2024 Hypatia Contest
Solutions
(Grade 11)
Thursday, April 4, 2024
(in North America and South America)
Friday, April 5, 2024
(outside of North American and South America)
Thursday, April 4, 2024
(in North America and South America)
Friday, April 5, 2024
(outside of North American and South America)
©2024 University of Waterloo
Of the \(4050\) trucks sold, \(32\%\) were white or \(\dfrac{32}{100}\cdot4050=1296\) were white.
Solution 1:
Of the \(4050\) trucks sold, \(24\%\) were grey or \(\dfrac{24}{100}\cdot4050=972\) were
grey.
Since \(\dfrac14\) of the grey trucks
sold were electric, then \(\dfrac14\cdot972=243\) trucks sold were
both grey and electric.
Solution 2:
Since \(24\%\) of the trucks sold
were grey, and \(\dfrac14\) of those
were electric, then \(\dfrac{24}{100}\cdot\dfrac14=\dfrac{6}{100}\)
(or \(6\%\)) were both grey and
electric.
Thus, of the \(4050\) trucks sold,
\(\dfrac{6}{100}\cdot4050=243\) were
both grey and electric.
Solution 1:
Of the \(4050\) trucks sold, \(44\%\) were black or \(\dfrac{44}{100}\cdot4050=1782\) were
black.
Thus, the total number of black trucks, sold and unsold, was \(1782+k\), and the total number of trucks,
sold and unsold, was \(4050+k\).
Since \(46\%\) of all trucks, sold and
unsold, were black, then \(\dfrac{1782+k}{4050+k}=\dfrac{46}{100}\).
Solving, we get \[\begin{align*}
\dfrac{1782+k}{4050+k}&=\dfrac{46}{100} \\
\dfrac{1782+k}{4050+k}&=\dfrac{23}{50} \\
50(1782+k)&=23(4050+k) \\
89\,100+50k&=93\,150+23k \\
27k&=4050\\
k&=150\end{align*}\] and so there were \(150\) unsold trucks, all of which were
black.
Solution 2:
Of the \(4050\) trucks sold, \(44\%\) were black and so \(100\%-44\%=56\%\) were not black.
Therefore, \(\dfrac{56}{100}\cdot4050=2268\) trucks sold
were not black.
Since all unsold trucks were black, then there were \(2268\) trucks, sold and unsold, that were
not black.
Since \(46\%\) of all trucks, sold and
unsold, were black, then \(100\%-46\%=54\%\) of all trucks, sold and
unsold, were not black.
The total number of trucks, sold and unsold, was \(4050+k\) and \(54\%\) of these trucks were not black, thus
\(\dfrac{2268}{4050+k}=\dfrac{54}{100}\).
Solving, we get \[\begin{align*}
\dfrac{2268}{4050+k}&=\dfrac{54}{100} \\
\dfrac{2268}{4050+k}&=\dfrac{27}{50} \\
50(2268)&=27(4050+k) \\
113\,400&=109\,350+27k \\
4050&=27k\\
k&=150\end{align*}\] and so there were \(150\) unsold trucks, all of which were
black.
Evaluating, we get \(f(132)=132+1+3+2=138\).
Suppose that \(n\) is equal to
the 3-digit positive integer \(abc\).
Then \(f(n)=f(abc)=100a+10b+c+a+b+c=101a+11b+2c\).
Since \(f(n)=175\), then \(101a+11b+2c=175\).
It cannot be the case that \(a \geq
2\), since if we had \(a \geq
2\), then \(101a \geq 202\)
which is too large, noting that \(11b+2c\) is always at least \(0\).
Therefore, \(a < 2\) which means
that \(a = 1\).
When \(a = 1\), we get \(101+11b+2c=175\) or \(11b+2c=74\).
It cannot be the case that \(b \geq
7\), since if we had \(b \geq
7\), then \(11b \geq 77\) which
is too large, noting that \(2c\) is
always at least \(0\).
Therefore, \(b < 7\). If \(b=6\), then \(66+2c=74\) or \(2c=8\), and so \(c=4\).
If \(b\leq5\), then \(11b\leq55\), and so \(2c\geq74-55=19\), which is not possible
since \(c\leq9\).
We can confirm that \(f(164)=164+1+6+4=175\), and so \(n=164\).
Suppose that \(n\) is equal to
the 3-digit positive integer \(pqr\).
Then \(f(pqr)=100p+10q+r+p+q+r\), and
so \(101p+11q+2r=204\).
If \(p\geq3\), then \(101p\geq303\), and so \(p=1\) or \(p=2\).
If \(p=1\), then \(101+11q+2r=204\) or \(11q+2r=103\).
