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MOCK AMC 9/09 ANSWERS & SOLUTIONS

Answer Key 1. E 2. B 3. B 4. E 5. A 6. B 7. C 8. A 9. C 10. B 11. E 12. D 13. E 14. B 15. A 16. A 17. E 18. D 19. D 20. C 21. C 22. A 23. B 24. D 25. D

1.

Note that 0 = 1 ? 2 ? 4 + 5 = 7 ? 8 ? 10 + 11 = . . . = 2005 ? 2006 ? 2008 + 2009. Clearly, the ?nal sum will be 2010 which is more than 1.

2.

We go straight to base 9. First we expand all the powers of 3: N = 317 + 316 + 315 + 311 + 310 + 39 + 35 + 34 + 33 . A little rearranging gives that quantity as: N = 4 · 316 + .3 · 314 + 4 · 310 + 3 · 38 + 4 · 34 + 3 · 32 . Now, having each exponent, we go directly to base 9: N = 4304304309 . If follows that the digital sum of N is 21.

3.

90 · 110 Let the market have value x at the beginning. In two days, the market will thus have value x · 100 100 , or 99% of its original value. Using this pattern, we take 99% to the ?fth power, which yields a ?nal answer of about 95%x.

4.

2 The probability that you don’t pass either of the mocking AMCs is ( 74 75 ) . Subtract that from one to get the probability of you attending the AIMEtator. Using di?erence of squares, it’s easy to compute p as 149.

5.

We have a nice system of equations. 4x + 6r 4x + 18r = = 50 86

Solving yields x = 8, r = 3. Thus the prices of the balls in cents are 8, 11, 14, 17, 20, 23, 26, and they cost a total of $1.19. If George buys all the balls except for the two cheapest, he will have no money left over.

1 2 x

6.

From double angle, we ?nd that x ? 1 = 2 desired product is clearly 3 · 2 = 6.

? 1. Solving this equation for x we ?nd that x =

√ 3

2. The

7.

The sum of squares from 1 to n is n · (n + 1) · (2n + 1) , 6 while the sum of cubes from 1 n is n(n + 1) 2

2

.

Plugging in 2008 and 2009 respectively, we get p = 72 · 41 · 13 · 103 · 22 · 251 and q = 5 · 3 · 67 · 41 · 72 . Hence the two terms share 2 prime factors.

11 2

8.

Pick’s theorem quickly tell us that the minimal area of the 13-gon must be

giving m · n = 22, as desired.

9.

Clearly the other root is 2 ? 3i, which makes f a parabola centered at x = 2. By Vieta’s, the sum of the roots of f is 4. Re?ecting the graph, shifting it up, and shrinking it by 1/2 does not change the vertex. However, shifting it left does change the vertex to x = 0. The two roots are now opposites of each other, with sum 0

5 There exist many ways to do this problem, but here’s the fastest way. We can compute GC and CF as √ . 2 Since we are dealing with regular polygons, we know that HGF E will itself be a square like ABCD. Since we are seeking its area, if we could simply compute a side length, we could square it. This gives the idea of the law of cosines. Using the law of cosines on GCF with GCF = 150? , we ?nd that GF 2 = area of square √ 25 3 EF GH = 2 + 25

10.

11.

1 Plug in 21 x for x, and we get the equation 3f (2x) + f ( x ) = ? log2 x. Solving this system of equations, we ?nd that ?8f (2x) = 1 + 4 log2 x. Plugging in 16 gives 8f (32) = ?17.

12.

It is easy to determine that the area of the bullseye is 1 1 2 2 ) + p · ( 80 ? p = 161 of darts is equal is ( 81 81 ) =

1 81

that of the board. The probability p that the amount

13.

Realize (x + 2)(x2 + 7x + 13) = (x + 3)3 ? 1. Two digit numbers whose cubes have remainder 1 mod 9 include 10, 13, 16, 19 . . . , for a total sum of: 1605.

c Let BC = a, AC = b, and AB = c. We therefore have BD = 3 . Since bc 3a BC BD AC CD ).

14.

BCD ?

BAC , we can use triangle

bc 3a ?c

(use the fact that = Therefore the area of triangle ABC is ratios to ?nd that CD = bc2 1 2 2 √ . which means that 3a = ab, or c = 3a . Thus, sin ∠CAB = a c = 3 15. First, reduce all those cube roots to simpler expressions!

3

2

=

ab 2 ,

√ √ 54 + 2 675 = 3 + 3 10 + √ 108 = 1 + 1 2+4+ √ 3

3

Thus, we only have to ?nd f (2): f (2) = 4 + The repeating fraction is √ √ 1 6 + 40 2 y= =? y + 6y ? 1 =? y = = ?3 + 10 6+y 2 Thus, f (2) = 1 + √ 10.

1 6+···

16.

The roots of z n?1 + z n?2 + . . . + z + 1 are the non real roots of unity that belong to the equation z n ? 1 = 0 (we are using an odd z ). So, we have that

8

(x ? zk ) = x8 + x7 + . . . + x + 1

k=1

where zk = cis 2kπ for k ∈ {1, 2, . . . , 8}. Thus, the quantity we seek is none other than 9 38 + 34 + . . . + 3 + 1 = 9841.

2009

17.

Note that 2009 = 41 · 49 ≡ ?1 · 497 ≡ ?7 (mod 42). Therefore we must ?nd the remainder when ? 72009 is divided by 42. Note that 7 · 497 = 49 ≡ 7 (mod 42). Therefore 72009 ≡ 7 mod 42, and 2009 20092009 ≡ ?7 ≡ 35 (mod 42). Alternatively, you can take (mod 6), from which you get ?1.

