# MMT China Individual Round Solutions January 5, 2019

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BmMT ChinaIndividual Round SolutionsJanuary 5, 201912. Four boxes of cookies and one bag of chips cost exactly 1000 jelly beans. Five bags of chipsand one box of cookies cost less than 1000 jelly beans. If both chips and cookies cost a wholenumber of jelly beans, what is the maximum possible cost of a bag of chips?Answer: 156Solution: Let x be price of a box of cookies, and let y be the price of a bag of chips. Then4x y 1000, and x 5y 1000. Solving for y, we get y 300019 . Since y must be a whole1000 yis an integer, which occurs if and onlynumber, y 157. We must also have that x 4if y is a multiple of 4. Thus the maximum value of y is the largest multiple of 4 less than orequal to 157. Hence, the answer is 156 .13. June is making a pumpkin pie, which takes the shape of a truncated cone, as shown below.The pie tin is 18 inches wide at the top, 16 inches wide at the bottom, and 1 inch high. Howmany cubic inches of pumpkin filling are needed to fill the pie?Answer:217π3Solution: We claim that the pumpkin pie is a cone with a base of radius 9 and a height of9, minus a cone with a base of radius 8 and a height of 8. To see this, we focus on the conefrom which the pumpkin pie was truncated. Let A be the center of the top face of the pie,let B be the center of the bottom face of the pie, and let X be the tip of the cone. Since thepumpkin pie is 1 inch high, we have AX BX 1. By similar triangles, we have9AXBX 1 .8BXBXSolving for BX, we get BX 8 and AX 9 as claimed. Thus, the volume of the pumpkinpie is11217ππ · 92 · 9 π · 82 · 8 333cubic inches.14. For two real numbers a and b, let a#b ab 2a 2b 6. Find a positive real number xsuch that (x#7)#x 82.Answer: 6Solution: We compute x#7 7x 2x 14 6 5x 8. Thus,(5x 8)#x 5x2 8x 10x 16 2x 6 5x2 20x 22.Therefore 5x2 20x 22 82 so x2 4x 12 0 which has positive solution x 6 .15. Find the sum of all positive integers n such thatn2 20n 51n2 4n 3

BmMT ChinaIndividual Round SolutionsJanuary 5, 2019is an integer.Answer: 26Solution: We see that(n 3)(n 17)n 17n2 20n 51 2n 4n 3(n 3)(n 1)n 1since n 3 6 0 when n is positive. Thus we want to find all integers n so that n 1 dividesn 17. Setting m n 1, this is equivalent to finding the values m 2 so that m dividesm 16. This occurs if and only if m divides 16, giving m 2, 4, 8, 16. These correspond ton 1, 3, 7, 15, giving a sum of1 3 7 15 26 .16. Let ABC be a right triangle with hypotenuse AB such that AC 36 and BC 15. Asemicircle is inscribed in ABC as shown, such that the diameter XC of the semicircle lieson side AC and that the semicircle is tangent to AB. What is the radius of the semicircle?Answer: 10 Solution: By the Pythagorean theorem, AB 362 152 39. Let O be the centerof the semicircle, and let D be the point of tangency of the semicircle and AB. By SSAcongruence of right triangles, right triangles ODB and OCB are congruent, and therefore,DB CB 15, and AD AB DB 24. We can also see that triangles AOD and ABCrare similar. Thus 2436 15 , giving r 10 .17. Let a and b be relatively prime positive integers such that the product ab is equal to the leastcommon multiple of 16500 and 990. If 16500and 990ab are both integers, what is the minimumvalue of a b?Answer: 599Solution: Note that 16500 22 · 3 · 53 · 11, and 990 2 · 32 · 5 · 11, so the least commonmultiple is L 22 · 32 · 53 · 11. Since a and b must be relatively prime, one of a or b mustbe divisible by 53 . If b was divisible by 53 , then 990/b would not be an integer, so a mustbe divisible by 53 . Similarly, we can show that a is divisible by 22 and b is divisible by 32 .Therefore, the only two possibilities are:a 22 · 53 · 11, b 32or a 22 · 53 , b 32 · 11.We see that a b is minimized in the second case, giving a b 599 .18. Let x be a positive real number so that x Answer: 21 51x 1. Compute x8 1.x8Solution 1: We see that x2 x 1 0, so that taking the positive solution to the quadraticequation gives 1 5x .2

