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Interesting problems from the AMATYC Student Math League Exams 2011 (February 2011, #1) After Ed eats 20% of a pie and Ahn eats 40% of a pie, Ed has twice as much left as Ahn. Find Ed’s original amount of pie as a percentage of Ahn’s original amount. Let E be Ed’s original amount of pie, and A be Ahn’s original amount of pie. .8E 2 .6 A E 2 .6 3 150% . A .8 2 So the correct answer is D) 150. (February 2011, #2) The expression a # b ab2 a for integers a, b 0 . If a # b #3 250 , find a b . a # b #3 250 9 ab2 a ab2 a 250 ab 2 a 25 a b 2 1 25 . The factor pairs of 25 are 1, 25 and 5, 5. This means that a 5 and b 2 . So the correct answer is B) 7. (February 2011, #3) Alicia always climbs steps 1, 2, or 4 at a time. For example, she climbs 4 steps by 1-1-1-1, 1-1-2, 1-2-1,2-1-1,2-2, or 4. In how many ways can she climb 10 steps? Only 1’s Ten 1’s 10! 1 10! Only 2’s Five 2’s 5! 1 5! 1’s and 2’s: eight 1’s and one 2 six 1’s and two 2’s Four 1’s and three 2’s Two 1’s and four 2’s 9! 9 8!1! 8! 28 6!2! 7! 35 4!3! 6! 15 2!4! 1’s and 4’s: six 1’s and one 4 two 1’s and two 4’s 7! 7 6!1! 4! 6 2!2! 2’s and 4’s: three 2’s and one 4 one 2 and two 4’s 4! 4 3!1! 3! 3 1!2! 1’s, 2’s, and 4’s: four 1’s, one 2, and one 4 two 1’s, two 2’s, and one 4 6! 30 4!1!1! 5! 30 2!2!1! This gives 1 1 9 28 35 15 7 6 4 3 30 30 . OR The number of different ways to get to the 5th step is equal to the number of different ways to get to the 4th step plus the number of different ways to get to the 3rd step plus the number of different ways to get to the 1st step. This is so because the 1st step plus 4 steps gets you to the 5th step, the 3rd step plus 2 steps gets you to the 5th step, and the 4th step plus 1 step gets you to the 5th step. In general, an an1 an2 an4 . With the starting values of a1 1, a2 2, a3 3, a4 6 , let’s try this new scheme to get the rest. a5 a4 a3 a1 6 3 1 10 , a6 a5 a4 a2 10 6 2 18 , a7 a6 a5 a3 18 10 3 31, a8 a7 a6 a4 31 18 6 55 , a9 a8 a7 a5 55 31 10 96, a10 a9 a8 a6 96 55 18 169 So the correct answer is E) 169. [See the section on Sets and Counting] (February 2011, #4) The sum of six consecutive positive integers beginning at n is a perfect cube. The smallest such n is 2. Find the sum of the next two smallest such n’s. n n 1 n 2 n 3 n 4 n 5 6n 15 . Since this must be an odd number, we’ll only consider odd cubic numbers: 27, 125, 343, …. 27 gives the value 2, 6 2 15 27 . 729 gives the value 119. From the list of answer choices, we can check 679 119 560, 680 119 561, 681 119 562, 682 119 563, 683 119 564 to see if they generate a cubic number. 6 560 15 3375 153 . So the correct answer is A) 679. (February 2011, #5) The sum of the infinite geometric series S is 6, and the sum of the series whose terms are the squares of the terms of S is 15. Find the sum of the infinite geometric series with the same first term and opposite common ratio as S. a ar ar ar ar 2 3 4 a , a2 a2r 2 a2r 4 a2r 6 a2r8 1 r a ar ar ar ar 2 3 4 a 1 r a2 1 r 2 a 1 r 15 5 . 