Since \(r \leq 9\), then \(2r \leq 18\) and so \(11q \geq 103 - 18 = 85\).
Therefore, \(q=8\) or \(q=9\).
If \(q=8\), then \(88+2r=103\) or \(2r=15\), which is not possible since \(r\) is an integer.
If \(q=9\), then \(99+2r=103\) or \(2r=4\), and so \(r=2\).
In this case, \(n=192\) and we can
confirm that \(f(192)=192+1+9+2=204\).
If \(p=2\), then \(202+11q+2r=204\) or \(11q+2r=2\).
The only possible solution to \(11q+2r=2\) is \(q=0\) and \(r=1\).
In this case, \(n=201\) and we can
confirm that \(f(201)=201+2+0+1=204\).
Therefore, if \(f(n)=204\), then the
possible values of \(n\) are \(192\) and \(201\).
To determine the coordinates of \(F\), we find the point of intersection of
the line through \(A\) and \(C\) and the line through \(B\) and \(E\).
The line through \(A(0,0)\) and \(C(12,12)\) has slope \(\frac{12-0}{12-0}=1\).
Since it passes through \((0,0)\), this
line has equation \(y=x\).
The line through \(B(12,0)\) and \(E(0,6)\) has slope \(\frac{6-0}{0-12}=-\frac12\).
Since it passes through \((0,6)\), this
line has equation \(y=-\frac12x+6\).
To determine the \(x\)-coordinate of
the point of intersection, \(F\), we
solve \(x=-\frac12x+6\), which gives
\(\frac32x=6\) or \(3x=12\), and so \(x=4\).
Since \(F\) lies on the line with
equation \(y=x\), then the coordinates
of \(F\) are \((4,4)\).
Solution 1:
Consider \(\triangle AEF\) as having
base \(AE=6\).
Then \(\triangle AEF\) has height equal
to the perpendicular distance from \(F\) to \(AE\), which is 4, the \(x\)-coordinate of \(F\).
The area of \(\triangle AEF\) is thus
\(\frac12\cdot6\cdot4=12\).
Solution 2:
We can determine the area of \(\triangle
AEF\) by subtracting the area of \(\triangle AFB\) from the area of \(\triangle AEB\).
Consider \(\triangle AFB\) as having
base \(AB=12\).
Then \(\triangle AFB\) has height equal
to the perpendicular distance from \(F\) to \(AB\), which is 4, the \(y\)-coordinate of \(F\).
The area of \(\triangle AFB\) is thus
\(\frac12\cdot12\cdot4=24\).
The area of \(\triangle AEB\) is \(\frac12\cdot12\cdot6=36\), and so the area
of \(\triangle AEF\) is \(36-24=12\).
To determine the area of quadrilateral \(GDEF\), our strategy will be to subtract
the area of \(\triangle AEF\) and the
area of \(\triangle CDG\) from the area
of \(\triangle ACD\).
We need to find the area of \(\triangle
CDG\) still, which means finding the coordinates of \(G\).
We can find the coordinates of \(G\) by
determining the intersection of the line through \(A\) and \(C\) with the given circle.
Thus, we proceed by finding the equation of the circle.
Since the circle has diameter \(EB\),
then its centre is the midpoint of \(EB\), which is \(\left(\frac{0+12}{2},\frac{6+0}{2}\right)\)
or \((6,3)\).
The diameter has length \(EB=\sqrt{(12-0)^2+(0-6)^2}\) or \(EB=\sqrt{180}\), which simplifies to \(EB=6\sqrt{5}\).
Thus the radius of the circle is \(r=\frac12\cdot6\sqrt{5}=3\sqrt{5}\), and so
the circle has equation \((x-6)^2+(y-3)^2=(3\sqrt{5})^2\) or \((x-6)^2+(y-3)^2=45\).
Suppose the \(x\)-coordinate of \(G\) is \(g\).
Since \(G\) lies on the line with
equation \(y=x\), then the coordinates
of \(G\) are \((g,g)\).
The point \(G\) also lies on the
circle, and thus the coordinates of \(G\) satisfy the equation of the
circle.
That is, \((g-6)^2+(g-3)^2=45\), and
solving for \(g\), we get \[\begin{align*}
g^2-12g+36+g^2-6g+9&=45 \\
2g^2-18g+45&=45\\
2g^2-18g&=0\\
2g(g-9)&=0\end{align*}\] and so \(g=0\) or \(g=9\).
Since \(G\) is distinct from \(A\), then \(g=9\) and \(G\) has coordinates \((9,9)\).
We may now determine the area of \(\triangle
CDG\).
