Difference between revisions of "2020 AIME I Problems/Problem 9"
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== Solution == | == Solution == | ||
+ | |||
+ | <asy> | ||
+ | size(12cm); | ||
+ | for (int x = 1; x < 18; ++x) { | ||
+ | draw((x, 0) -- (x, 9), dotted); | ||
+ | } | ||
+ | for (int y = 1; y < 9; ++y) { | ||
+ | draw((0, y) -- (18, y), dotted); | ||
+ | } | ||
+ | |||
+ | draw((0, 0) -- (18, 0) -- (18, 9) -- (0, 9) -- cycle); | ||
+ | |||
+ | pair b1, b2, b3; | ||
+ | pair c1, c2, c3; | ||
+ | pair a1, a2, a3; | ||
+ | b1 = (3, 0); b2 = (12, 0); b3 = (16, 0); | ||
+ | c1 = (0, 2); c2 = (0, 4); c3 = (0, 8); | ||
+ | a1 = b1 + c1; a2 = b2 + c2; a3 = b3 + c3; | ||
+ | |||
+ | draw(b1 -- a1 -- c1); | ||
+ | draw(b2 -- a2 -- c2); | ||
+ | draw(b3 -- a3 -- c3); | ||
+ | |||
+ | dot(a1); dot(a2); dot(a3); | ||
+ | label("$a_1$", a1, NE); | ||
+ | label("$a_2$", a2, NE); | ||
+ | label("$a_3$", a3, NE); | ||
+ | label("$b_1$", b1, S); | ||
+ | label("$b_2$", b2, S); | ||
+ | label("$b_3$", b3, S); | ||
+ | label("$c_1$", c1, W); | ||
+ | label("$c_2$", c2, W); | ||
+ | label("$c_3$", c3, W); | ||
+ | |||
+ | </asy> | ||
First, prime factorize <math>20^9</math> as <math>2^{18} \cdot 5^9</math>. Denote <math>a_1</math> as <math>2^{b_1} \cdot 5^{c_1}</math>, <math>a_2</math> as <math>2^{b_2} \cdot 5^{c_2}</math>, and <math>a_3</math> as <math>2^{b_3} \cdot 5^{c_3}</math>. | First, prime factorize <math>20^9</math> as <math>2^{18} \cdot 5^9</math>. Denote <math>a_1</math> as <math>2^{b_1} \cdot 5^{c_1}</math>, <math>a_2</math> as <math>2^{b_2} \cdot 5^{c_2}</math>, and <math>a_3</math> as <math>2^{b_3} \cdot 5^{c_3}</math>. |
Revision as of 13:10, 13 March 2020
Problem
Let be the set of positive integer divisors of Three numbers are chosen independently and at random with replacement from the set and labeled and in the order they are chosen. The probability that both divides and divides is where and are relatively prime positive integers. Find
Solution
First, prime factorize as . Denote as , as , and as .
In order for to divide , and for to divide , , and . We will consider each case separately. Note that the total amount of possibilities is , as there are choices for each factor.
We notice that if we add to and to , then we can reach the stronger inequality . Therefore, if we pick integers from to , they will correspond to a unique solution, forming a 1-1 correspondence.This is also equivalent to applying stars and bars on distributing the powers of 2 and 5 through differences. The amount of solutions to this inequality is .
The case for ,, and proceeds similarly for a result of . Therefore, the probability of choosing three such factors is Simplification gives , and therefore the answer is .
-molocyxu
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.