AHSMC 2010 Part I

The first math contest of this year! The AHSMC, has 16 multiple choice problems. 20 free marks, then 5 marks for every correct answer, 2 marks for every blank answer, and 0 for every wrong answer, so possible scores range from 20 to 100. The contest is written in 80 minutes, without a calculator.

The answers I got were: bacb bedd ccbe acbe. I’ll post my solutions here.

Problem 1

The number of positive integers $n$ such that $4n$ has 2 digits is

(a) 21; (b) 22; (c) 23; (d) 24; (e) 25

As $4n$ is between 10 and 99, $3 \leq n \leq 24$, giving 22 values. The answer is (b).

Problem 2

A 4×6 plot of land is divided into 1×1 lots by fences parallel to the edges (with fences along the edges too). The total length of the fences is

(a) 58; (b) 62; (c) 68; (d) 72; (e) 96

Count the vertical fences separately from the horizontal fences. So let’s suppose the grid is 4 columns and 6 rows, then we have 5 vertical fences and 7 horizontal fences; each vertical fence is 6 units long and each horizontal fence 4 units long.

Therefore the total length is $5 \times 6 + 7 \times 4$ or $58$. The answer is (a).

Problem 3

The GCD of two positive numbers is 1, and the LCM is 10. If neither of them are 10, their sum is

(a) 3; (b) 6; (c) 7; (d) 11; (e) none of these

Obviously the numbers are 2 and 5, as no other two coprime numbers both divide into 10. So their sum is 7. The answer is (c).

Problem 4

How many non-negative solutions $(x,y)$ are there to the equation $3x+2y=27$?

(a) 4; (b) 5; (c) 8; (d) 9; (e) 10

Solving for x, we get $x = \frac{27-2y}{3}$ or $x = 9 - \frac{2y}{3}$. So in order for x to be non-negative, both $3 | 2y$ and $\frac{2y}{3} \leq 9$, so then $y \leq 13$. Of the numbers y between 0 and 13 inclusive, 5 of them are divisible by 3. The answer is (b).

Problem 5

The sequence 1,2,3,4,6,7,8,9,… is obtained by deleting multiples of 5 from the positive integers. What is the 2010th term?

(a) 2511; (b) 2512; (c) 2513; (d) 2514; (e) none of these

Notice that the 4th term is 4, the 5th term is 9, the 12th term is 14, and so on. So the pattern is that the $4n \mathrm{th}$ term is $5n-1$.

Therefore term 2008 would be $5(502) - 1 = 2509$; then term 2009 is 2511 (skipping 2510) and term 2010 is 2512. The answer is (b).

Problem 6

5 people in a building are on floors 1, 2, 3, 21, and 40. In order to minimize their total travel distance, what floor should they get together on?

(a) 18; (b) 19; (c) 20; (d) 21; (e) none of these

Suppose we choose floor 19. Then the total distance is 18+17+16+2+21 or 74. If we choose 17, the total distance is 17+16+15+3+22 or 73, which is smaller.

In fact we can repeat this several times: at floor 3, the total is 2+1+0+18+37 or only 58 floors in total. The answer is (e).

Problem 7

9 holes are arranged in a 3×3 configuration. Two pigeons each choose a hole at random (possible the same one). The probability that they choose two holes on the opposite side of an interior wall is

(a) $\frac{1}{18}$; (b) $\frac{1}{9}$; (c) $\frac{4}{27}$; (d) $\frac{8}{27}$; (e) $\frac{1}{3}$

Let the first pigeon choose a random hole. Then we split the problem into 3 cases:

• If it’s in one of the 4 corners, then the next pigeon has a $\frac{2}{9}$ chance of landing in the correct spot, so the probability here is $\frac{4}{9} \times \frac{2}{9}$.
• If it’s in one of the 4 edges, then the next pigeon has a $\frac{3}{9}$ chance of landing in the correct spot. This is a probability of $\frac{4}{9} \times \frac{3}{9}$.
• If it’s in the center hole, then the next pigeon may land in 4 possible places, so the probability here is $\frac{1}{9} \times \frac{4}{9}$.

The total is $\frac{4}{9} \times \frac{2}{9} + \frac{4}{9} \times \frac{3}{9} + \frac{1}{9} \times \frac{4}{9}$ which is equal to $\frac{8}{27}$. The answer is (d).

Problem 8

The set of real x where $\frac{1}{x} \leq -3 \leq x$ is:

(a) $\{x \leq - \frac{1}{3}\}$; (b) $\{-3 \leq x \leq - \frac{1}{3}\}$; (c) $\{ -3 \leq x < 0 \}$; (d) $\{ - \frac{1}{3} \leq x < 0 \}$; (e) none of these

I solved this graphically:

Problem 9

In quadrilateral $ABCD$, $AB || DC$, $DC = 2AB$, $\angle ADC = 30^\circ$, $BCD = 50^\circ$. M is the midpoint of CD. The measure of $\angle AMB$ is

(a) 80; (b) 90; (c) 100; (d) 110; (e) 120

Because $DC || AB$ and $DM = AB = MC$, both ABMD and ABCM are parallelograms. Opposite angles in parallelograms are equal, so $\angle MAB = 50^\circ$, $\angle MBA = 30^\circ$. Thus $\angle AMB = 100^\circ$. The answer is (c).

Problem 10

We construct isosceles but non-equilateral triangles with integer side lengths between 1 and 9 inclusive. The number of such non-congruent triangles is

(a) 16; (b) 36; (c) 52; (d) 61; (e) none of these

Let the sides be a, b, c with $a \geq b \geq c$. There are 2 cases, one where $a=b$ and the other when $b=c$.

