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 Image may be NSFW.
Clik here to view. such that Image may be NSFW.
Clik here to view. has 2 digits is

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

As Image may be NSFW.
Clik here to view. is between 10 and 99, Image may be NSFW.
Clik here to view. $3 \leq n \leq 24$ , 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 Image may be NSFW.
Clik here to view. $5 \times 6 + 7 \times 4$ or Image may be NSFW.
Clik here to view.. 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 Image may be NSFW.
Clik here to view. (x,y) are there to the equation Image may be NSFW.
Clik here to view. 3x+2y=27 ?

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

Solving for x, we get Image may be NSFW.
Clik here to view. $x = \frac{27-2y}{3}$ or Image may be NSFW.
Clik here to view. $x = 9 - \frac{2y}{3}$ . So in order for x to be non-negative, both Image may be NSFW.
Clik here to view. 3 | 2y and Image may be NSFW.
Clik here to view. $\frac{2y}{3} \leq 9$ , so then Image may be NSFW.
Clik here to view. $y \leq 13$ . 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 Image may be NSFW.
Clik here to view. $4n \mathrm{th}$ term is Image may be NSFW.
Clik here to view. 5n-1 .

Therefore term 2008 would be Image may be NSFW.
Clik here to view. 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) Image may be NSFW.
Clik here to view. $\frac{1}{18}$ ; (b) Image may be NSFW.
Clik here to view. $\frac{1}{9}$ ; (c) Image may be NSFW.
Clik here to view. $\frac{4}{27}$ ; (d) Image may be NSFW.
Clik here to view. $\frac{8}{27}$ ; (e) Image may be NSFW.
Clik here to view. $\frac{1}{3}$

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

The total is Image may be NSFW.
Clik here to view. $\frac{4}{9} \times \frac{2}{9} + \frac{4}{9} \times \frac{3}{9} + \frac{1}{9} \times \frac{4}{9}$ which is equal to Image may be NSFW.
Clik here to view. $\frac{8}{27}$ . The answer is (d).

Problem 8

The set of real x where Image may be NSFW.
Clik here to view. $\frac{1}{x} \leq -3 \leq x$ is:

(a) Image may be NSFW.
Clik here to view. $\{x \leq - \frac{1}{3}\}$ ; (b) Image may be NSFW.
Clik here to view. $\{-3 \leq x \leq - \frac{1}{3}\}$ ; (c) Image may be NSFW.
Clik here to view. $\{ -3 \leq x < 0 \}$ ; (d) Image may be NSFW.
Clik here to view. $\{ - \frac{1}{3} \leq x < 0 \}$ ; (e) none of these

I solved this graphically:

Image may be NSFW.
Clik here to view.

The answer is (d).

Problem 9

In quadrilateral Image may be NSFW.
Clik here to view. ABCD , Image may be NSFW.
Clik here to view. AB || DC , Image may be NSFW.
Clik here to view. DC = 2AB , Image may be NSFW.
Clik here to view. $\angle ADC = 30^\circ$ , Image may be NSFW.
Clik here to view. $BCD = 50^\circ$ . M is the midpoint of CD. The measure of Image may be NSFW.
Clik here to view. $\angle AMB$ is

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

Image may be NSFW.
Clik here to view.

Because Image may be NSFW.
Clik here to view. DC || AB and Image may be NSFW.
Clik here to view. DM = AB = MC , both ABMD and ABCM are parallelograms. Opposite angles in parallelograms are equal, so Image may be NSFW.
Clik here to view. $\angle MAB = 50^\circ$ , Image may be NSFW.
Clik here to view. $\angle MBA = 30^\circ$ . Thus Image may be NSFW.
Clik here to view. $\angle AMB = 100^\circ$ . 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 Image may be NSFW.
Clik here to view. $a \geq b \geq c$ . There are 2 cases, one where Image may be NSFW.
Clik here to view. a=b and the other when Image may be NSFW.
Clik here to view. b=c .

First, the case Image may be NSFW.
Clik here to view. a=b . If Image may be NSFW.
Clik here to view. c=1 , a can be from 2 to 9; if Image may be NSFW.
Clik here to view. 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 Image may be NSFW.
Clik here to view. b=c . We have Image may be NSFW.
Clik here to view. a > 2b so for Image may be NSFW.
Clik here to view. a=9 we have b = 1, 2, 3, 4, and if Image may be NSFW.
Clik here to view. 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) Image may be NSFW.
Clik here to view. $2^{2^{2^{2^3}}}$ ; (b) Image may be NSFW.
Clik here to view. $2^{2^{2^{3^2}}}$ ; (c) Image may be NSFW.
Clik here to view. $2^{2^{3^{2^2}}}$ ; (d) Image may be NSFW.
Clik here to view. $2^{3^{2^{2^2}}}$ ; (e) Image may be NSFW.
Clik here to view. $3^{2^{2^{2^2}}}$

Immediately we know that B>A because Image may be NSFW.
Clik here to view. 3^2 > 2^3 . Next, B>C because 512 > 81. Comparing B and D, we compare Image may be NSFW.
Clik here to view. $2^{512}$ with Image may be NSFW.
Clik here to view. $3^{16}$ . Obviously B is bigger.