6 2 So the correct answer is B) 2.5. [See the section on Algebraic Formulas] a2 1 r2 (February 2011, #11) Multiplying the corresponding terms of a geometric and an arithmetic sequence yields 96, 180, 324, 567, …. Find the next term of the new sequence. a, ar , ar 2 , ar 3 , b, b d , b 2d , b 3d , , so we get ab 96, ar b d 180, ar 2 b 2d 324, ar 3 b 3d 567 . Assuming that a and b are whole numbers, 96 25 3 , let’s try a 4 and b 24 . This leads to the system 4r 24 d 180,4r 2 24 2d 324,4r 3 24 3d 567 , which has as a solution d 6 and 4,6,9, 272 , 814 , 3 r . These values lead to the sequences 2 24,30,36,42,48, 96,180,324,567,972, . and the product sequence So the correct answer is B) 972. [See the section on Algebraic Formulas] (February 2011, #12) If log x y log y x 2.9 and xy 128 , find x y . Assuming that x and y are whole numbers, then since 128 26 , the possible values for x and y are x 1, y 128 , x 2, y 64 , x 4, y 32 , x 8, y 16 . For the pair x 4, y 32 , you get log 4 32 log32 4 5 2 29 2.9 . 2 5 10 So the correct answer is B) 36. (February 2011, #13) The equation a5 b2 c 2 2011(a, b, c positive integers) has a solution in which two of the three numbers are prime. Find the value of the nonprime number. The possible values of a are 1, 2, 3, and 4. For a 1, we get b2 c 2 2010 . For a 2 , we get b2 c 2 1979 . For a 3 , we get b2 c 2 1768 . For a 4 , we get b2 c 2 987 . We can eliminate a 1 and a 4 . So now we need to check a 1 and a 3 . For a 1, 2010 382 566 , which is not a square, 2010 402 410 , which is not a square, 2010 422 246 , which is not a square, 2007 442 71 , which is not a square, 2007 462 109 . For a 3 , 1768 382 324 , which is 182 , but 18 is not a prime. 1768 402 402 , which is not a square, 1768 422 4 , which is 22 , so we get 3, 2, and 42. So the correct answer is C) 42. (February 2011, #14) A palindrome is a number like 121 or 1551 which reads the same from right to left and from left to right. How many 4-digit palindromes are divisible by 17? 4-digit palindromes are of the form abba, where a is 1,2,3,4,5,6,7,8,or 9 and b is 0,1,2,3,4,5,6,7,8,9. Now abba 1000a 100b 10b a 1001a 110b 11 91a 10b , so we can just examine the 4-digit numbers which are multiples of both 17 and 11, and hence just multiples of 187. We can skip multiples of 10. 187 6 1122 187 19 3553 187 31 5797 187 42 7854 187 7 1309 187 21 3927 187 32 5984 187 43 8041 187 8 1496 187 22 4114 187 33 6171 187 44 8228 187 9 1683 187 23 4301 187 34 6358 187 45 8415 187 11 2057 187 24 4488 187 35 6545 187 46 8602 187 12 2244 187 25 4675 187 36 6732 187 47 8789 187 13 2431 187 26 4862 187 37 6919 187 48 8976 187 14 2618 187 27 5049 187 38 7106 187 49 9163 187 15 2805 187 28 5236 187 39 7293 187 51 9537 187 17 3179 187 29 5423 187 41 7667 187 52 9724 And 187 53 9911 . So the correct answer is B) 4. (February 2011, #16) The increasing sequence of positive integers a1 , a2 , a3 , satisfies the equation an2 an an1 for all n 1 . If a7 160 , find a8 . The sequence is a1 , a2 , a1 a2 , a1 2a2 ,2a1 3a2 ,3a1 5a2 ,5a1 8a2 ,8a1 13a2 , . We know that 5a1 8a2 160 , which means that a2 must be a multiple of 5. If we go with a2 5 , then a1 14 , but it doesn’t work. a8 8a1 13a2 8 16 13 10 258 . So the correct answer is B) 258. If we go with a2 10 , then a1 16 , it works. (October 2011, #1) If the standard order of operations is reversed (that is, additions and subtractions are done first and exponentiation is done last), what is the value of 2 3 ^ 2 3 ? 23^ 2 3 23^ 5 6^5 7776 So the correct answer is E) 7776. (October 2011, #2) The price of a stock rose 20% on Monday, fell 10% on Tuesday, and 1 increased by on Wednesday. By what percent did the price rise from Monday to 6 Wednesday? 