First, the case $a=b$. If $c=1$, a can be from 2 to 9; if $c=2$ then a can be from 3 to 9, and so on. So the possibilities are 8+7+6+…+1 = 36.

Next, the case $b=c$. We have $a > 2b$ so for $a=9$ we have b = 1, 2, 3, 4, and if $a=8$ then b = 1, 2, 3, and so on. Then the possibilities are 4+3+3+2+2+1+1 or 16.

The combined possibilities are 36+16 = 52. The answer is (c).

Problem 11

Which of the following is the largest?

(a) $2^{2^{2^{2^3}}}$; (b) $2^{2^{2^{3^2}}}$; (c) $2^{2^{3^{2^2}}}$; (d) $2^{3^{2^{2^2}}}$; (e) $3^{2^{2^{2^2}}}$

Immediately we know that B>A because $3^2 > 2^3$. Next, B>C because 512 > 81. Comparing B and D, we compare $2^{512}$ with $3^{16}$. Obviously B is bigger.

Finally we compare B with E. $B=2^{2^{512}}$ and $E = 3^{2^{16}}$. But B can be written as $4^{2^{511}}$ which is obviously bigger. The answer is (b).

Problem 12

A gold number is one expressible in the form $ab + a + b$ for positive integers a and b. The number of gold numbers between 1 and 20 inclusive is

(a) 8; (b) 9; (c) 10; (d) 11; (e) 12

Write $ab + a + b$ as $a(b+1) + b$. Take this modulo $b+1$, so $n \equiv b \mod b+1$. Then $n+1 \equiv 0 \mod b+1$ or $b+1 | n+1$, with $b+1 > 1$. Now if $n+1$ is composite this is possible, but if n+1 is prime then this is impossible (if $b=n$ then $a=0$, a contradiction). Therefore a gold number is any number that’s not one less than a prime.

Below 20, the primes are 2, 3, 5, 7, 11, 13, 17, 19, so 8 numbers between 1 and 20 are not gold numbers. Then 12 are gold numbers. The answer is (e).

Problem 13

In tetrahedron ABCD, edges DA, DB, DC are perpendicular. If $DA=1$, $DB=DC=2$, then the radius of a sphere passing through A, B, C, D is:

(a) $\frac{3}{2}$; (b) $\frac{\sqrt{5}+1}{2}$; (c) $\sqrt{3}$; (d) $\sqrt{2} +\frac{1}{2}$; (e) none of these

Put the tetrahedron on a 3D cartesian grid with D being at $(0,0,0)$, A at $(1,0,0)$, B at $(0,2,0)$, and C at $(0,0,2)$. The equation of a sphere is $(x-a)^2 + (y-b)^2 + (z-c)^2 = r^2$, and since we know four points on the sphere, we can get four equations:

$a^2 + b^2 + c^2 = r^2$

$(1-a)^2 + b^2 + c^2 = r^2$

$a^2 + (2-b)^2 + c^2 = r^2$

$a^2 + b^2 + (2-c)^2 = r^2$

Solving for a, b, c, we get $a = \frac{1}{2}$, $b = 1$, $c = 1$. So the radius, or distance from origin is $\sqrt{(\frac{1}{2})^2 + 1^2 + 1^2}$ or $\frac{3}{2}$. The answer is (a).

Problem 14

Let $f(x) = x^2$, $g(x) = x^4$. We apply f and g alternatively: $f(x)$, $g(f(x))$, $f(g(f(x)))$, etc. When we apply f 50 times and g 49 times, the answer is $x^n$ where n is

(a) 148; (b) 296; (c) $2^{148}$; (d) $2^{296}$; (e) none of these

Rather than looking at the numbers themselves, we look at the exponents. Then $f(x)$ multiplies the exponent by 2 and $g(x)$ multiplies the exponent by 4. Applying f 50 times and g 49 times gives an exponent of $2^{50} \cdot 4^{49}$ or $2^{148}$. The answer is (c).

Problem 15

Triangle ABC has area 1. X and Y are on AB such that $XY = 2AX$, and Z is a point on AC such that $XZ || YC$ and $YZ || BC$. The area of $XYZ$ is

(a) $\frac{1}{27}$; (b) $\frac{2}{27}$; (c) $\frac{1}{9}$; (d) $\frac{2}{9}$; (e) $\frac{1}{3}$

As $XZ || YC$, it follows that triangles $AXZ$ and $AYC$ are similar, and $AY : AX = 3:1$ so $AC:AZ = 3:1$ and $AB:AY = 3:1$.

The area of $ABC$ is 1, so $AYC$ is $\frac{1}{3}$ and $AYZ$ is $\frac{1}{9}$, and $XYC$ is $\frac{2}{27}$. The answer is (b).

Problem 16

The number of integers n for which $2n+1 | n^3 -3n + 2$ is

(a) 3; (b) 4; (c) 5; (d) 6; (e) 8

Notice the left side is always odd, and the right side is always even. Therefore, it is equivalent to count solutions to $2n+1 | 8(n^3 - 3n + 2)$.

Now $8(n^2 - 3n + 2) = 8 n^3 - 24n + 16$ and by long division we have $2n+1 | (2n+1) (4n^2 - 2n + 11) + 27$, or $2n + 1 | 27$.

27 has 4 positive factors (1, 3, 9, 27) and 4 negative factors, all odd. Thus there are 8 solutions. The answer is (e).