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

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==Solution==
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==Solution 1==
  
 
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>. Using difference of cubes, we would have <math>(a-b)(a^{2}+ab+b^{2})=31106</math>. Since we know <math>a-b</math> is equal to <math>2</math>, <math>(a-b)(a^{2}+ab+b^{2})</math> would become <math>2(a^{2}+ab+b^{2})=31106</math>. Simplifying more, we would get <math>a^{2}+ab+b^{2}=15553</math>.
 
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>. Using difference of cubes, we would have <math>(a-b)(a^{2}+ab+b^{2})=31106</math>. Since we know <math>a-b</math> is equal to <math>2</math>, <math>(a-b)(a^{2}+ab+b^{2})</math> would become <math>2(a^{2}+ab+b^{2})=31106</math>. Simplifying more, we would get <math>a^{2}+ab+b^{2}=15553</math>.
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Notice how if we had 1 to both sides, the left side would become a perfect square trinomial: <math>x^{2}+2x+1=5184</math> which is <math>(x+1)^{2}=5184</math>. Since 2 is too small to be a valid number, the two primes must be odd, so x+1 is even the middle of them. Conveniently enough, <math>5184=72^{2}</math> so the two numbers are 71 and 73. The next prime number is 79, and 7+9=16 so the answer is  <math>\boxed{\textbf{(E) }16}</math>.
 
Notice how if we had 1 to both sides, the left side would become a perfect square trinomial: <math>x^{2}+2x+1=5184</math> which is <math>(x+1)^{2}=5184</math>. Since 2 is too small to be a valid number, the two primes must be odd, so x+1 is even the middle of them. Conveniently enough, <math>5184=72^{2}</math> so the two numbers are 71 and 73. The next prime number is 79, and 7+9=16 so the answer is  <math>\boxed{\textbf{(E) }16}</math>.
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~Trex226
  
 
==Solution 1==
 
==Solution 1==

Revision as of 20:08, 17 November 2022

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, so x+1 is even 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 1

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

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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