal of square
\(OABC\), so the side length of the
square is \(\dfrac{OB}{\sqrt{2}}=\dfrac{1}{\sqrt{2}}\).
Thus, the area of \(OABC\) is \(\left(\dfrac{1}{\sqrt{2}}\right)^2=\dfrac{1}{2}\).
Quarter circle \(OAC\) has radius \(OA=\dfrac{1}{\sqrt{2}}\), so its area is is
\(\dfrac{1}{4}\pi\left(\dfrac{1}{\sqrt{2}}\right)^2=\dfrac{\pi}{8}\).
Thus, the area of the "largest" shaded region is \(\dfrac{1}{2}-\dfrac{\pi}{8}\).
Suppose the total area of all shaded regions is \(x\), and suppose the total area of all
shaded regions except the largest is \(y\).
We have just computed \(x-y=\dfrac{1}{2}-\dfrac{\pi}{8}\), and we
can use a notion of similarity to find another relationship between
\(x\) and \(y\).
More precisely, since there are infinitely many shaded regions, if we
were to look separately at the part of the diagram that is within
quarter circle \(OAC\), it would look
like a scaled version of the entire diagram.
Specifically, the following relationship holds: \[\frac{y}{x} = \frac{\text{Area of }
OAC}{\text{Area of } OPQ}\] The area of \(OAC\) was computed as \(\dfrac{\pi}{8}\) above, and the area of
\(OPQ\) is \(\dfrac{\pi}{4}\). Therefore \[\frac{y}{x} = \frac{\frac{\pi}{8}}{\frac{\pi}{4}}
= \frac{1}{2}\] which can be rearranged to get \(y=\dfrac{x}{2}\). This can be substituted
into \(x-y=\dfrac{1}{2}-\dfrac{\pi}{8}\) to get
\(x-\dfrac{x}{2} =
\dfrac{1}{2}-\dfrac{\pi}{8}\) or \(x=1-\dfrac{\pi}{4}=\dfrac{4-\pi}{4}\).
Answer: \(\dfrac{4-\pi}{4}\)
Adding the two equations gives \(\sin^2
x + \cos ^2 x + \cos^2 y - \sin^2 y = \dfrac{160}{144}\).
Using the Pythagorean identity, we have \(\sin^2 x + \cos^2 x = 1\) and \(\cos^2 y = 1-\sin^2 y\), and we can
substitute into the equation above to get \(2-2\sin^2 y = \dfrac{160}{144}\).
Simplifying this equation gives \(\sin^2 y =
\dfrac{64}{144} = \dfrac{4}{9}\).
Since \(y\) is between \(0\degree\) and \(90\degree\), \(\sin y >0\), and so we must have that
\(\sin y =
\sqrt{\dfrac{4}{9}}=\dfrac{2}{3}\).
If we instead subtract the two given equations, we obtain \(\sin^2 x - \cos^2 x + \cos^2 y + \sin^2 y =
\dfrac{18}{144}\).
Substituting using the Pythagorean identity again gives \(2\sin^2 x = \dfrac{18}{144}\) or \(\sin^2 x = \dfrac{1}{16}\).
The angle \(x\) is also between \(0\degree\) and \(90\degree\), so \(\sin x >0\), so \(\sin x =
\sqrt{\dfrac{1}{16}}=\dfrac{1}{4}\).
Answering the question, we have \(\sin x +
\sin y = \dfrac{1}{4} + \dfrac{2}{3} = \dfrac{11}{12}\).
Answer: \(\dfrac{11}{12}\)
Let \(AB=x\), \(AD=y\), and \(AC=z\).
Applying the cosine law to \(\triangle
ADC\), we have \(5^2=y^2+z^2 -
2yz\cos\angle CAD\). Using \(\cos\angle
CAD= \dfrac{12}{13}\), we can simplify to \(25 = y^2+z^2-\dfrac{24yz}{13}\).