Consider \(\triangle CDG\) as having
base \(CD=12\).
Then \(\triangle CDG\) has height equal
to the perpendicular distance from \(G\) to \(CD\), which is \(12-9=3\), since \(CD\) lies along the line \(y=12\) and the \(y\)-coordinate of \(G\) is 9.
The area of \(\triangle CDG\) is thus
\(\frac12\cdot12\cdot3=18\).
The area of \(\triangle ACD\) is half
the area of square \(ABCD\) or \(\frac12\cdot12^2=72\).
From part (b), the area of \(\triangle
AEF\) is \(12\), and so the area
of \(GDEF\) is \(72-18-12=42\).
Each Hewitt number, \(H\), can
be written as \(H=(n-1)^3+n^3+(n+1)^3\)
where \(n\) is an integer and \(n\geq2\).
(We chose \(H=(n-1)^3+n^3+(n+1)^3\)
instead of \(H=n^3+(n+1)^3+(n+2)^3\),
since the quadratic term and constant term subtract out when simplified,
as shown below.)
Expanding and simplifying, we get \[\begin{align*}
H&=(n-1)^3+n^3+(n+1)^3 \\
&=n^3-3n^2+3n-1+n^3+n^3+3n^2+3n+1 \\
&=3n^3+6n \\
&=3n(n^2+2)\end{align*}\] A Hewitt number is divisible by
\(10\) exactly when its units digit is
equal to \(0\).
If for example the units digit of \(n\)
is \(9\), then the units digit of \(3n\) is \(7\), the units digit of \(n^2+2\) is \(3\), and so the units digit of \(H=3n(n^2+2)\) is \(1\) (since \(7\times3\) has units digit \(1\)).
For each possible units digit of \(n\),
we determine the units digit of \(H\)
in the table below.
| Units digit of \(\boldsymbol{n}\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) | \(8\) | \(9\) |
|---|---|---|---|---|---|---|---|---|---|---|
| Units digit of \(\boldsymbol{3n}\) | \(0\) | \(3\) | \(6\) | \(9\) | \(2\) | \(5\) | \(8\) | \(1\) | \(4\) | \(7\) |
| Units digit of \(\boldsymbol{n^2+2}\) | \(2\) | \(3\) | \(6\) | \(1\) | \(8\) | \(7\) | \(8\) | \(1\) | \(6\) | \(3\) |
| Units digit of \(\boldsymbol{H=3n(n^2+2)}\) | \(0\) | \(9\) | \(6\) | \(9\) | \(6\) | \(5\) | \(4\) | \(1\) | \(4\) | \(1\) |
Thus, for a Hewitt number to be divisible by \(10\), the units digit of \(n\) must be \(0\), and so \(n\) must be divisible by \(10\).
When \(n=10\), \(H=3(10)(10^2+2)=3060\) which is less than
\(10\,000\).
When \(n=20\), \(H=3(20)(20^2+2)=24\,120\) which lies
between \(10\,000\) and \(100\,000\).
When \(n=30\), \(H=3(30)(30^2+2)=81\,180\) which lies
between \(10\,000\) and \(100\,000\).
When \(n\geq40\), \(H\geq3(40)(40^2+2)=192\,240\) which is
greater than \(100\,000\).
Thus, there are \(2\) Hewitt numbers
between \(10\,000\) and \(100\,000\) that are divisible by \(10\).
From part (a), each Hewitt number can be written as \(H=3n(n^2+2)\) where \(n\) is an integer and \(n\geq2\).
Since \(216=2^3\cdot3^3\), then a
Hewitt number is divisible by \(216\)
exactly when \(3n(n^2+2)\) is divisible
by \(2^3\cdot3^3\), or exactly when
\(n(n^2+2)\) is divisible by \(2^3\cdot3^2\).
That is, we need \(n(n^2+2)\) to be
divisible by \(2^3=8\) and by \(3^2=9\).
We begin by considering what is required for \(n(n^2+2)\) to be divisible by \(8\).
If \(n(n^2+2)\) is divisible by \(8\), then \(n(n^2+2)\) is divisible by \(2\) and thus is even.
If \(n\) is odd, then \(n(n^2+2)\) is odd, and so \(n\) must be even.
Since \(n\) is even, then \(n=2a\) for some positive integer \(a\), and so \(n^2+2=4a^2+2\) which is \(2\) more than a multiple of \(4\) and so \(n^2+2\) is not divisible by \(4\) (but it is divisible by \(2\)).
Since \(n(n^2+2)\) is divisible by
\(8\) and \(n^2+2\) contains exactly one factor of
\(2\), then \(n\) must be divisible by \(4\).
