Mock Aime III 2010

8/6/2019 Mock Aime III 2010

1/6

Mock AIME III

By:

Zhero

Brut3Forc3

andersonw

dysfunctionalequations

April 1, 2010


2/6

Negative AIME 1

-0. Find the value of 645.

-1. Find the sum of all distinct integers n such that n|243.-2. Find the number of ways a teacher can distribute 1 piece of candy among 243 hungry schoolchildren sitting in a row

such that no child receives more than one piece of candy, and that no two children that sit next to each other may bothreceive candy.

-3. We have a triangle abc with sides ab, bc, ca with lengths , , and , respectively, so that + + = . The measuresof angles bac, acb, and cba are A, B, and C, respectively, and the perimeter of the triangle with sides AB, BC

and CA is 2

3. Let , with center I, be the circumcircle ofabc, and let , with center O, be the incircle ofabc. Let

be the circle tangent to , , and bc, let be the circle tangent to , , and ca, and let be the circle tangent to , and ac. Let , , and be tangent to at , , and , respectively. Find the largest integer less than or equal tothe largest possible value of the area of.

-4. Let

f(x,y,z) =xyz + 49(x + y + z) + 1337

xy + yz + zx + 240.

Let g(1) = f(1, 1, 1), and let g(n) = f(g(n 1), n , n) for n 2. Find g(2010).-5. Let f(x) = x2. Find the last three digits of f(f(f((((f(((((((1 + 1f((1 + f(1)1f(11f((1 + f(1)) + 1) + 11f(f(f(1) +

1)) + 1) + 11) + f(11)) + 1)1f(1 + f(1))))))) + 1) + 1)) + 1) + f(1 + f(1 + f(1)))f(1 + f(1))) + 1) + 1) + 1.

-6. Let ABC be a triangle with AC = 5 3, BC = 2. Let A, B be points outside of triangle ABC such thatABC = 15,ACB = 75,ACB = 60,BAC = 30. Given that AB can be expressed in the form a

b + c

where a,b,c are positive integers and b is not divisible by the square of any prime, find a + b + c.

-7. In ABC, let AB = 9, BC = 13, and CA = 14. Let G, H, I, and O be the centroid, orthocenter, incenter, andcircumcenter of ABC, respectively. Let G1 be the centroid of quadrilateral HIOG, and let G1 be the isogonaconjugate of G1 with respect to ABC. Let line 1 be the polar of G1 with respect to the circumcircle ofABC, let2 be the polar of I with respect to ABC, and let X be the point of intersection of 1 and 2. Let A, B, and C bethe inverse images of points A, B, and C, with respect to an inversion about X with radius 12. Given that the area o

ABC can be expressed in the form ab

c, where b is a positive integer not divisible by the square of any prime and a

and c are relatively prime positive integers, find a + b + c.

-8. Let n be the largest number for which 13372420108242010

3nis a positive integer. Find the last three digits of the value o

this integer.

-9. Given that the root to x 3 = 0 is mn

, where m and n are complex numbers such that m = a + bi,n = c + di, where i isdefined as the square root of4, and a,b,c,d are complex numbers such that a = e + fi,b = g + hi,c = i +ji,d = k + liwhere e,f,g,h ,j,k,l are positive integers such that g,j,k and l are not divisible by any prime and e is no greater thanh, and that (|a2 + b2 + c2 + d2 + e2 + f2 + g2 + h2 + i2 + j2 + k2 + l2 + m2 + n2|)2 = y, where y is a positive integerfind the leftmost three digits of y.

-10. Let x be a positive real number satisfying x34x2 + 3x5 = 0. Ifx = a+3

b+cd

e+

3

f+g

h

j

k, where a,b,c,d,e,f,g,h ,j,k

are integers with d,h > 0, gcd(a, k) = gcd(b,c,e) = gcd(f , g , h) = 1, gcd(b, c) and gcd(f, g) not divisible by the cube oany prime. d, h not divisible by the square of any prime, and g > c, find the remainder when abce + d + f+ g + h +j + k

is divided by 1000.

-11. Given that the value of n=1

n4 + 40n3 + 225n2 + 248n + 4

n6 + 19n5 + 101n4 + 103n3 + 19n2 5n + 7can be expressed in the form p

q, where p and q are relatively prime positive integers, find the last three digits of p + q

-12. Find the sum of all positive integers k such that there exist positive integers a, b, c, d, e, f, g, h, i, j, l, m, n, o, p, qr, s, t, u, v, w, x, y, and z such that


3/6

Negative AIME 2

wz + h + j q = 0(7gk + 7k 1)(h + j) + h z = 0

112k(7k 1)3(n + 1)2 + 1 f2 = 02n + p + q + z e = 0

e3(e + 2)(a + 1)2 + 1 o2 = 0(a2 1)y2 + 1 x2 = 0

16r2y4(a2 1) + 1 u2 = 0n + l + v y = 0

(a2 1)l2 + 1 m2 = 0ai + 7k 1 l i = 0

a + u2(u2 a)2 1 (n + 4dy)2 + 1 (x + cu)2 = 0p + l(a n 1) + b(2an + 2a n2 2n 2)m = 0

q + y(ap 1) + s(2ap + 2ap2 2p 2) x = 0z + pl(a

p) + t(2ap

p2

1)

pm = 0.

