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Mock AIME Series Thomas Mildorf November 24, 2005 The following are five problem sets designed to be used for preparation for the American Invi-tation Math Exam. Part of my philosophy is that one should train by working problems that are more difficult than one is likely to encounter, so I have made these mock contests ex... | Mildorf Mock AIME.pdf |
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1 Mock AIME 1: Problems 1. Let Sdenote the sum of all of the three digit positive integers with three distinct digits. Compute the remainder when Sis divided by 1000. 2. If x2+y2-30x-40y+ 242= 0, then the largest possible value ofy xcan be written asm n, where mandnare relatively prime, positive integers. Determine m+n... | Mildorf Mock AIME.pdf |
10. ABCDEFG is a regular heptagon inscribed in a unit circle centered at O. lis the line tangent to the circumcircle of ABCDEFG at A, and Pis a point on lsuch that △AOP is isosceles. Letpdenote value of AP·BP·CP·DP·EP·FP·GP. Determine the value of p2. 11. Let Sdenote the value of the sum 668∑ n=0(-1)n(2004 3n) Determin... | Mildorf Mock AIME.pdf |
2 Mock AIME 1: Answers 1. 680 2. 161 3. 110 4. 801 5. 118 6. 504 7. 320 8. 177 9. 576 10. 113 11. 006 12. 338 13. 443 14. 352 15. 141 5 | Mildorf Mock AIME.pdf |
3 Mock AIME 1: Solutions 1. Let Sdenote the sum of all of the three digit positive integers with three distinct digits. Compute the remainder when Sis divided by 1000. Answer: 680. Consider independently the sums from each digit in ABC. Each digit {1,2,..., 9} appears as A exactly 9 ·8 = 72 times since exactly 9 and 8 ... | Mildorf Mock AIME.pdf |
Answer: 801. Note that S= 70+71+···+72004= (70+71+72+73)(70+74+···+72000)+72004. But 1 + 7 + 49 + 343 = 400, so that when we divide Sby 1000 we care only about∑500 k=074k modulo 10 and the extra term 72004. Since the sum contains 74k= (2401)k≡1 (mod 10) for 501 values of k, (70+ 71+ 72+ 73)(70+ 74+···+ 72000)≡400·501≡4... | Mildorf Mock AIME.pdf |
a8=a7+a6+a5= 81 + 44 + 24 = 149 a9=a8+a7+a6= 149 + 81 + 44 = 274 a10=a9+a8+a7= 274 + 149 + 81 = 504 7. Let Ndenote the number of permutations of the 15-character string AAAABBBBBCCCCCC such that None of the first four letters is an A. (4) None of the next five letters is a B. (5) None of the last six letters is a C. (6) ... | Mildorf Mock AIME.pdf |
9. p,q, and rare three non-zero integers such that p+q+r= 26 and 1 p+1 q+1 r+360 pqr= 1 Compute pqr. Answer: 576. Consider the following algebra: 1 p+1 q+1 r+360 pqr= 1 pq+qr+rp+ 360 = pqr 359 = pqr-(pq+qr+rp) + (( p+q+r)-26)-1 385 = ( p-1)(q-1)(r-1) Now consider the factorization 385 = 5 ·7·11. Each term in the produc... | Mildorf Mock AIME.pdf |
Letω3= 1 with ω̸= 1. We have f(1) + f(ω) +f(ω2) 3=(1-1)2004+ (ω-1)2004+ (ω2-1)2004 3 =1 32004∑ n=0(2004 n) ·(-1)n·( 12004-n+ω2004-n+ (ω2)2004-n) =668∑ n=0(-1)n(2004 3n) where the last step follows in part from the fact that the only integers nfor which 1n+ωn+ω2n is non-zero are multiples of three, where the expression ... | Mildorf Mock AIME.pdf |
13. A sequence {Rn}n≥0obeys the recurrence 7 Rn= 64-2Rn-1+ 9Rn-2for any integers n≥2. Additionally, R0= 10 and R1=-2. Let S=∞∑ i=0Ri 2i Scan be expressed asm nfor two relatively prime positive integers mandn. Determine the value of m+n. Answer: 443. We have R0= 10, R1=-2 and R2=1 7·(64-2(-2) + 9(10)) =158 7. We solve f... | Mildorf Mock AIME.pdf |
