2022 AMC 10B Problems/Problem 13: Difference between revisions

Revision as of 19:05, 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 $79$ $\implies \boxed{\textbf{(E) }16}$

@@ Line 2: / Line 2: @@
 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>