1.2 S .9 7 1.26S 6 So the correct answer is B) 26. (October 2011, #3) The system of equations ax by 8 and ax by 20 has the solution x, y 2,3 . Find a b . Plugging in 2,3 leads to 2a 3b 8 2a 3b 20 get that b 2 . So the correct answer is D) 9. . Adding the two equations leads to a 7 , so we also (October 2011, #4) The positive integers a, b, and c satisfy a6 b2 c 2 2011 . a bc. Find The possible values of a are 1, 2, and 3. These lead to b2 c 2 2010 with b c 49 to 53 , b2 c 2 1947 with b c 48 to 52 , and b2 c 2 1282 with b c 47 to 51 With a 3 , you get b 1 c 1282 b 2 35.79106 2 35.74913 3 35.67913 4 35.58089 5 35.4542 6 35.29873 7 35.1141 8 34.89986 9 34.65545 10 34.38023 11 34.07345 12 33.73426 13 33.36165 14 32.95451 15 32.51154 16 32.03123 17 31.5119 18 30.95158 19 30.34798 20 29.69848 21 29 So the correct answer is D) 53. The previous table can be generated on a TI-83 calculator by pressing the ‘Y =’ key, entering 1282 X , pressing the ‘TABLE’ key, and scrolling down. 2 (October 2011, #5) Different shades of pink, red, and white can be made by mixing whole numbers of quarts of red and white paint. Shades are different if the ration of red to white paint is different. Find the number of different possible shades that can be made from at most 4 quarts of red and 5 quarts of white paint. The total number of ratios is 4 5 20 , but some of them are equivalent. 1:1, 2:2, 3:3, and 4:4 are equivalent. 1:2 and 2:4 are equivalent. 2:1 and 4:2 are equivalent. This gets us down to only 15 different ratios. So the correct answer is A) 15. [See the section on Sets and Counting] (October 2011, #6) The function y f x has zeros 2 and 6. Find the zeros of y 3 f 2 2 x . 2 2 x 2 x 2 and 2 2 x 6 x 2 So the correct answer is A) 2, 2 . (October 2011, #7) One population P1 t grows exponentially at the same rate that another population P2 t decays exponentially. If the populations were both equal to P on Jan. 1 2009, how will the populations be related on Jan. 1 2012? P1 t Pe at and P2 t Pe at , so P1 t P2 t P 2 . So the correct answer is B) P1 t P2 t P 2 . (October 2011, #8) For b c 0 , both x2 bx 8 and x 2 cx 8 factor over the integers. Find b c . It must be that x 2 bx 8 x 8 x 1 x 2 9 x 8 and b x cx 8 x 2 x 4 x 6 x 8 2 2 c So the correct answer is C) 3. (October 2011, #9) Ed drives from San Mateo to Atascadero, a distance of 197.5 mi. He starts driving at a constant speed and reduces his speed by 5 mph after each half hour of driving. If the trip takes 3 hr 20 min, how far did he travel in the first 2 hours? 1 2 R 12 R 5 12 R 10 12 R 15 12 R 20 12 R 25 13 R 30 197.5 10 3 R 952 197.5 20 R 285 1185 R 73.5 1 2 73.5 12 73.5 5 12 73.5 10 12 73.5 15 132 So the correct answer is B) 132. (October 2011, #10) Sun fills her 10 liter radiator with 20% antifreeze and 80% water. She removes some of the mixture and replaces it with antifreeze. If the radiator is now one quarter antifreeze, how many liters of the original mixture did she remove? 2 .2 x x 1 10 4 .8 x 2 1 10 4 3.2 x 8 10 3.2 x 2 x .625 So the correct answer is D) .625. (October 2011, #11) How many numbers with no more than six digits can be formed using only the digits 1 through 7, with no digit used more than once in a given number? One-digit numbers: 7 Two-digit numbers: 7 6 42 Three-digit numbers: 7 6 5 210 Four-digit numbers: 7 6 5 4 840 Five-digit numbers: 7 6 5 6 3 2520 Six-digit numbers: 7 6 5 4 3 2 5040 So the correct answer is E) 8659. [See the section on Sets and Counting] (October 2011, #12) The lines with equations 2 x 3 y 24 and 3x 2 y 6 are symmetric with respect to a line with equation y mx b with m 0 . Find m b . 