ALTERNATE SOLUTION (Due to Yoni Levy) Divide the given by 2n, obtaining 7Rn 2n=64 2n-Rn-1 2n-1+9 4Rn-2 2n-2. Let us sum this equation from 2 to∞. That is, ∞∑ n=27Rn 2n=64 2n-Rn-1 2n-1+9 4Rn-2 2n-2 7∞∑ n=2Rn 2n= 64∞∑ n=2-∞∑ n=1Rn 2n+9 4n∑ n=0Rn 2n 7(S-9) = 32-(S-10) +9 4S This equation can readily be solved for S=420 23... | Mildorf Mock AIME.pdf |
14. Wally's Key Company makes and sells two types of keys. Mr. Porter buys a total of 12 keys from Wally's. Determine the number of possible arrangements of Mr. Porter's 12 new keys on his keychain (Where rotations are considered the same and any two keys of the same type are identical. ) Answer: 352. Suppose that the ... | Mildorf Mock AIME.pdf |
Obviously, there is one string with 0 B's, and there is one string with a single B that has 12 rotational positions. The case with 2 B's is a question of how far apart the B's are, which has 6 possibilities. The case with 5 B's has a rotation iff all of the string is all B's, which is a contradiction as there are only 5... | Mildorf Mock AIME.pdf |
2·32·162·k2= 1058 = 2 ·232 k=±23 48 So we have cos A=11 12,cos B=7 16, and cos C=-1 24, which imply sin A=√ 23 12,sin B=3 16√ 23, and sin C=5 24√ 23 respectively. Finally, [ABC ] = 2R2sin Asin Bsin C= 2·162·(√ 23 12)(3 16√ 23)(5 24√ 23) =115√ 23 3 which gives an answer of 115 + 23 + 3 = 141. 15 | Mildorf Mock AIME.pdf |
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4 Mock AIME 2: Problems 1. Compute the largest integer ksuch that 2004kdivides 2004!. 2. xis a real number with the property that x+1 x= 3. Let Sm=xm+1 xm. Determine the value of S7. 3. In a box, there are 4 green balls, 4 blue balls, 2 red balls, a brown ball, a white ball, and a black ball. These balls are randomly d... | Mildorf Mock AIME.pdf |
11. α, β, and γare the roots of x(x-200)(4 x+ 1) = 1. Let ω= tan-1(α) + tan-1(β) + tan-1(γ) The value of tan( ω) can be written asm nwhere mandnare relatively prime positive integers. Determine the value of m+n. 12. ABCD is a cyclic quadrilateral with AB= 8, BC = 4, CD = 1, and DA= 7. Let Oand Pdenote the circumcenter ... | Mildorf Mock AIME.pdf |
5 Mock AIME 2: Answers 1. 012 2. 843 3. 437 4. 604 5. 255 6. 351 7. 0004 8. 704 9. 280 10. 727 11. 167 12. 589 13. 071 14. 152 15. 529 4400 was also accepted, due to allegedly ambiguous wording. 19 | Mildorf Mock AIME.pdf |
6 Mock AIME 2: Solutions 1. Compute the largest integer ksuch that 2004kdivides 2004!. Answer: 012. The number of 2's in the prime factorization of 2004! is⌊2004 21⌋ +⌊2004 22⌋ +⌊2004 23⌋ +···= 1002 + 501 + 250 + ···>1000. There are 2 2's in the prime factorization of 2004; hence (22)k|2004! for all integers k≤250. Sim... | Mildorf Mock AIME.pdf |
Answer: 437. Note that the probability we want is equivalent to the probability that among 6 balls drawn out simultaneously, no two have the same color. This can be accomplished only by choosing exactly one of each color, which leaves 4 ·4·2·1·1·1 possibilities out of(13 6) total possibilities. Hence, the desired proba... | Mildorf Mock AIME.pdf |
[AFP ] + [FBP ] = 126 + 63 = 189. We have 24 + k 189=48 k 48·189 = k2+ 24k k=-24±√ 242+ 4·48·189 2=-12±√ 122+ 48·189 =-12±12√ 1 + 63 =-108,84 We take k= 84 since it represents an area. Now,AE EC=7 2and CD DB=4 7. By Menelaus' the-orem,BF FAAP PDDC DB=-1 (Ceva and Menelaus use the convention of directed distances, where... | Mildorf Mock AIME.pdf |
