2019 AIME II Problems/Problem 3
Contents
Problem 3
Find the number of -tuples of positive integers that satisfy the following systems of equations:
Solution 1
As 71 is prime, , , and must be 1, 1, and 71 (in some order). 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
Solution 2
We know that any two consecutive numbers are coprime. Using this, we can figure out that and . then has to be 71. Now we have two equations left. and . To solve these we just need to figure out all of the factors. Doing the prime factorization of and , we find that they have and factors, respectively. The answer is
~Hithere22702
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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