Applying the Pythagorean Theorem to \(\triangle ABC\), we have \(z^2=x^2+9^2\). Applying the Pythagorean
Theorem to \(\triangle ABD\), we have
\(y^2=x^2+4^2\).
Since \(y\) and \(z\) are side lengths, they are positive,
which means \(z=\sqrt{x^2+81}\) and
\(y=\sqrt{x^2+16}\).
Substituting into the equation obtained earlier by using the cosine law,
we get \[25 =
x^2+16+x^2+81-\dfrac{24\sqrt{x^2+16}\sqrt{x^2+81}}{13}\]
Rearranging slightly gives the equation \(13(36+x^2) =
12\sqrt{x^2+16}\sqrt{x^2+81}\).
Squaring both sides gives \(13^2(36^2+72x^2+x^4) =
12^2(x^4+97x^2+36^2)\), which can be rearranged to get \[(13^2-12^2)x^4 + (72\times 13^2 - 97\times
12^2)x^2 + (13^2-12^2)36^2\] Doing some calculator-free
calculations, we have \(13^2-12^2 =
169-144=25\), and \[\begin{align*}
72\times 13^2 - 97\times 12^2 &= 12(6\times 169 - 97\times 12) \\
&= 12(1014 - 1164) \\
&= -12\times 150 \\
&= -72\times 25\end{align*}\] and so the quartic equation in
\(x\) becomes \(25x^4 - 72\times 25x^2+25\times 36^2 = 0\),
and after dividing by \(25\), we get
\(x^4-72x^2+36^2=0\).
This quartic factors as \((x^2-36)^2=0\). Hence, \(x^2-36=0\), so \(x=\pm 6\), but \(x\) is a side length, so \(x=6\).
Answer: \(6\)
Let \(A = \log_y(x)\). Noting
\(A \neq 0\), by the change of base
formula, we have \[\frac{1}{A} =
\dfrac{1}{\log_y(x)} = \frac{1}{\frac{\log_{10}(x)}{\log_{10}(y)}} =
\frac{\log_{10}(y)}{\log_{10}(x)} = \log_x(y)\] Using this, the
equation \(3\log_y(x)+3\log_x(y)=10\)
becomes \(3A+\dfrac{3}{A} = 10\).
Multiplying through by \(A\) and
rearranging gives \(3A^2-10A+3=0\).
Factoring, we have \((3A-1)(A-3)=0\),
and so \(A=\log_y(x)=3\) or \(A=\log_y(x)=\dfrac{1}{3}\).
These two possibilities lead to \(y^3=x\) or \(\sqrt[3]{y}=x\), the latter of which is
equivalent to \(x^3=y\).
We have shown that either \(x^3=y\) or
\(y^3=x\).
If \(x^3=y\), then we can multiply
by \(x\) to get \(x^4=xy\), so \(x^4=144\) by the first given equation.
Since \(x\) is positive, this means
\(x=\sqrt[4]{144}=\sqrt{12}\). Again
using \(xy=144\), we get \(y=\dfrac{144}{\sqrt{12}}=12\sqrt{12}\), so
\(x+y=12\sqrt{12}+\sqrt{12}=13\sqrt{12}=26\sqrt{3}\).
Nearly identical reasoning shows that if \(y=x^3\), then \(x+y=26\sqrt{3}\).
Answer: \(26\sqrt{3}\)
The condition that \(f(c)=g(c)=0\) means \(c^2-2c-p=0\) and \(c^3-5c^2-q=0\).
Multiplying \(c^2-2c-p=0\) by \(c\) gives \(c^3-2c^2-pc = 0\).
Subtracting \(c^3-5c^2-q=0\) from \(c^3-2c^2-pc = 0\) gives the equation \(3c^2-pc+q=0\).
Multiplying \(c^2-2c-p=0\) by \(3\) gives \(3c^2-6c-3p=0\).
Subtracting \(3c^2-pc+q=0\) from \(3c^2-6c-3p=0\) gives the equation \((p-6)c-3p-q=0\) or \((p-6)c = 3p+q\).