Because φ(125) = 100, we have 4200320022001 ≡4320022001 (mod 125). We are interested in 320022001(mod 100). We play the same card again, that is, φ(100) = 40 so that 320022001≡322001(mod 100). We are also interested in 22001(mod 40). Clearly, 8 divides 22001so that 22001≡0 (mod 8). We also have 22001≡2 (mod 5) by Ferma... | Mildorf Mock AIME.pdf |
AB·25x+AE·40x= 132 ·39x 25AB+ 40AE = 39 ·132 15AB+ 24AE =39·132·3 5 Finally, [ ABE ] =1 2·(15AB+24AE) =1 239·132·3 5=7722 5. Therefore, the answer is 722+5 = 727. 11. α, β, and γare the roots of x(x-200)(4 x+ 1) = 1. Let ω= tan-1(α) + tan-1(β) + tan-1(γ) The value of tan( ω) can be written asm nwhere mandnare relativel... | Mildorf Mock AIME.pdf |
Hence D′ABis a right triangle and the circumradius of ABCD is√ 65 2. Now, by similar triangles, we have AP:BP:CP:DP= 56 : 32 : 4 : 7. Let AP= 56xso that AC= 60x and BD= 39x. Ptolemy's theorem applied to ABCD yields 60 x·39x= 1·8 + 4·7 = 36 from which x2=1 65. Now we apply Stewart's theorem to triangle BOD and cevian OP... | Mildorf Mock AIME.pdf |
The first ksuch that ∆k(n) is constant for all integers nmust be at least k= 9; hence Pis at least 9th degree. Since Pis of minimal degree, we may assert that ∆9(n) is constant. We may now retrace our subtractions to find ∆0(11). Specifically, ∆0P(11) = ∆0(10) + ∆1(10) = ∆0(10) +( ∆1(9) + ∆2(9)) =··· =9∑ k=0∆k(10-k) =9∑ k... | Mildorf Mock AIME.pdf |
exactly two of {1,2,3,4}are D's and the other two are O's, then there are 6 possible type arrangements. x+y=nhasn-1 solutions in positive integers, hence this subcase has 6·3·4 = 72 possible strings. Finally, if there is one slot filled with D's and three filled with O's, then there are 4 type arrangements. x+y+z=nhas(n-... | Mildorf Mock AIME.pdf |
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7 Mock AIME 3: Problems 1. Three circles are mutually externally tangent. Two of the circles have radii 3 and 7. If the area of the triangle formed by connecting their centers is 84, then the area of the third circle iskπfor some integer k. Determine k. 2. Let Ndenote the number of 7 digit positive integers have the pr... | Mildorf Mock AIME.pdf |
Determine the remainder obtained when Nis divided by 1000. 9. ABC is an isosceles triangle with base AB. Dis a point on ACand Eis the point on the extension of BDpast Dsuch that ∠BAE is right. If BD= 15, DE = 2, and BC= 16, then CDcan be expressed asm n, where mandnare relatively prime positive integers. Determine m+n.... | Mildorf Mock AIME.pdf |
8 Mock AIME 3: Answers 1. 196 2. 435 3. 601 4. 071 5. 936 6. 121 7. 049 8. 472 9. 225 10. 058 11. 035 12. 164 13. 115 14. 468 15. 041 31 | Mildorf Mock AIME.pdf |
9 Mock AIME 3: Solutions 1. Three circles are mutually externally tangent. Two of the circles have radii 3 and 7. If the area of the triangle formed by connecting their centers is 84, then the area of the third circle iskπfor some integer k. Determine k. Answer: 196. Let rdenote the radius of the third circle. Then the... | Mildorf Mock AIME.pdf |