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

Problem 12

A gold number is one expressible in the form Image may be NSFW.
Clik here to view. 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 Image may be NSFW.
Clik here to view. ab + a + b as Image may be NSFW.
Clik here to view. a(b+1) + b . Take this modulo Image may be NSFW.
Clik here to view. b+1 , so Image may be NSFW.
Clik here to view. $n \equiv b \mod b+1$ . Then Image may be NSFW.
Clik here to view. $n+1 \equiv 0 \mod b+1$ or Image may be NSFW.
Clik here to view. b+1 | n+1 , with Image may be NSFW.
Clik here to view. b+1 > 1 . Now if Image may be NSFW.
Clik here to view. n+1 is composite this is possible, but if n+1 is prime then this is impossible (if Image may be NSFW.
Clik here to view. b=n then Image may be NSFW.
Clik here to view. 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 Image may be NSFW.
Clik here to view. DA=1 , Image may be NSFW.
Clik here to view. DB=DC=2 , then the radius of a sphere passing through A, B, C, D is:

(a) Image may be NSFW.
Clik here to view. $\frac{3}{2}$ ; (b) Image may be NSFW.
Clik here to view. $\frac{\sqrt{5}+1}{2}$ ; (c) Image may be NSFW.
Clik here to view. $\sqrt{3}$ ; (d) Image may be NSFW.
Clik here to view. $\sqrt{2} +\frac{1}{2}$ ; (e) none of these

Put the tetrahedron on a 3D cartesian grid with D being at Image may be NSFW.
Clik here to view. (0,0,0) , A at Image may be NSFW.
Clik here to view. (1,0,0) , B at Image may be NSFW.
Clik here to view. (0,2,0) , and C at Image may be NSFW.
Clik here to view. (0,0,2) . The equation of a sphere is Image may be NSFW.
Clik here to view. (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:

Image may be NSFW.
Clik here to view. a^2 + b^2 + c^2 = r^2

Image may be NSFW.
Clik here to view. (1-a)^2 + b^2 + c^2 = r^2

Image may be NSFW.
Clik here to view. a^2 + (2-b)^2 + c^2 = r^2

Image may be NSFW.
Clik here to view. a^2 + b^2 + (2-c)^2 = r^2

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

Problem 14

Let Image may be NSFW.
Clik here to view. f(x) = x^2 , Image may be NSFW.
Clik here to view. g(x) = x^4 . We apply f and g alternatively: Image may be NSFW.
Clik here to view. f(x) , Image may be NSFW.
Clik here to view. g(f(x)) , Image may be NSFW.
Clik here to view. f(g(f(x))) , etc. When we apply f 50 times and g 49 times, the answer is Image may be NSFW.
Clik here to view. x^n where n is

(a) 148; (b) 296; (c) Image may be NSFW.
Clik here to view. $2^{148}$ ; (d) Image may be NSFW.
Clik here to view. $2^{296}$ ; (e) none of these

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

Problem 15

Triangle ABC has area 1. X and Y are on AB such that Image may be NSFW.
Clik here to view. XY = 2AX , and Z is a point on AC such that Image may be NSFW.
Clik here to view. XZ || YC and Image may be NSFW.
Clik here to view. YZ || BC . The area of Image may be NSFW.
Clik here to view. XYZ is

(a) Image may be NSFW.
Clik here to view. $\frac{1}{27}$ ; (b) Image may be NSFW.
Clik here to view. $\frac{2}{27}$ ; (c) Image may be NSFW.
Clik here to view. $\frac{1}{9}$ ; (d) Image may be NSFW.
Clik here to view. $\frac{2}{9}$ ; (e) Image may be NSFW.
Clik here to view. $\frac{1}{3}$

Image may be NSFW.
Clik here to view.

As Image may be NSFW.
Clik here to view. XZ || YC , it follows that triangles Image may be NSFW.
Clik here to view. AXZ and Image may be NSFW.
Clik here to view. AYC are similar, and Image may be NSFW.
Clik here to view. AY : AX = 3:1 so Image may be NSFW.
Clik here to view. AC:AZ = 3:1 and Image may be NSFW.
Clik here to view. AB:AY = 3:1 .

The area of Image may be NSFW.
Clik here to view. ABC is 1, so Image may be NSFW.
Clik here to view. AYC is Image may be NSFW.
Clik here to view. $\frac{1}{3}$ and Image may be NSFW.
Clik here to view. AYZ is Image may be NSFW.
Clik here to view. $\frac{1}{9}$ , and Image may be NSFW.
Clik here to view. XYC is Image may be NSFW.
Clik here to view. $\frac{2}{27}$ . The answer is (b).

Problem 16

The number of integers n for which Image may be NSFW.
Clik here to view. 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 Image may be NSFW.
Clik here to view. 2n+1 | 8(n^3 - 3n + 2) .

Now Image may be NSFW.
Clik here to view. 8(n^2 - 3n + 2) = 8 n^3 - 24n + 16 and by long division we have Image may be NSFW.
Clik here to view. 2n+1 | (2n+1) (4n^2 - 2n + 11) + 27 , or Image may be NSFW.
Clik here to view. 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).

Image may be NSFW.
Clik here to view.

AHSMC 2010 Part I

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

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