-13. Solve the following 14 problems. The answer to each will be an integer. Use the following key to convert your numbersinto a message which will give you further instructions. If any of your answers have more than one digit, simply usethe last digit for converting purposes.

0 1 2 3 4 5 6 7 8 9D I V A S B O N E R

Problem the first:Alex and Anderson independently each pick one of the seven deadly sins to commit. In how many distinct ways canthey do this?

Problem the second:Let {an} be a sequence such that a1 = 1, an+1 = an + 1. Find a18.Problem the third:We call a scalene triangle with integer side lengths almost isosceles if its two shortest side lengths differ by 1. Thesmallest (in terms of area) almost isosceles right triangle has side lengths 3,4,5. Find the smallest side length of thenext smallest almost isosceles right triangle.

Problem the fourth:The Ulam sequence is defined as follows: U1 = 1, U2 = 2, and for n > 2, Un is defined to be the smallest integer thatcan be expressed as the sum of two distinct earlier terms in exactly one way. Find the eleventh term in this sequence.

Problem the fifth:There are four distinguishable hamsters. For lunch, each has the choice of 7 distinguishable meals. How many wayscan they eat lunch together?

Problem the sixth:Timmy is excited because he is only 1 year away from becoming 18, the legal age of adulthood. How old is Timmy?

Problem the seventh:In 1989, L. Ming showed that there are only 4 distinct numbers which are both Fibonacci numbers and triangularnumbers. Find the largest of these 4 numbers.

Problem the eighth:What is the smallest positive integer n such that 1/2 + 1/3 + 1/7 + 1/n < 1?

Problem the ninth:5 letters are written to different addresses, and 5 matching envelopes are prepared. In how many ways can the letterbe placed in the envelopes so that no letter is in a correct envelope?

Problem the tenth:Find the largest integer x such that 16x divides (10!)2004.


4/6

Negative AIME 3

Problem the eleventh:Let x be the integer two below one hundred, and let y be the number of two-digit primes. Find x y.Problem the twelfth:Hammy the hamster is 8 years old. If he dies in 3 years, how old (in years) will he be when he dies?

Problem the thirteenth:What is the only prime which satisfies the property that it is equal to the sum of the digits (in base ten) of its cube?

Problem the fourteenth:How many three-digit numbers (with no leading zeroes) with three distinct digits satisfy the property that if you writethe number in reverse, this reversed number is larger?

-14. In Billbobland, currency is in the form of bobs, and comes in bill (paper) form. 16 friends named Bobb, Bobby-Bob, BillBobby, Bill-Bob, Bob, Bobby-Bill, Bob-Billy, Billy-Bobby, Billy, Bob-Bill, Billy-Bob, Bob-Bob, Bill-Bobby, Bill-Bobband Frank each have some nonnegative integer amount of bills worth 1 bob each. If the number of bills that Bob-Bilhas is equal to the 3 times the number of bills that Billy-Bobby has plus 7 times the number of bobs that Frank has;Bobbs bob amount is equal to Bobby-Bobs number of bills plus 2 times the amount of bobs Frank has; Bobby gaining1337 bills would cause him to have 42 times the amount of bobs as Bill-Bobby; Bob has an amount of bobs equivalentto 67 bobs plus the sum of Bill-Bobs, Bobby-Bills, Bobbys, and Bob-Billys bobs; the combined amount of 350 timesBill-Bobbs bills plus 10 times Bob-Bobs bobs plus Bill-Bobs number of bills is 9001 over Bob-Billys number of bills;Bill-Bob has an amount of bills equal to 4 bills plus the amount of bobs that Bob-Billy and Bobby have combined;