3 =( ζ2 1+ζ2 2+ζ2 3) (ζ1+ζ2+ζ3) =( ζ3 1+ζ3 2+ζ3 3) +∑ Symζ2 1ζ2 = 7 +∑ Symζ2 1ζ2 =⇒∑ Symζ2 1ζ2=-4 1 = ( ζ1+ζ2+ζ3)3=( ζ3 1+ζ3 2+ζ3 3) + 3∑ Symζ2 1ζ2+ 6ζ1ζ2ζ3 = 7-12 + 6 ζ1ζ2ζ3 =⇒ζ1ζ2ζ3= 1 But now, ( x-ζ1)(x-ζ2)(x-ζ3) =x3-(x2+x+ 1). If we write Sn=ζn 1+ζn 2+ζn 3, we have Sn+3=Sn+2+Sn+1+Sn. With this recursion, we find S4=... | Mildorf Mock AIME.pdf |
=1√ 29800∑ n=1√ n+ 1-√ n-1 =1√ 2·(√ 9801 +√ 9800-√ 1-√ 0) = 70 + 49√ 2 And it follows that the answer is 70 + 49 + 2 = 121. 7. ABCD is a cyclic quadrilateral that has an inscribed circle. The diagonals of ABCD in-tersect at P. If AB= 1, CD= 4, and BP:DP= 3 : 8, then the area of the inscribed circle of ABCD can be expre... | Mildorf Mock AIME.pdf |
is positive for all of its partial sums. Thus, there are 128 ordered 8-tuples corresponding to this subset. Continuing in this fashion: {5,6,7} 64. {1,2,7,8} 0. {1,3,6,8} 128. {1,4,5,8} 128. {2,3,5,8} 128. {1,4,6,7} 128. {2,3,6,7} 128. {2,4,5,7} 128. {3,4,5,6} 128. {1,2,3,4,8} 128. {1,2,3,5,7} 128. {1,2,4,5,6} 128. Whe... | Mildorf Mock AIME.pdf |
11. ABC is an acute triangle with perimeter 60. Dis a point on BC. The circumcircles of triangles ABD and ADC intersect ACand ABat Eand Frespectively such that DE= 8 and DF= 7. If ∠EBC ∼=∠BCF, then the value of AE AFcan be expressed asm n, where mand nare relatively prime positive integers. Compute m+n. Answer: 035. Si... | Mildorf Mock AIME.pdf |
Determine the remainder obtained when Sis divided by 1000. Answer: 115. We note the combinatorial identity k(n k) =n(n-1 k-1) and write k2(2005 k) = 2005 k(2004 k-1) = 2005( (k-1)(2004 k-1) +(2004 k-1)) = 2005( 2004(2003 k-2) +(2004 k-1)) Employing this result, S=(2 3)2005 ·2005∑ k=1k2 2k(2005 k) =(2 3)2005 ·20052005∑ ... | Mildorf Mock AIME.pdf |
But inversion is its own inverse transformation. Hence, PC= 18 and PB= 12. The inversive distance formula gives BC=R2·B′C′ PB′·PC′=18·12 24= 9. Finally, the area of PBC may be found via Heron's formula: K=√ 39 221 215 23 2=9√ 455 4. The answer is therefore 455 + 9 + 4 = 468. 15. Let Ω denote the value of the sum 40∑ k=... | Mildorf Mock AIME.pdf |
39 | Mildorf Mock AIME.pdf |
10 Mock AIME 4: Problems 1. For how many positive integers n >1 is it possible to express 2005 as the sum of ndistinct positive integers? 2. a1, a2,... is a sequence of real numbers where anis the arithmetic mean of the previous n-1 terms for n > 3 and a2004= 7. b1, b2,... is a sequence of real numbers in which bnis th... | Mildorf Mock AIME.pdf |
13. x,y, and zare distinct non-zero integers such that-7≤x, y, z ≤7. Compute the number of solutions ( x, y, z ) to the equation 1 x+1 y+1 z=1 x+y+z 14. In triangle ABC,BC= 27, CA = 32, and AB= 35. Pis the unique point such that the perimeters of triangles BPC, CPA, and APB are equal. The value of AP+BP+CPcan be expres... | Mildorf Mock AIME.pdf |
11 Mock AIME 4: Answers 1. 061 2. 180 3. 103 4. 667 5. 075 6. 002 7. 434 8. 919 9. 035 10. 595 11. 346 12. 240 13. 504 14. 171 15. 039 42 | Mildorf Mock AIME.pdf |
12 Mock AIME 4: Solutions 1. For how many positive integers n >1 is it possible to express 2005 as the sum of ndistinct positive integers? Answer: 061. The sum of ndistinct positive integers is at least 1 + 2 + 3 + ···+n=n(n+1) 2, but because we can exchange nwith n+kfor any integer k≥0, the sum of ndistinct positive i... | Mildorf Mock AIME.pdf |
