# Difference between revisions of "2019 AIME II Problems/Problem 3"

Scrabbler94 (talk | contribs) (add solution) |
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+ | ==Problem 3== | ||

Find the number of <math>7</math>-tuples of positive integers <math>(a,b,c,d,e,f,g)</math> that satisfy the following systems of equations: | Find the number of <math>7</math>-tuples of positive integers <math>(a,b,c,d,e,f,g)</math> that satisfy the following systems of equations: | ||

− | \begin{align*} | + | <cmath>\begin{align*} |

abc&=70,\\ | abc&=70,\\ | ||

cde&=71,\\ | cde&=71,\\ | ||

efg&=72. | efg&=72. | ||

− | \end{align*} | + | \end{align*}</cmath> |

+ | |||

+ | ==Solution== | ||

+ | As 71 is prime, <math>c</math>, <math>d</math>, and <math>e</math> must be 1, 1, and 71 (up to ordering). However, since <math>c</math> and <math>e</math> are divisors of 70 and 72 respectively, the only possibility is <math>(c,d,e) = (1,71,1)</math>. Now we are left with finding the number of solutions <math>(a,b,f,g)</math> satisfying <math>ab = 70</math> and <math>fg = 72</math>, which separates easily into two subproblems. The number of positive integer solutions to <math>ab = 70</math> simply equals the number of divisors of 70 (as we can choose a divisor for <math>a</math>, which uniquely determines <math>b</math>). As <math>70 = 2^1 \cdot 5^1 \cdot 7^1</math>, we have <math>d(70) = (1+1)(1+1)(1+1) = 8</math> solutions. Similarly, <math>72 = 2^3 \cdot 3^2</math>, so <math>d(72) = 4 \times 3 = 12</math>. | ||

+ | |||

+ | Then the answer is simply <math>8 \times 12 = \boxed{096}</math>. | ||

+ | |||

+ | -scrabbler94 |

## Revision as of 15:16, 22 March 2019

## Problem 3

Find the number of -tuples of positive integers that satisfy the following systems of equations:

## Solution

As 71 is prime, , , and must be 1, 1, and 71 (up to ordering). However, since and are divisors of 70 and 72 respectively, the only possibility is . Now we are left with finding the number of solutions satisfying and , which separates easily into two subproblems. The number of positive integer solutions to simply equals the number of divisors of 70 (as we can choose a divisor for , which uniquely determines ). As , we have solutions. Similarly, , so .

Then the answer is simply .

-scrabbler94