Difference between revisions of "2019 AIME II Problems/Problem 3"
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==Solution 2== | ==Solution 2== | ||
− | We know that any two consecutive numbers are coprime. Using this, we can figure out that <math>c=1</math> and <math>e=1</math>. <math>d</math> then has to be 71. Now we have two equations left. <math>ab=70</math> and <math>fg=72</math>. To solve these we just need to figure out all of the factors. Doing the prime factorization of <math>70</math> and <math>72</math>, we find that they have <math>8</math> and <math>12</math> factors, respectively. The answer is <math>8 \times 12=\boxed{096}</math> | + | We know that any two consecutive numbers are coprime. Using this, we can figure out that <math>c=1</math> and <math>e=1</math>. <math>d</math> then has to be <math>71</math>. Now we have two equations left. <math>ab=70</math> and <math>fg=72</math>. To solve these we just need to figure out all of the factors. Doing the prime factorization of <math>70</math> and <math>72</math>, we find that they have <math>8</math> and <math>12</math> factors, respectively. The answer is <math>8 \times 12=\boxed{096}</math> |
~Hithere22702 | ~Hithere22702 | ||
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==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=II|num-b=2|num-a=4}} | {{AIME box|year=2019|n=II|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:29, 3 June 2020
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 . 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.