Answer: 075. We apply the algebra 13x2+[ 52( y+1 2)2-13]-20x( y+1 2) = 563 13(x 2)2 + 13( y+1 2)2-10(x 2)( y+1 2) = 144 13x2 0+ 13y2 0-10x0y0= 144 where x0=x 2andy0=y+1 2. It is clear that the transformation ( x, y)(x0, y0) shifts ( x, y) up half of a unit and then scales this image by a factor of1 2along the xdirectio... | Mildorf Mock AIME.pdf |
Answer: 919. Let p(n) be the probability that the atom is safely contained if released from (n,0). In this notation, P(-3) = 0, P(7) = 1. Now, since the particle is twice as likely to move right as it is likely to move left, P(n) =2 3P(n+ 1) +1 3P(n-1) or equivalently P(n+ 1) =3P(n)-P(n-1) 2for-2≤n≤6. Let P(-2) = r. Th... | Mildorf Mock AIME.pdf |
11. 10 lines and 10 circles divide the plane into at most ndisjoint regions. Compute n. Answer: 346. Any arrangement of 10 lines and 10 circles can be constructed in any order. Ten lines such that no two are parallel and no three have a common intersection divide the plane into 1 + (1 + 2 + 3 + ···+ 10) = 56 regions. N... | Mildorf Mock AIME.pdf |
14. In triangle ABC,BC= 27, CA = 32, and AB= 35. Pis the unique point such that the perimeters of triangles BPC, CPA, and APB are equal. The value of AP+BP+CPcan be expressed asp+q√r s, where p, q, r, and sare positive integers such that there is no prime divisor common to p, q, and s, and ris not divisible by the squa... | Mildorf Mock AIME.pdf |
that MS= 2√ 2, from which UV2=1 4( 2UM2+ 2US2-MS2) = 20042+ 20052-2≡39 (mod 1000). 48 | Mildorf Mock AIME.pdf |
49 | Mildorf Mock AIME.pdf |
13 Mock AIME 5: Problems 1. The length of a diagonal connecting opposite vertices of a rectangular prism is 47. Determine its volume, given that one of its dimensions is 2 and that the other two dimensions differ by√ 2005. 2. Compute the sum of the prime divisors of 12+ 22+ 32+···+ 20052. 3. x,y, and zare positive integ... | Mildorf Mock AIME.pdf |
Letγ= min ( |x1|,|x2|,|x3|), where |a+bi|=√ a2+b2andi=√-1. Determine the value ofγ6-15γ4+γ3+ 56γ2. 12. ABC is a scalene triangle. The circle with diameter ABintersects BCat D, and Eis the foot of the altitude from C. Pis the intersection of ADand CE. Given that AP= 136, BP= 80, and CP= 26, determine the circumradius of... | Mildorf Mock AIME.pdf |
14 Mock AIME 5: Answers 1. 200 2. 680 3. 001 4. 028 5. 057 6. 233 7. 046 8. 013 9. 707 10. 253 11. 056 12. 085 13. 024 14. 392 15. 037 52 | Mildorf Mock AIME.pdf |
15 Mock AIME 5: Solutions I have provided full solutions for only the last two problems, which are extraordinarily difficult. See if you can solve the others!6 1. The length of a diagonal connecting opposite vertices of a rectangular prism is 47. Determine its volume, given that one of its dimensions is 2 and that the ot... | Mildorf Mock AIME.pdf |
11. x1, x2, and x3are complex numbers such that x1+x2+x3= 0 x2 1+x2 2+x2 3= 16 x3 1+x3 3+x3 3=-24 Letγ= min ( |x1|,|x2|,|x3|), where |a+bi|=√ a2+b2andi=√-1. Determine the value ofγ6-15γ4+γ3+ 56γ2. 12. ABC is a scalene triangle. The circle with diameter ABintersects BCat D, and Eis the foot of the altitude from C. Pis t... | Mildorf Mock AIME.pdf |
15. Let O= (0,0) and A= (14,0) denote the origin and a point on the positive x-axis respectively. B= (x, y) is a point not on the line y= 0. These three points determine lines l1, l2, and l3. Let P1,..., P ndenote all of the points that are equidistant from lifori= 1,2,3. Let Qj denote the distance from Pjto the liforj... | Mildorf Mock AIME.pdf |
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