2022 AMC 10B Problems/Problem 13

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

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

Page

Toolbox

Search