BmMT ChinaLet Fn xn Individual Round Solutions1xn .January 5, 2019We see that (x x 1 )Fn Fn 1 Fn 1 . Since 11 5 1 5 x 5x22we therefore find for n 2Fn 1 We have F1 1 and F2 3 52 3 52 5Fn Fn 1 .5, so we can recursively computeF3 4 F4 3 5F5 11 F6 8 5F7 29 F8 21 5 .Solution 2: As above we find x 1 52so that x x 1 x2 x 2 x x 14x x 42 x x 2 5. Thus we can compute 2 3 2 2 2 7.Hencex8 x 8 x4 x 4 x2 x 2 x x 1 x x 1 7 · 3 · 1 · 5 21 5 .19. Six people sit around a round table. Each person rolls a standard 6-sided die. If no twopeople sitting next to each other rolled the same number, we will say that the roll is valid.How many different rolls are valid?Answer: 15630Solution: Label each person a number between 1 and 6. It is useful to consider persons 1,3, and 5.Case 1: 1, 3, 5 rolled the same numberThere are 6 numbers they could have rolled, and 53 125 possibilities for the other 3 people.This gives 6 · 53 750 rolls in this case.Case 2: 1, 3, 5 rolled 2 different numbersThere are 3 ways to pick who rolled the number different than the other two, and then 6 · 5ways to pick the two numbers. Then there are 5 · 4 · 4 ways to pick what players 2, 4, and 6rolled. This gives 3 · 6 · 52 · 42 7200 rolls.Case 3: 1, 3, 5 rolled distinct numbersThere are 6 · 5 · 4 ways to pick what players 1, 3, and 5 rolled. Then there are 43 ways topick the other rolls. This gives 6 · 5 · 44 7680 rolls.All together, we have 750 7200 7680 15630 valid rolls.1120. Given that 31 0.a1 a2 a3 a4 a5 · · · an (that is, 31can be written as the repeating decimalexpansion 0.a1 a2 · · · an a1 a2 · · · an a1 a2 · · · ), what is the minimum value of n?Answer: 15

BmMT ChinaIndividual Round SolutionsSolution 1: If n is such thathave131January 5, 2019has a repeating decimal expansion of this form, then we1a n3110 1where a is the number with digits a1 a2 · · · an . This occurs if and only if 10n 1 31a forsome a, which occurs if and only if 10n 1 (mod 31). Thus we want to find the smallest nso that 10n 1 (mod 31).By Fermat’s Little Theorem, 1030 1 (mod 31), so that we must have n dividing 30. Butwe can compute101 10(mod 31)210 7(mod 31)103 10 · 102 70 852(mod 31)310 10 · 10 56 2563 2105 2(mod 31)10 (10 ) 64 2101510(mod 31)2 (10 ) 25 625 5510 10 · 10(mod 31) 5 · 25 125 1(mod 31)Hence, the minimal k is 15 .Solution 2: Note that n is the minimal positive integer such that if we shift the decimal1expansion of 31by n places, then the decimal part of the resulting real number would be the1. But shifting by n places is the same as multiplying by 10n .same as the decimal part of 3111Thus, when we multiply 31by 10n , the mixed fraction we get is a 31for some a. Hence11 a 313110n 31a 110n ·10n 1(mod 31)So we must find the minimal n such that 10n 1 (mod 31). We may now proceed as insolution 1.

BmMT China Individual Round Solutions January 5, 2019 8.How many integers nwith 1 n 100 have exactly 3 positive divisors? Answer: 4 Solution: An integer has exactly 3 positive divisors if and only if it is the square of a prime.

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