Bob-Bobs number of bills is equal to the sum of Billy-Bob and Billys number of bobs; the amount of bills that Bobbhas plus twice the amount of bobs that Bobby-Bob has is equal to 3 times the amount of bills that Bill has; 3 timesthe amount of bobs Bobby-Bob has is equal to the sum of Bills amount of bills and Bobbs amount of bills; Frankhas no bills nor bobs (which are, admittedly, the same) at all; if Bobby-Bills amount of bills had 13 times Franksamount of bobs subtracted from it, it would be equal to 8 times the amount of bills that Bob-Billy has plus 2 additionabobs; the sum of Billy-Bobbys amount of bobs and Billys amount of bobs is equal to the sum of Bob-Bills amountof bobs and Franks amount of bobs; 3 times the amount of bills that Bobby-Bill has plus the amount of bobs thatBill-Bob has is equal to the sum of Bobs bobs and Bob-Billys bobs; 13 times Billy-Bobbys number of bobs plus 3times Billys number of bills plus 7 times Bob-Bills number of bobs equals 120; 2 times Bobbys number of bills plus 2times Bill-Bobs number of bobs plus 17 times Franks number of bills equals 4 bobs plus the number of bills Bobby-Billhas; and the 4 times the number of bills Bob-Billy possesses plus the sum of Bobbs amount of bills, Billy-Bobs amountof bills, and Bobs amount of bills equals 3 times Bob-Bobs number of bobs plus Bill-Bobbys number of bills plusBill-Bobbs number of bobs plus 2 times Bill-Bobs number of bills plus 3 times Bobbys number of bobs plus Billysnumber of bills plus 1 lone bob; then, how many bobs do Bobb, Bobby-Bob, Bill, Bobby, Bill-Bob, Bob, Bobby-BillBilly-Bobby, Billy, Bob-Bill, Billy-Bob, Bob-Bob, Bill-Bobby, Bill-Bobb, and Frank have combined?


5/6

Negative AIME 4

-15. Let a,b,c,d,e,f, g,h, i, j, k,l ,m,n,o,p,q,r,s ,t ,u,v,w,x, y,z, be functions such that

a(x) = 1x + 2x + 3x + 4x + 5x 15x + 89b(x) = 80x2 + 160x + 80

c(x) = 69x 69d(x) = 1 + 3x + 3x2 + 7

e(x) = 2x2 278x + 9669f(x) = x 2g(x) = 2 xh(x) = 37x + 13

i(x) = 1 + 33x + 7x2 + 1337x3

j(x) = 1337x1337

k(x) = x

l(x) = 1337x

m(x) = 1 + 2x2 + 3x3 + 4x4 + 5x5

n(x) = x + 42

o(x) = x 42p(x) = x 1

x1

q(x) = x 2x

2

r(x) =

x

5

s(x) = 42x

t(x) = x + 537

u(x) = 13x + 37

v(x) = 964 + 931x

w(x) = x2

x(x) = ln ex

y(x) = |x|z(x) = |x|

(x) = 70 x

Let f1, f2, f3, f4, f5, f6 be functions such that

f1(X) = n(e(v(e(r( (g(o(n(n(a( (g(i(v(e( (y(o(u( (u(p(X))))))))))))))))))))))),

f2(X) = n(e(v(e(r( (g(o(n(n(a( (l(e(t( (y(o(u( (d(o(w(n(X)))))))))))))))))))))))),

f3(X) = n(e(v(e(r( (g(o(n(n(a( (r(u(n( (a(r(o(u(n(d( (a(n(d( (d(e(s(e(r(t( (y(o(u(X))))))))))))))))))))))))))))))))))))

f4(X) = n(e(v(e(r( (g(o(n(n(a( (m(a(k(e( (y(o(u( (c(r(y(X)))))))))))))))))))))))),f5(X) = n(e(v(e(r( (g(o(n(n(a( (s(a(y( (g(o(o(d(b(y(e(X))))))))))))))))))))))),

f6(X) = n(e(v(e(r( (g(o(n(n(a( (t(e(l(l( (a( (l(i(e( (a(n(d( (h(u(r(t( (y(o(u(X))))))))))))))))))))))))))))))))))),

and let chorus(x) = f6(f5(f4(f3(f2(f1(x)))))).


6/6

Negative AIME 5

Let

a1 = w(e(r(e( (n(o( (s(t(r(a(n(g(e(r(s( (t(o( (l(o(v(e(42))))))))))))))))))))))))),

a2 = y(o(u( (k(n(o(w( (t(h(e( (r(u(l(e(s( (a(n(d( (s(o( (d(o( (i(a1)))))))))))))))))))))))))))))),

a3 = a( (f(u(l(l( (c(o(m(m(i(t(m(e(n(t(s( (w(h(a(t( (i(m( (t(h(i(n(k(i(n(g( (o(f(a2))))))))))))))))))))))))))))))))))))

a4 = y(o(u( (w(o(u(l(d(n(t( (g(e(t( (t(h(i(s( (f(r(o(m( (a(n(y( (o(t(h(e(r( (g(u(y(a3)))))))))))))))))))))))))))))))))))

