AMC 10 2012AReview

Each non-leap year has 365 days, 365 = 1 (m 7) 1 extra day Each leap year has 366 days, 366 = 2 (m 7) 2 extra daysFor 200 years, there are 200 / 4 = 50 possible leap years; but year 1900 is not a leap years, so there are 49 leap years

Total extra days = (200 – 49) * 1 + 49 * 2 = 249

Again, 249 = 4 (m 7) 4 extra days

This means that, 2/7/2012 has 4 extra weekdays from the date Dickens was born.

4 days before Tuesday is Friday. Answer: (A)

Suppose A, B, C, running at speed of 4.4, 4.8, 5.0 meters/sec.Note that A lose .4 meters per second behind B C gains .2 meters per second ahead of BThus every 500/.4 = 1250 seconds, A lose a full circle behind Band every 500/.2 = 2500 seconds, C gains a full circle ahead of B

The LCM(1250, 2500) = 2500 would be the time A, B, C meet again where A falls two full circles behind B, and C gains one full circle ahead of B.

Answer: 2500 (C)

a3 – b3 = (a – b) (a2 + ab + b2)

a3 – b3 / (a – b)3 = (a2 + ab + b2) / (a2 - 2ab + b2 ) = 73/3

3a2 + 3ab + 3b2 = 73a2 - 146 ab + 73b2

70a2 - 149ab + 70b2 = 0(7a – 10b) (10a – 7b) = 0

Hence a = 10/7 b or a = 7/10 b

Since a > b > 0, we can only have a = 10/7 b a/b = 10/7

Since a and b are relatively prime, we get a = 10, b = 7

Answer: a – b = 10 – 7 = 3 (C)

Note that the enclosed area is: area(hexagon) + 6 of the area(120-fan) – 3 * area(120-fan)

Area(hexagon) = 6 * (area of equilateral triangle of size 2) = 6 * (1/2 * 3 * 2) = 6 3Area(120-fan) = 1/3 *12 * = /3 Answer = 63 + /3 * 3 = 63 + (E)

Let Vp be speed of Paula, and V be speed of helpers.Let X be the # of minutes for lunch break(480 – X) (Vp + V) = 50% ------- (1)(372 – X) V = 24% ------- (2)(672 – X) Vp = 26% ------- (3)From (2)+(3): (372 – X) V + (672 – X)Vp = 50% ----- (4)From (1), (4): (372 – X) V + (672 – X)Vp = (480 – X) (Vp + V) 372 V + 672 Vp = 480 Vp + 480 V 192 Vp = 108 V V/Vp = 16/9 ----- (5)From (2)/(3): (372 – X)/(672 – X) (V/Vp) = 24/26 (372 – X)/(672 – X) * 16/9 = 24/26 (372 – X)/(672 – X) = 27/52 25 X = 372 *52 – 672 * 27 = 1200 X = 48 (D)

We know the center piece doesn’t change color. So the probability for the center to be black is ½. Consider the 4 corners: 7 ways to succeed

If there are 4 blacks, always succeed, and 1 way to paint so.If there are 3 blacks, always succeed, and 4 way to paint so.If there are 2 blacks, only succeed if opposite blocks are white. And there are 2 ways to paint so.If there are 1 blacks, will never succeed.

Similarly we can find 7 ways to make all non-corner blocks black.Total ways to succeed (for the 8 non-center block): 7 * 7 = 49Total ways to paint 8 non-center block: 28= 256Answer: ½ * 49/256 = 49/512 (A)

Calculate the coordinates of E, F, G, H:

a b

c

d

E

F

G

H

E=(½, 0, 3/2); F=(½, 0, 0), G=(0,1,0), H=(0,1,3/2)

Hence |EF|=|GH|=3/2; |FG|=|HE|=5/2Also EF // GH; FG // HE

Now Vector(EF) = (0, 0, -3/2)Now Vector(FG) = (-1/2, 1, 0)

Vector(EF) * Vector (FG) = 0 * (-1/2) + 0*1 + (-3/2) * 0 = 0Hence EF FG EFGH is a rectangle

Calculating the dot-product of EF, FG:

Area(EFGH) = 3/2 * 5/2 = 35/4 (C)

Sum of the first m odd integers = 1 + 3 + … + 2(m-1) + 1 = m2

Sum of the first n even integers = 2 + 4 + … + 2n = n (n + 1)m2 = n (n + 1) + 212 n2 + n + 212 – m2 = 0n = (-1 + (12 – 4(212 – m2))/2 = (-1 + (m2 - 847))/2 ---- (1)For n to be an integer, m2 – 847 must be the square of an odd int.Assume m2 – 847 = k2, where K is an odd integerm2 – k2 = 847 = 7 * 112

(m + k) * (m – k)= 7 * 112

Since (m+k), (m-k) are integers, and (m-k) < (m+k), only possible values for m-k are: 1, 7, 11Solving for m – k = 1, 7, 11 m + k = 847, 121, 77We get m = 424, 64, 44; k = 423, 57, 33From (1): n = 211, 28, 16. Answer: 211+28+16 = 255 (B)

Let N be the # of friends each one can have. We have 1 N 4If N = 1, we can form 3 pairs of friend. For the 1st person, he has 5 choices, then his friend has 1 choice.The 3rd person will then have 3 choices, & his friend has 1 choice.The 5th and 6th person will only has 1 choiceTotal choice for N =1: 5 * 3 = 15

If N = 2, we can form 2 rings of 3-friends or 1 ring of 6-friends. For the 1 ring of 6-friends, we can have 5*4*3*2/2 = 60 waysFor the 2 rings of 3-friends, we can have 5*4/2 = 10 ways to split into 2 groups, and each group only one way to form the friend.Total choices for N = 2: 60 + 10 = 70

Note that N=3 is the same as N = 2, and N=4 is the same as N =1Total ways = 15 + 70 + 70 + 15 = 170 (B)

a2 - b2 - c2 + ab = 2011 ----- (1)a2 +3b2 + 3c2 – 3ab – 2ac – 2bc = -1997 ----- (2)(1) + (2): 2a2 +2b2 + 2c2 – 2ab – 2ac – 2bc = 14(a - c)2 + (a - b)2 + (b - c)2 = 14Since a b c, (a-c) (a-b), and (a-c) (b-c)(a – c) is the largest compared to (a-b) and (b-c)Note that (a - c)2 , (a - b)2 , (b - c)2 are all positive integers,

And we have either a – b = 1 & b – c = 2 or a – b = 2 & b – c = 1

(a-c) can only be 3!

Putting: c=a-3, b=a-1 into (1) a = 2021/7 -- invalid Solving: c=a-3, b=a-2 into (1) a = 2024/8 = 253 (E)

For 0 x, y, z n, the total possible space is n3

Assume 0 x y z n, total possible space will be n3/6(since there are 6 permutations of x, y, z)For no two of x, y, z are within 1 unit, we have:y – x > 1 and z – y > 1 x < y - 1 and y < z - 1 Let y’ = y - 1 x < y’ and y’ + 1 < z - 1 Let z’ = z – 2 x < y’ and y’ < z’

Hence we get: 0 x y’ z’ n - 2

Again, for 0 x, y’, z’ n-2, the total possible space is (n-2)3

Thus for 0 x y’ z’ n-2, the total possible space is (n-2)3/6

Hence we need (n-2)3/6 / (n3/6) > ½ (n-2)3/n3 > ½ smallest n = 10