Next, we consider what is required for \(n(n^2+2)\) to be divisible by \(9\).
If \(n(n^2+2)\) is divisible by \(9\), then at least one of the following
must be true:
\(n\) is divisible by \(3\) and \(n^2+2\) is divisible by \(3\), or
\(n\) is divisible by \(9\), or
\(n^2+2\) is divisible by \(9\).
Assume that \(n\) is divisible by
\(9\), and thus divisible by \(3\).
Then \(n=3b\) for some positive integer
\(b\), and so \(n^2+2=9b^2+2=3(3b^2)+2\) which is \(2\) more than a multiple of \(3\) and so \(n^2+2\) is not divisible by \(3\), and thus not divisible by \(9\).
This tells us that if \(n\) is
divisible by \(3\), then \(n^2+2\) is not divisible by \(3\), and so (i) cannot be true.
Further, if \(n\) is divisible by \(9\), then \(n^2+2\) is not divisible by \(9\), and so exactly one of (ii) or (iii) is
true.
Summarizing, we get \(n(n^2+2)\) is divisible by \(8\) and by \(9\) (and thus a Hewitt number is divisible by \(216\)) exactly when \(n\) is divisible by \(4\) and by \(9\), or when \(n\) is divisible by \(4\) and \(n^2+2\) is divisible by \(9\).
Case 1: \(n\) is divisible by \(4\) and by \(9\)
Since \(4\) and \(9\) share no common divisor larger than
\(1\), then \(n\) is divisible by 4 and by 9 exactly when
\(n\) is divisible by \(4\cdot9=36\).
In this case, \(n=36k\) for positive
integers \(k\).
The first Hewitt number occurs when \(n=2\), and so the 2024th Hewitt number
occurs when \(n=2025\).
That is, \(2\leq n\leq 2025\) or \(2\leq 36k \leq 2025\), and so \(\frac{2}{36}\leq k \leq
\frac{2025}{36}\).
Since \(k\) is an integer and \(\frac{2025}{36}=56.25\), then \(1\leq k \leq 56\).
Thus in this case, 56 of the smallest 2024 Hewitt numbers are divisible
by 216.
Case 2: \(n\) is divisible by \(4\) and \(n^2+2\) is divisible by \(9\)
Since \(n\) is divisible by \(4\), then \(n=4m\) for some positive integer \(m\), and so \(n^2+2=16m^2+2\) is divisible by \(9\).
For some non-negative integers \(q\)
and \(r\), where \(0\leq r\leq 8\), every positive integer
\(m\) can be written as \(m=9q+r\), depending on its remainder, \(r\), when divided by \(9\).
Since \(16m^2+2\) must be divisible by
\(9\), then each of the following
equivalent expressions must also be divisible by \(9\): \[\begin{align*}
16m^2+2&=16(9q+r)^2+2\\
&=16(9^2q^2 +2\cdot9qr+r^2)+2\\
&=16(9^2q^2 +2\cdot9qr)+16r^2+2\\
&=9\cdot16(9q^2 +2qr)+16r^2+2\end{align*}\] which is
divisible by \(9\) exactly when \(16r^2+2\) is divisible by \(9\).
That is, \(n\) is divisible by \(4\) and \(n^2+2=16m^2+2\) is divisible by \(9\) exactly when \(16r^2+2\) is divisible by \(9\), where \(r\) is the remainder when \(m\) is divided by \(9\).
For each of the possible remainders \(0\leq r
\leq 8\), we may determine the remainder when \(16r^2+2\) is divided by \(9\).
For example, when \(r=2\), \(16r^2+2=16(2)^2+2=66\) leaves remainder
\(3\) when divided by \(9\).
Similarly, we determine the remainder when \(16r^2+2\) is divided by \(9\) for each of the possible values of
\(r\):
| Value of \(\boldsymbol{r}\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) | \(8\) |
|---|---|---|---|---|---|---|---|---|---|
| Remainder when \(\boldsymbol{16r^2+2}\) is divided by \(\boldsymbol{9}\) | \(2\) | \(0\) | \(3\) | \(2\) | \(6\) | \(6\) | \(2\) | \(3\) | \(0\) |
Thus, \(16r^2+2\) is divisible by
\(9\) when \(r=1\) or when \(r=8\).
Summarizing, we get that \(n\) is
divisible by \(4\) and \(n^2+2\) is divisible by \(9\) exactly when \(n=4m=4(9q+1)\) or when \(n=4(9q+8)\) for non-negative integers \(q\).