If \(p-6\) is not \(0\), then \(c =
\dfrac{3p+q}{p-6}\), but since \(p\) and \(q\) are both rational, the quantity on the
right of this equation must be rational. We are given that \(c\) is irrational, so we conclude that
\(p-6=0\), which is equivalent to \(p=6\).
It follows from \((p-6)c = 3p+q\) that
\(3p+q=0\) as well, so \(q=-3p=-3(6)=-18\).
The question is to find the real solutions to the equation \(x^3-5x^2-q=0\) or \(x^3-5x^2+18=0\).
The cubic polynomial in this equation factors as \((x-3)(x^2-2x-6)=0\) (one way to find this
factorization is to use the Rational Roots Theorem). By the quadratic
formula, the roots of \(x^2-2x-6=0\)
are \(1\pm \sqrt{7}\), so the roots are
\(3\), \(1+\sqrt{7}\), and \(1-\sqrt{7}\).
Answer: \(3\), \(1+\sqrt{7}\), \(1-\sqrt{7}\)
A regular dodecagon can be circumscribed by a circle. Let \(O\) denote the center of this circle, which we also call the center of the dodecagon. Draw segments \(O A\), \(O D\), and \(O H\), as shown in the diagram below:
Note that the dodecagon is composed of the \(12\) congruent triangles \(\triangle OAB\), \(\triangle OBC\), \(\dots\), and \(\triangle OLA\). We have that \[\angle AOB = \angle BOC = \cdots = \angle LOA =
\frac{1}{12} 360\degree = 30\degree\] In particular, we have that
\(\angle AOD = 3(30\degree) =
90\degree\), \(\angle DOH =
4(30\degree) = 120\degree\), and \(\angle HOA =~ 150\degree\). Let \(r\) denote the radius of the dodecagon, by
which we mean that \(r= OA\).
We will repeatedly use that the area of a triangle is \(\frac{1}{2} ab \sin \theta\) where \(a\), \(b\)
are side lengths of the triangle, and \(\theta\) is the angle between the sides of
lengths \(a\) and \(b\).
The area of the full dodecagon is \[\mathop{\mathrm{area}}(ABCDEFGHIJKL) = 12 \times
\dfrac{1}{2} r^2 \sin (30\degree) = 3r^2\] On the other hand, we
have that \[\begin{align*}
\mathop{\mathrm{area}}(\Delta ADH) &= \mathop{\mathrm{area}}(\Delta
ADO) + \mathop{\mathrm{area}}(\Delta DHO) +
\mathop{\mathrm{area}}(\Delta HAO) \\
&= \frac{1}{2} r^2 \sin ( 90\degree) + \frac{1}{2} r^2 \sin (
120\degree) + \frac{1}{2} r^2 \sin ( 150\degree) \\
&=\dfrac{1}{2}r^2 [\sin (90\degree) + \sin(120\degree) + \sin
(150\degree)] \\
&= \dfrac{1}{4}r^2 (3 + \sqrt{3})\end{align*}\] and we
conclude that \[\frac{\mathop{\mathrm{area}}(\Delta
ADH)}{\mathop{\mathrm{area}}(ABCDEFGHIJKL)} = \dfrac{\frac{1}{4}r^2 (3
+ \sqrt{3})}{3r^2} = \dfrac{3+\sqrt{3}}{12}\]
Answer: \(\dfrac{3+\sqrt{3}}{12}\)
Draw the line \(PD\) and
consider the figures \(ABCDP\) and
\(PAQD\).
The figure \(ABCDP\) is a square
pyramid with base \(ABCD\), and \(PAQD\) is a triangular pyramid with base
\(AQD\).
Observe that the entire pyramid \(ABCDE\) is composed of these two pyramids
along with the figure \(EABPQ\).
Suppose, in general, that we have two pyramids, Pyramid 1 and Pyramid
2. Pyramid 1 has base area \(B_1\) and
height \(h_1\) from its base, and that
Pyramid 2 has base area \(B_2\) and
height \(h_2\) from its base.