3x 2 y 6 2 x 3 y 24 The line that we want must bisect the obtuse angle between the two given lines. The acute angle between the lines can be determined from the formula 23 32 m1 m2 5 1 5 tan , so the obtuse angle would be . We need 180 tan 1 m1m2 1 23 23 12 12 half of this angle for the bisector, and this would be 90 12 tan 1 5 . And we need to add this 12 5 2 2 angle to tan 1 , giving us 90 12 tan 1 tan 1 . The slope of the line we want is 12 3 3 5 2 the tangent of this angle , tan 90 12 tan 1 tan 1 . From some trig identities, we get 12 3 1 5 1 cos tan 2 12 1 2 2 1 5 1 5 1 1 tan 90 2 tan tan tan cot tan 5 2 sin tan 1 3 12 12 3 3 12 1 5 1 . The line 1 cos tan 5 2 1 5 2 1 tan 90 12 tan 1 tan tan 1 1 23 cot 12 tan 1 3 1 512 12 12 3 sin tan 12 passes through the point of intersection of the given lines which is 6,12 , so an equation for the line is y 12 x 6 y x 18 . So the correct answer is D) 19. [See the section on Trigonometric Formulas] OR Lines with reciprocal slopes are bisected by lines with slope of 1 and 1: yx y mx y x 1 1 m y 1 x m 1 m 1 The lines with equations 2 x 3 y 24 and 3x 2 y 6 have reciprocal slopes, so the bisecting line with positive slope must have slope of 1. [See the section on Equations of Lines] (October 2011, #13) A square of area 45 is inscribed in a circle C. Find the area of a square inscribed in a semicircle of circle C. (Inscribed means having all 4 vertices on a given figure.) From November 2003, #13: Square ABCD is inscribed in circle O, and its area is a. Square EFGH is inscribed in a semicircle of O. What is the area of square EFGH? H G D C E F A B The area of an inscribed square in a circle of radius r, is 2r 2 . r 2 r 45 2r The area of an inscribed square in a semicircle of radius r, is r 4r 2 . 5 s s 2 So if a 2r 2 , then the area of the square inscribed inside the semicircle is 2a . 5 Using this, we get that the area of the square inscribed inside the semicircle is 2 45 18 . 5 So the correct answer is B) 18. (October 2011, #14) The left edge of a dollar bill is folded against the bottom edge to form an isosceles right triangle at the left edge. The new left edge is again folded against the bottom edge. A vertex of the new triangle is the upper right corner of the bill. If a dollar bill is 157 mm long, find its width to the nearest millimeter. 157 W W 2W 157 W W 157 W 2W From the labeled diagram, it must be that 2W W 157 2 1 0 . So W 157 So the correct answer is C) 65. 2W 157 W 2 1 65.031... . 2 W 2 W 2 W 2 157 . This simplifies into (October 2011, #15) Five boxes are placed inside an empty box. Each of the 5 new boxes is either left empty or has 5 new boxes placed inside it. This process is repeated until there are 18 boxes containing other boxes. Find the number of empty boxes. Here are 18 boxes containing other boxes, and there are 73 empty boxes. So the correct answer is A) 73. (October 2011, #16) Al, Bo, Cy, and Di are to receive math, physics, chem, and bio awards. Al thinks Di will win bio, Bo thinks Cy will win chem., Cy thinks Al won’t win mth, and Di thinks Bo will win physics. The math and bio winners are both right, and the other winners are both wrong. Who wins the math award? Let’s try to make 2 of them right and 2 of them wrong: Al and Bo are right, Cy and Di are wrong. Al Bo Cy Di chem bio Not possible. Al and Cy