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

Revision as of 00:05, 15 June 2023

Problem

The positive difference between a pair of primes is equal to $2$ , and the positive difference between the cubes of the two primes is $31106$ . What is the sum of the digits of the least prime that is greater than those two primes?

$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 13 \qquad\textbf{(E)}\ 16$

Solution 1

Let the two primes be $a$ and $b$ . We would have $a-b=2$ and $a^{3}-b^{3}=31106$ . Using difference of cubes, we would have $(a-b)(a^{2}+ab+b^{2})=31106$ . Since we know $a-b$ is equal to $2$ , $(a-b)(a^{2}+ab+b^{2})$ would become $2(a^{2}+ab+b^{2})=31106$ . Simplifying more, we would get $a^{2}+ab+b^{2}=15553$ .

Now let's introduce another variable. Instead of using $a$ and $b$ , we can express the primes as $x+2$ and $x$ where $a$ is $x+2$ and b is $x$ . Plugging $x$ and $x+2$ in, we would have $(x+2)^{2}+x(x+2)+x^{2}$ . When we expand the parenthesis, it would become $x^{2}+4x+4+x^{2}+2x+x^{2}$ . Then we combine like terms to get $3x^{2}+6x+4$ which equals $15553$ . Then we subtract 4 from both sides to get $3x^{2}+6x=15549$ . Since all three numbers are divisible by 3, we can divide by 3 to get $x^{2}+2x=5183$ .

Notice how if we had 1 to both sides, the left side would become a perfect square trinomial: $x^{2}+2x+1=5184$ which is $(x+1)^{2}=5184$ . Since $2$ is too small to be a valid number, the two primes must be odd, therefore $x+1$ is the number in the middle of them. Conveniently enough, $5184=72^{2}$ so the two numbers are $71$ and $73$ . The next prime number is $79$ , and $7+9=16$ so the answer is $\boxed{\textbf{(E) }16}$ .

~Trex226

Solution 2

Let the two primes be $a$ and $b$ , with $a$ being the smaller prime. We have $a - b = 2$ , and $a^3 - b^3 = 31106$ . Using difference of cubes, we obtain $a^2 + ab + b^2 = 15553$ . Now, we use the equation $a - b = 2$ to obtain $a^2 - 2ab + b^2 = 4$ . Hence, $a^2 + ab + b^2 - (a^2 - 2ab + b^2) = 3ab = 15553 - 4 = 15549$ $ab = 5183.$ Because we have $b = a+2$ , $ab = (a+1)^2 - (1)^2$ . Thus, $(a+1)^2 = 5183 + 1 = 5184$ , so $a+1 = 72$ . This implies $a = 71$ , $b = 73$ , and thus the next biggest prime is $79$ , so our answer is $7 + 9 = \boxed{\textbf{(E) }16}$

~mathboy100

Solution 3 (Estimation)

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, $31106=2(p^{2}+pq+q^{2})\approx 6p^{2}\\p^2\approx \tfrac{31106}{6}\approx 5200 Recall that$ 70^2=4900 $and$ 80^2=6400 $. It follows that our primes must be only marginally larger than$ 70 $, where we conveniently find$ p=73, q=71 $The least prime greater than these two primes is$ 79 $7 + 9$ \implies \boxed{\textbf{(E) }16}$

~BrandonZhang202415 ~SwordOfJustice (small edits)

Video Solution 1

https://youtu.be/FQn9XOPKlbw

~Education, the Study of Everything

Video Solution by Interstigation

https://youtu.be/yLFiSmLJJ5A

@@ Line 41: / Line 41: @@
 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 \tfrac{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
+Recall that </math>70^2=4900<math> and </math>80^2=6400<math>. It follows that our primes must be only marginally larger than </math>70<math>, where we conveniently find </math>p=73, q=71<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> 7 + 9 </math>\implies \boxed{\textbf{(E) }16}$
-The least prime greater than these two primes is <math>79</math> <math>\implies \boxed{\textbf{(E) }16}</math>
 ~BrandonZhang202415
+~SwordOfJustice (small edits)
 ==Video Solution 1==

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