We can compute the ratio of the volumes of Pyramid 1 and Pyramid 2 in
terms of the ratios of their base areas and heights. Specifically, we
have \[\begin{align*}
\frac{\text{volume of Pyramid 1}}{\text{volume of Pyramid 2}} &=
\frac{\frac{1}{3}B_1h_1}{\frac{1}{3}B_2h_2} \\
&= \frac{B_1h_1}{B_2h_2}\end{align*}\]
Let \(R\) be the point in \(ABCD\) so that \(ER\) is perpendicular to the plane \(ABCD\), and let \(T\) be the point in \(ABCD\) such that \(PT\) is perpendicular to the plane \(ABCD\) (think about why \(CTR\) is a line).
We have that \(\triangle ERC\) is
similar to \(\triangle PTC\), and so
\(\frac{PT}{ER}=\frac{PC}{EC}\).
Using that \(PC=EC-EP\) and that \(\frac{EP}{EC}=\frac{1}{3}\), we get \[\frac{PT}{ER}=\frac{PC}{EC} =
\frac{EC-EP}{EC}=1-\frac{EP}{EC}=1-\frac{1}{3}=\frac{2}{3}\] The
pyramids \(EABCD\) and \(PABCD\) have a common base \(ABCD\) (so their bases have the same area),
and their heights are \(ER\) and \(PT\), respectively. Using the result from
earlier, we have \[\dfrac{\text{volume of
$PABCD$}}{\text{volume of $EABCD$}} =
\dfrac{PT}{ER}=\dfrac{2}{3}\] and since we are given that the
volume of \(EABCD\) is \(28\), we get that the volume of \(PABCD\) is \(28\times\dfrac{2}{3}=\dfrac{56}{3}\).
Since \(EABCD\) is a regular
pyramid, the figure \(EACD\) has half
the volume of \(EABCD\), so \(EACD\) has a volume \(\dfrac{28}{2}=14\).
Figure \(EACD\) is a pyramid with base
\(\triangle ADE\) and vertex \(C\). As well, pyramid \(PADQ\) has base \(\triangle ADQ\), so the base of \(PADQ\) is contained in the base of \(EACD\), which is \(\triangle ADE\).
Consider \(\triangle ADE\) with base
\(ED\) and \(\triangle ADQ\) with base \(DQ\).
These two triangles have the same height, \(h\), from \(A\). Therefore, the ratio of their areas is
\[\dfrac{\text{area $\triangle
ADQ$}}{\text{area $\triangle ADE$}} =
\dfrac{\frac{1}{2}(DQ)h}{\frac{1}{2}(ED)h}=\dfrac{DQ}{ED}=\dfrac{ED-EQ}{ED}=1-\dfrac{EQ}{ED}=1-\dfrac{1}{2}=\dfrac{1}{2}\]
We have now shown that pyramids \(PADQ\) and \(EACD\) have bases that are in the ratio
\(1:2\). Using that \(\dfrac{EP}{EC}=\dfrac{1}{3}\) in a way
similar to earlier, we also get that their altitudes (from \(P\) to \(ADQ\) and \(C\) to \(ADE\), respectively), are in the ratio
\(1:3\).
Using the fact about volumes of pyramids from earlier, we have that
\[\dfrac{\text{volume $PADQ$}}{\text{volume
$EACD$}}=\frac{1}{2}\times\dfrac{1}{3}=\dfrac{1}{6}\] and since
the volume of \(EACD\) is \(14\), we have that the volume of \(PADQ\) is \(\dfrac{14}{6}=\dfrac{7}{3}\).
We have now computed that the volume of \(PABCD\) is \(\dfrac{56}{3}\) and the volume of \(PADQ\) is \(\dfrac{7}{3}\), so the volume of figure \(EABPQ\) is \(28-\dfrac{56}{3}-\dfrac{7}{3}=7\).
Answer: \(7\)
Throughout this solution, when we mention the moment, we mean the moment at which Ash tells a colleague that he already tested Program 6. We will consider separately the possibilities that Program 7 was written before the moment and that Program 7 has not yet been written.
