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:

The answer is (d).

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).

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