a5 = i( (j(u(s(t( (w(a(n(n(a( (t(e(l(l( (y(o(u( (h(o(w( (i(m( (f(e(e(l(i(n(g(a4)))))))))))))))))))))))))))))))))))),a6 = g(o(t(t(a( (m(a(k(e( (y(o(u( (u(n(d(e(r(s(t(a(n(d(a5))))))))))))))))))))))))),

a7 = chorus(a6),

a8 = w(e(v(e( (k(n(o(w(n( (e(a(c(h( (o(t(h(e(r( (f(o(r( (s(o(( (l(o(n(g(a7)))))))))))))))))))))))))))))))))),

a9 = y(o(u(r( (h(e(a(r(t(s( (b(e(e(n( (a(c(h(i(n(g( (b(u(t(a8))))))))))))))))))))))))))),

a10 = y(o(u(r(e( (t(o(o( (s(h(y( (t(o( (s(a(y( (i(t(a9))))))))))))))))))))))),

a11 = i(n(s(i(d(e( (w(e( (b(o(t(h( (k(n(o(w( (w(h(a(t(s( (b(e(e(n( (g(o(i(n(g( (o(n(a10))))))))))))))))))))))))))))))))))

a12 = w(e( (k(n(o(w( (t(h(e( (g(a(m(e( (a(n(d( (w(e(r(e( (g(o(n(n(a( (p(l(a(y( (i(t(a11)))))))))))))))))))))))))))))))))

a13 = a(n(d( (i(f( (y(o(u( (a(s(k( (m(e( (h(o(w( (i(m( (f(e(e(l(i(n(g(a12)))))))))))))))))))))))))))))))),

a14 = d(o(n(t( (t(e(l(l( (m(e( (y(o(u(r(e( (t(o(o( (b(l(i(n(d( (t(o( (s(e(e(a13))))))))))))))))))))))))))))))))))),

a15 = chorus(a14),a16 = chorus(a15),

a17 = o(o(h( (g(i(v(e( (y(o(u( (u(p(a16))))))))))))))),

a18 = o(o(h( (g(i(v(e( (y(o(u( (u(p(a17))))))))))))))),

a19 = n(e(v(e(r( (g(o(n(n(a( (g(i(v(e( (n(e(v(e(r( (g(o(n(n(a( (g(i(v(e(a18))))))))))))))))))))))))))))))))),

a20 = g(i(v(e( (y(o(u( (u(p(a19))))))))))),

a21 = n(e(v(e(r( (g(o(n(n(a( (g(i(v(e( (n(e(v(e(r( (g(o(n(n(a( (g(i(v(e(a20))))))))))))))))))))))))))))))))),

a22 = g(i(v(e( (y(o(u( (u(p(a21))))))))))),

a23 = w(e(v(e( (k(n(o(w(n( (e(a(c(h( (o(t(h(e(r( (f(o(r( (s(o(( (l(o(n(g(a22)))))))))))))))))))))))))))))))))),

a24 = y(o(u(r( (h(e(a(r(t(s( (b(e(e(n( (a(c(h(i(n(g( (b(u(t(a23))))))))))))))))))))))))))),

a25 = y(o(u(r(e( (t(o(o( (s(h(y( (t(o( (s(a(y( (i(t(a24))))))))))))))))))))))),

a26 = i(n(s(i(d(e( (w(e( (b(o(t(h( (k(n(o(w( (w(h(a(t(s( (b(e(e(n( (g(o(i(n(g( (o(n(a25))))))))))))))))))))))))))))))))))

a27 = w(e( (k(n(o(w( (t(h(e( (g(a(m(e( (a(n(d( (w(e(r(e( (g(o(n(n(a( (p(l(a(y( (i(t(a26)))))))))))))))))))))))))))))))))

a28 = i( (j(u(s(t( (w(a(n(n(a( (t(e(l(l( (y(o(u( (h(o(w( (i(m( (f(e(e(l(i(n(g(a27)))))))))))))))))))))))))))))))))))),

a28 = g(o(t(t(a( (m(a(k(e( (y(o(u( (u(n(d(e(r(s(t(a(n(d(a28))))))))))))))))))))))))),

a30 = chorus(a29),

a31 = chorus(a30),

a32 = chorus(a31)

Compute t(h(e( (q(u(i(c(k( (b(r(o(w(n( (f(o(x( (j(u(m(p(e(d( (o(v(e(r( (t(h(e( (l(a(z(y( (d(o(g(a32)))))))))))))))))))))))))))))))))))))))))).

Happy April Fools Day! As you may have noticed, the problems arent quite... standard. But each one is completelypossible. Some of these problems are really wacky, and we hope youll take some time to twist your mind, actually dothe test, and participate in a strange competition. Well recognize the high scorers as well as any notable responses werecieve. Have fun!

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Mock Aime III 2010