For the smallest \(2024\) Hewitt
numbers, \(2\leq n\leq 2025\) or \(2\leq 4(9q+1) \leq 2025\), and so \(0\leq q \leq 56\) (since \(q\) is a non-negative integer).
In this case, \(57\) of the smallest
\(2024\) Hewitt numbers are divisible
by \(216\).
Similarly, \(2\leq 4(9q+8) \leq 2025\),
and so \(0\leq q \leq 55\).
In this case, \(56\) of the smallest
\(2024\) Hewitt numbers are divisible
by \(216\).
Of the smallest \(2024\) Hewitt numbers, \(56+57+56=169\) are divisible by \(216\).
From part (a), each Hewitt number is given by \(H=3n(n^2+2)\) where \(n\) is an integer and \(n\geq2\).
If \(S\) is the sum of two distinct
Hewitt numbers, then \(S=3n(n^2+2)+3m(m^2+2)\) for some integers
\(m\) and \(n\) and for which we may assume that \(2\leq m<n\).
If there are two distinct Hewitt numbers whose sum is equal to \(9\cdot2^k\) for some positive integer \(k\), then we get the following equivalent
equations \[\begin{align*}
S&=3n(n^2+2)+3m(m^2+2)\\
9\cdot2^k&=3n(n^2+2)+3m(m^2+2)\\
3\cdot2^k&=n(n^2+2)+m(m^2+2)\\
3\cdot2^k&=n^3+m^3+2n+2m\\
3\cdot2^k&=(n+m)(n^2-nm+m^2)+2(n+m)\\
3\cdot2^k&=(n+m)(n^2-nm+m^2+2)\end{align*}\] and so if there
are two distinct Hewitt numbers whose sum is equal to \(9\cdot2^k\), then\((n+m)(n^2-nm+m^2+2)=3\cdot2^k\) for some
positive integers \(k,m,n\) where \(2\leq m<n\).
If \(m\) and \(n\) have different parity (one is even and
the other is odd), then \(n+m\) is
odd.
Also, \(n^2-nm+m^2+2\) is odd since
exactly one of \(n^2\) or \(m^2\) is odd and the remaining three terms
in the sum are even.
In this case, \(n+m\) and \(n^2-nm+m^2+2\) are both odd, and so their
product is odd.
However, \(3\cdot2^k\) is even for all
positive integers \(k\). Therefore,
\(m\) and \(n\) must have the same parity (they must
both be odd or they must both be even).
Case 1: \(m\) and \(n\) are both odd
If \(m\) and \(n\) are both odd, then \(n^2-nm+m^2+2\) is odd.
Since the only odd factors of \(3\cdot2^k\) are 1 and 3, then \(n^2-nm+m^2+2\) must equal 1 or 3.
However, if \(m\) and \(n\) are both odd with \(2\leq m<n\), then \(m\geq 3\) and \(n\geq 5\) and \(n-m \geq 2\).
Therefore, \[\begin{align*}
n^2-nm+m^2+2&=n(n-m)+m^2+2\\
&\geq 5(2)+3^2+2\\
&=21\end{align*}\] and so \(n^2-nm+m^2+2\) cannot equal \(1\) or \(3\).
Therefore, \(m\) and \(n\) cannot both be odd.
Case 2: \(m\) and \(n\) are both even
If \(m\) and \(n\) are both even, then \(m=2a\) and \(n=2b\) for some integers \(a\) and \(b\) with \(1\leq
a<b\).
Substituting and simplifying, we get \[\begin{align*}
3\cdot2^k&=(n+m)(n^2-nm+m^2+2)\\
3\cdot2^k&=(2b+2a)(4b^2-4ab+4a^2+2)\\
3\cdot2^k&=4(b+a)(2b^2-2ab+2a^2+1)\\
3\cdot2^{k-2}&=(b+a)(2b^2-2ab+2a^2+1)\end{align*}\]
Since the right side is the product of two integers, then \(k\geq2\).The factor \(2b^2-2ab+2a^2+1\) is one more than a
multiple of \(2\), and thus is odd and
so it must be equal to \(1\) or \(3\).
Since \(1\leq a<b\), then \(a\geq 1\), \(b\geq2\) and \(b-a \geq 1\), and so \[\begin{align*}
2b^2-2ab+2a^2+1&=2b(b-a)+2a^2+1\\
&\geq 4(1)+2(1^2)+1\\
&=7\end{align*}\] and so \(2b^2-2ab+2a^2+1\) cannot equal \(1\) or \(3\).
Therefore, \(m\) and \(n\) cannot both be even.
We can conclude that there cannot be two distinct Hewitt numbers whose sum is equal to \(9\cdot2^k\) for some positive integer \(k\).