If Program 7 was written before the moment, then all programs have
been written. The number of possible ordered sequences of programs that
Ash will test is equal to the number of possible lists of programs that
remain to be tested.
Note that the list of programs that are ready to be tested is always in
increasing order, so in this case, we really only need to count the
number of possibilities for the set of programs that have not yet been
tested (since each set can occur in only one possible order).
As it turns out, every possible subset of \(\{1,2,3,4,5,7\}\) could remain. For
example, for the subset \(\{3,4,7\}\)
to be remaining, the following could have happened.
Willow wrote Programs \(1\) and \(2\) before Ash started testing.
Ash finished testing Program \(2\) and started testing Program \(1\) all before Willow finished writing Program \(3\).
Willow finished programs \(3\), \(4\), \(5\), and \(6\) while Ash was testing Program \(1\).
Ash tested Program \(6\) and started testing Program \(5\) while Willow was writing Program \(7\).
Willow finished Program \(7\) and then Ash finished testing Program \(5\), at which time the moment started.
It is also possible for all programs to have been tested before the
moment, and it is also possible that only Program \(6\) has been tested before the moment. For
this to happen, it’s possible that Willow wrote the first 6 programs
before Ash arrived and finished program \(7\) before the moment, and the moment
occurs immediately after Ash finished testing Program \(6\).
Thus, for each of Programs \(1\)
through \(7\) (excluding \(6\), which is the only one we know has
definitely been tested before the moment), it either has already been
tested or it has not been tested yet. There are \(6\) programs in all and \(2\) options for each program, so there are
\(2^6=64\) possibilities for the
sequence of programs to be tested after the moment.
We now consider the possibility that Program \(7\) was not completed before the
moment.
Since Program \(6\) has been tested and
the programs are written in order, this means Program \(7\) is the only one that still needs to be
written after the moment.
Consider the list of programs waiting to be tested immediately after the
moment. We are assuming that Program \(7\) is not in the list. By reasoning
similar to that which was used in the previous case, the list could be
any subset of Programs \(1\) through
\(5\). Note, again, that for every
subset, there is only one possible order.
As Ash begins testing programs, Program \(7\) might be finished at any time. Suppose,
for example, that the list of programs ready to be tested immediately
after the moment is \(1\), \(3\), \(4\). Then the sequence of programs tested
in the afternoon could be any of \(4\),
\(3\), \(1\), \(7\)
or \(4\), \(3\), \(7\), \(1\)
or \(4\), \(7\), \(3\), \(1\), for a total of \(3\) possibilities. Note that \(7\), \(4\), \(3\), \(1\)
could be possible depending on exactly when Ash starts testing and when
Program \(7\) is completed, but this
sequence was accounted for in the previous case.
In general, if there are \(k\) programs
in the list immediately after the moment, then there are \(k\) possible sequences, for each of the
possible places in the sequence that Program \(7\) occurs, excluding the possibility that
Program \(7\) is tested first since we
have already counted these possibilities.
For each \(k\) from \(0\) through \(5\), there are \(\dbinom{5}{k}\) subsets of \(\{1,2,3,4,5\}\) of size \(k\) (and hence \(\dbinom{5}{k}\) possible lists of programs of length \(k\) ready to be tested after the moment), and then \(k\) ways to place Program \(7\).
Thus, in this case we get \[1\dbinom{5}{1}+2\dbinom{5}{2}+3\dbinom{5}{3}+4\dbinom{5}{4}+5\dbinom{5}{5}=1(5)+2(10)+3(10)+4(5)+5(1)=80\]
In total, there are \(64+80=144\) possible sequences of programs that could be tested after the moment.
Answer: \(144\)
(Note: Where possible, the solutions to parts (b) and (c) of each relay are written as if the value of \(t\) is not initially known, and then \(t\) is substituted at the end.)
Solution 1:
Listing the numbers, we have \(3\), \(5\), \(6\), \(9\), \(10\), \(12\), \(15\), \(18\), \(20\), \(21\), \(24\), \(25\), for a total of \(12\) integers.