2009

18.

The modular numbers are 5, 7, 11, 25, 35, 49, 55, . . .. Note that the sum of these numbers can be written as (1 + 5 + 52 + . . .)(1 + 7 + 72 + . . .)(1 + 11 + 112 + . . .) ? 1 Fortunately, the question asks for the reciprocals, producing easily quaniti?ed in?nite series: (1 + 5?1 + 5?2 + . . .)(1 + 7?1 + 7?2 + . . .)(1 + 11?1 + 11?2 + . . .) ? 1 5 4 7 6 11 10 ?1= 29 48

19.

This is a brutal PIE question. First, we know that the number of ways to get to (4, 4, 4) is 12 4 8 4 = 34650. (1)

The number of paths that pass through (1, 1, 1) is the same as those that pass through (3, 3, 3): 3! while the number of paths through (2, 2, 2) is 6 2 4 2

2

9 3

6 3

= 10080,

(2)

= 8100.

(3)

The number of paths that pass through two black holes is the same for all options: (3!)2 6 2 4 2 = 3240. (4)

Finally, the number that passes through all three black holes is merely (3!)4 = 1296. By PIE, the answer is (1) ? 2(2) ? (3) + 3(4) ? (5) = 14814. 20. Denote the desired some S . Working out the ?rst few terms, we have: S 20S 100S = = = 10 + 2 1 + + 1 101 2 101 3 101 + + + 1 102 6 102 7 102 + + + 3 103 14 103 17 103 + + + 7 104 34 104 41 104 + + + 17 105 82 105 99 105 + ··· + ··· + ··· (5)

Adding these individual parts together, we ?nd that S = 21.

9 79 ,

which gives R + G = 88.

Note that ABCDEF can be inscribed in a circle with radius 1. Now consider diagonals AD, BE, and CF. Since we are choosing G inside ABCDEF , then triangle ADG is either obtuse(with longest side AD), degenerate(G is on AD), or right(with hypotenuse AD). This means that either AG2 + GD2 < AD2 or AG2 + GD2 = AD2 , which is equivalent to AG2 + GD2 ≤ AD2 , which has equality when G happens to be one of A, B , C , D, E , or F . This means that AG2 + BG2 + CG2 + DG2 + EG2 + F G2 ≤ AD2 + BE 2 + CF 2 = 4 + 4 + 4 = 12, with equality when G happens to be one of A, B , C , D, E , or F . Therefore the maximum value of AG2 + BG2 + CG2 + DG2 + EG2 + F G2 is 12.

22.

Draw a perpendicular to CB from A and label their intersection F . Helps in realizing the relatively friendly side lengths, hence, we think of imposing a coordinate system. Drawing AF we note that one can labels C = (0, 0) than A = (8, 6). Now, using the angle bisector theorem, we can ?nd that D = 70 11 , 3 . Since we took C as (0, 0) the corresponding distance between C and E is thus CE = 88 + 70 22

2

+ 32 =? CE 2 =

7330 . 121

23.

Repetitiously apply the tangent addition formula on tan tan?1 a + tan?1 b + tan?1 c to ?nd that the quantity is none other than: a+b a + b + c ? abc 1?ab + c = . a+b 1 ? (ab + bc + ac) ·c 1?

1?ab

With this realization, the problem becomes quite elementary because we can now Viete’s to ?nd that √ √ m 29 5 2 2 2 tan w = 2 . Recall that 1 + tan θ = sec θ. Using our value of tan w, we can ?nd that sec w = n = 2 which yields m + n = 31, desired

24.

Note that (a + 2b)(b ? 3a + 4c) = 4ac ? 3a2 ? 5ab + 2b2 + 8bc Let’s make the substitution x = a + 2b, y = b ? 3a + 4c. Now, (xy + x + 2y ) = 491 =? (x + 2)(y + 1) = 493 = 29 · 17 Since a is odd, we can ensure that y = a + 2b. Now, we can test for values of (x, y, z ) that satisfy a + 2b = 25, b ? 3a + 4c = 16 or a + 2b = 15, b ? 3a + 4c = 28 We obtain ?ve solutions: (5, 11, 5), (13, 7, 12), (21, 3, 19), (1, 7, 6), (9, 3, 13) Of these, (13, 7, 12) satis?es the system of equations, giving a + b + c = 32 Divide both sides of the equation by 2 to ?nd that 3 tan?1 x + 2 tan?1 (3x) =

tan(3 tan x)+tan(2 tan (3x)) 1?tan(3 tan?1 x) tan(2 tan?1 (3x))

?1 ?1

25.

π 2,

now, apply the tangent

addition formula to ?nd that = tan which implies that: ?1 ?1 1 = tan(3 tan x) tan(2 tan (3x)). One can use the tangent addition formula and the double angle identities θ ?tan3 θ to prove that tan(3θ) = 3 tan 1?3 tan2 θ . Knowing this, one can express the previous quantity as:

3x?x3 2·3x 4 2 · 1? the quadratic formula produces 1?3x2 √ (3x1 ) . Cleaning yields 33x ? 30x + 1 = 0, and applying √ 15±8 3 15+8 3 2 x = 33 . We reject the positive solution by noting that 33 ≈ 1 which implies that x ≈ 1 making tan?1 x ≈ π 4 , which will not satisfy the original equation. Hence, we keep the negative one and ?nd that √ 15?8 3 2 x = 33 .

π 2

1=

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