are right, Bo and Di are wrong. Al Bo Cy Di chem math phys bio Al Bo Cy Di phys chem math bio Possible, but the math and bio winners aren’t both right. Al and Di are right, Bo and Cy are wrong. Al Bo phys Not possible. Cy Di bio Bo and Cy are right, Al and Di are wrong. Al Bo Cy Di bio math chem phys Al Bo phys bio Cy Di chem math Possible, but the math and bio winners aren’t both right. Bo and Di are right, Al and Cy are wrong. Al Bo Cy Di bio phys chem math Possible, but the math and bio winners aren’t both right. Cy and Di are right, Al and Bo are wrong. Al Bo Cy chem phys bio This is it. So the correct answer is D) Di. Di math (October 2011, #17) The digits 1 through 9 are separated into 3 groups of three digits, and the product of each group is found. Let P be the largest of the 3 products. Find the smallest possible value of P. 74 is not the product of 3 digits 1 through 9. 73 is not the product of 3 digits 1 through 9. 71 is not the product of 3 digits 1 through 9. So it’s down to either 70 or 72. 70 doesn’t work: 70 2 5 7 , and the other groups would consist of 9, ?, ? and 8, ? , ?, but no matter how you position the 1, 3, 4, and 6, you will get a product that’s larger than 70. With 72, you can get the groups 1, 8, 9 and 6, 3, 4 and 7, 2, 5 and the largest product is 72. So the correct answer is C) 72. (October 2011, #18) Out of 10 red chips and 15 green chips, 6 are placed into a bag, 10 into a box, and 9 into a bowl. In how many ways can the chips be distributed, if only the number of red and green chips in each container matters? Let r1 be the number of red chips in the bag, r2 be the number of red chips in the box, r3 be the number of red chips in the bowl. Let g1 be the number of green chips in the bag, g 2 be the number of green chips in the box, g3 be the number of green chips in the bowl. This leads to the system of equations r1 r2 r3 10 g1 g 2 g3 15 r1 g1 6 r2 g 2 10 r3 g 3 9 Which has solutions given by g3 t , g 2 s, g1 15 s t , r3 9 t , r2 10 s, r1 t s 9 , where s and t are nonnegative integers with 0 s t 9 . The number of different values of red chips and green chips in each container is equal to the number of different pairs of values of s and t. s t This is equal to 10 9 8 7 6 5 4 3 2 1 So the correct answer is D) 55. 10 11 55 . 2 (October 2011, #19) Square ABCD has side length 72. Let E be the midpoint of side AB , and let BD and CE intersect at G. Find the length of the altitude to BE in GEB . 72 B C 36 x G 72 - x 72 E 36 A 72 The area of the square can be calculated as area CBE area DAB area CGD area BGE . This leads to the equation 18 72 36 72 36 72 x 18 x 722 . This simplifies into 18 72 54 x x 24 . So the correct answer is D) 24. D (October 2011, #20) Let r be the positive real zero of P x 9 x5 7 x 2 9 . r 4 2r 9 kr 5k 1 can be expressed as the rational number a in lowest terms. Find b a b. Recall the geometric series formula: 1 x x 2 x3 x n1 1 ; x 1. 1 x Notice that 1 2 x 3x 2 4 x3 nx n1 1 x x 2 x3 x 4 1 1 x x 1 x x x 2 x3 x 4 x x x 2 3 4 x2 1 x x x 3 4 x4 The sum x3 1 x x4 1 x 1 x x2 x n1 1 x 1 x 1 x 1 x 1 1 x x 2 x3 x 4 1 x 1 ; x 1 2 1 x r 4 2r 9 r 4 1 2r 5 3r10 4r15 r4 2 1 r 5 kr 5 k 1 k r5 k 1 But r is the real zero of P x 9 x5 7 x 2 9 , so 9r 5 7r 2 9 0 r 5 1 79 r 2 . Plugging this into the previous equation leads to r 2r 4 9 kr 5 k 1 r4 7 9 r2 So the correct answer is E) 130. 2 81 . 49 [See the section on Algebraic Formulas]