Difference between revisions of "2022 AMC 10B Problems/Problem 13"

Revision as of 20:06, 17 November 2022

Let the two primes be $a$ and $b$ . We would have $a-b=2$ and $a^{3}-b^{3}=31106$

Let the two primes be $p$ and $q$ such that $p-q=2$ and $p^{3}-q^{3}=31106$

By the difference of cubes formula, $p^{3}-q^{3}=(p-q)(p^{2}+pq+q^{2})$

Plugging in $p-q=2$ and $p^{3}-q^{3}=31106$ ,

$31106=2(p^{2}+pq+q^{2})$

Through the givens, we can see that $p \approx q$ .

Thus, $31106=2(p^{2}+pq+q^{2})\approx 6p^{2}\\p^2\approx \dfrac{31106}{6}\approx 5200\\p\approx \sqrt{5200}\approx 72$

Checking prime pairs near $72$ , we find that $p=73, q=71$

The least prime greater than these two primes is $79$ $\implies \boxed{\textbf{(E) }16}$

~BrandonZhang202415

Revision as of 20:06, 17 November 2022 (view source) Brandonzhang202415 (talk \| contribs) (→‎Solution 2) ← Older edit		Revision as of 20:06, 17 November 2022 (view source) Brandonzhang202415 (talk \| contribs) (→‎Solution 2) Newer edit →
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	The least prime greater than these two primes is <math>79</math> <math>\implies \boxed{\textbf{(E) }16}</math>		The least prime greater than these two primes is <math>79</math> <math>\implies \boxed{\textbf{(E) }16}</math>
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	~BrandonZhang202415		~BrandonZhang202415