Difference between revisions of "2022 AMC 10B Problems/Problem 13"
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Let the two primes be <math>a</math> and <math>b</math>. We would have <math>a-b=2</math> and <math>a^{3}-b^{3}=31106</math> | Let the two primes be <math>a</math> and <math>b</math>. We would have <math>a-b=2</math> and <math>a^{3}-b^{3}=31106</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | Let the two primes be <math>p</math> and <math>q</math> such that <math>p-q=2</math> and <math>p^{3}-q^{3}=31106</math> | ||
+ | |||
+ | By the difference of cubes formula, <math>p^{3}-q^{3}=(p-q)(p^{2}+pq+q^{2})</math> | ||
+ | |||
+ | Plugging in <math>p-q=2</math> and <math>p^{3}-q^{3}=31106</math>, | ||
+ | |||
+ | <math>31106=2(p^{2}+pq+q^{2})</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 \dfrac{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> | ||
+ | |||
+ | The least prime greater than these two primes is <math>79</math> <math>\implies \boxed{\textbf{(E) }16}</math> |
Revision as of 19:05, 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