Difference between revisions of "2022 AMC 10B Problems/Problem 13"
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Through the givens, we can see that <math>p \approx q</math>. | Through the givens, we can see that <math>p \approx q</math>. | ||
− | Thus, <math>31106=2(p^{2}+pq+q^{2})\approx 6p^{2}\p^2\approx \ | + | Thus, <math>31106=2(p^{2}+pq+q^{2})\approx 6p^{2}\p^2\approx \tfrac{31106}{6}\approx 5200\p\approx \sqrt{5200}\approx 72</math> |
Checking prime pairs near <math>72</math>, we find that <math>p=73, q=71</math> | Checking prime pairs near <math>72</math>, we find that <math>p=73, q=71</math> |
Revision as of 19:08, 17 November 2022
Solution
Let the two primes be and . We would have and
Solution 2
Let the two primes be and such that and
By the difference of cubes formula,
Plugging in and ,
Through the givens, we can see that .
Thus,
Checking prime pairs near , we find that
The least prime greater than these two primes is
~BrandonZhang202415