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

(Solution 2)
m (Solution 2)
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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 \dfrac{31106}{6}\approx 5200\p\approx \sqrt{5200}\approx 72</math>
+
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 $a$ and $b$. We would have $a-b=2$ and $a^{3}-b^{3}=31106$

Solution 2

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

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