Solution 2:
The inequality \(1 \leq 3n \leq 25\) has \(8\) positive integer solutions, so there are \(8\) multiples of \(3\) between \(1\) and \(25\). Similarly, there are \(5\) multiples of \(5\) between \(1\) and \(25\). Note that only \(15\) is included in both these counts. Therefore our answer is \(8+5-1=12\). [You can imagine that, for example, if \(25\) was replaced with \(1000\), this solution method would be much more efficient than the first solution.]
The given condition means that we have \(t-x=y-t\), and therefore \(x+y=2t=2(12)=24\).
Setting equal the equations of the line and parabola gives us \(3x+10 = x^2 - 6x + t\), which is equivalent to \(x^2 - 9x + (t-10)=0\). Therefore we have that \[\begin{align*} x &= \dfrac{9 \pm \sqrt{81-4(t-10)}}{2} \\ &= \dfrac{9 \pm \sqrt{121-4t}}{2} \\ &= \dfrac{9 \pm \sqrt{121-4(24)}}{2} \\ &= \dfrac{9 \pm \sqrt{25}}{2} \\ &= \dfrac{9 \pm 5}{2} \end{align*}\] Therefore the smallest \(x\) value is \(\dfrac{9-5}{2} = 2\).
Answer: \((12,24,2)\)
There are \(90\) two-digit positive integers. Of these, \(9\) have repeated digits. Therefore there are \(90-9=81\) two-digit positive integers with two different digits.
Let \(x\) denote the cost of one ounce of coffee and \(y\) denote the cost of one ounce of tea. It is given that \(5x+6y=t\) and \(11x+12y=171\). We wish to determine \(y\). Multiplying the first equation by \(2\) gives us that \(10x+12y=2t\). Subtracting this from the second equation gives us that \(x=171-2t\). Using the first equation gives us that \[y = \dfrac{t-5x}{6} = \dfrac{t-5(171-2t)}{6} = \dfrac{11t-855}{6}=6\]
We translate \(2\) units to the
right by replacing \(x\) with \(x-2\). This gives us the parabola with
equation \[\begin{align*}
y &= (x-2)^2 + (m-t) (x-2) + 2m-1 \\
&= x^2 + (m-t-4)x + (4+2t-1)\end{align*}\] Therefore the new
\(y\)-intercept is \(4+2t-1 = 4+2(6)-1 = 15\).
(Note that the value of \(m\) didn’t
matter. With knowledge of the rules of the event - you are trying to
come up with a single number that doesn’t depend on \(m\) - you could also have realized that the
answer must not depend on \(m\), and
thus you could just assume \(m=0\).)
Answer: \((81,6,15)\)
A number is a multiple of \(12\)
if and only if it is a multiple of \(3\) and a multiple of \(4\). Note that \(2025\) is a multiple of \(3\) since the sum of its digits is
divisible by \(3\).
The next multiple of \(3\) is \(2028\).
Since the final two digits of \(2028\)
are \(28\), which is divisible by \(4\), the answer is \(2028\).
The volume of the green box is \(10 \times 10 \times 13\). The given condition tells us that \[\dfrac{t}{4} = \dfrac{n}{100} \times 10 \times 10 \times 13 = 13n\] which simplifies to \(n = \dfrac{t}{52} = \dfrac{2028}{52} = 39\).
Since these \(3\) triangles are
isosceles, and since \(2\) of their
\(3\) lengths are radii of the circle,
we conclude that each triangle has two occurrences of the angle \(t\degree\), and that this occurrence does
not occur at \(O\).
Let \(A\) denote the third internal
angle of each triangle, which we just noted must occur at \(O\).
We have that \(A+2t = 180\), so \(A=180-2t\).
Also, we have that \(3x+3A = 360\), and
therefore \[x=120-A = 120 - (180-2t) = 2t -
60 = 2(39)-60 = 18\]
Answer: \((2028,39,18)\)