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

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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>.
  
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Thus, <math>31106=2(p^{2}+pq+q^{2})\approx 6p^{2}\\p^2\approx \tfrac{31106}{6}\approx 5200
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Thus, <math>31106=2(p^{2}+pq+q^{2})\approx 6p^{2}\\p^2\approx \tfrac{31106}{6}\approx 5200</math>
  
−
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>
+
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>
  
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The least prime greater than these two primes is </math>79<math> </math>\implies 7 + 9 <math>\implies \boxed{\textbf{(E) }16}</math>
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The least prime greater than these two primes is <math>79</math> <math>\implies 7 + 9 </math>\implies \boxed{\textbf{(E) }16}$
  
 
~BrandonZhang202415
 
~BrandonZhang202415

Revision as of 23:08, 14 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 $79$ $\implies 7 + 9$\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

See Also

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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