Difference between revisions of "2022 AMC 10B Problems/Problem 13"
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− | Now let's introduce another variable. Instead of using <math>a</math> and <math>b</math>, we can express the primes as <math>x+2</math> and <math>x</math> where <math>a</math> is <math>x+2</math> and b is <math>x</math>. Plugging <math>x</math> and <math>x+2</math> in, we would have | + | Now let's introduce another variable. Instead of using <math>a</math> and <math>b</math>, we can express the primes as <math>x+2</math> and <math>x</math> where <math>a</math> is <math>x+2</math> and b is <math>x</math>. Plugging <math>x</math> and <math>x+2</math> in, we would have <math>(x+2)^{2}+x(x+2)+x^{2}</math>. When we expand the parenthesis, it would become <math>x^{2}+4x+4+x^{2}+2x+x^{2}</math>. Then we combine like terms to get <math>3x^{2}+6x+4</math> which equals <math>15553</math>. Then we subtract 4 from both sides to get <math>3x^{2}+6x=15549</math>. Since all three numbers are divisible by 3, we can divide by 3 to get <math>x^{2}+2x=5183</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>. | ||
==Solution 1== | ==Solution 1== |
Revision as of 20:06, 17 November 2022
Solution
Let the two primes be and . We would have and . Using difference of cubes, we would have . Since we know is equal to , would become . Simplifying more, we would get .
Now let's introduce another variable. Instead of using and , we can express the primes as and where is and b is . Plugging and in, we would have . When we expand the parenthesis, it would become . Then we combine like terms to get which equals . Then we subtract 4 from both sides to get . Since all three numbers are divisible by 3, we can divide by 3 to get .
Notice how if we had 1 to both sides, the left side would become a perfect square trinomial: which is . 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, so the two numbers are 71 and 73. The next prime number is 79, and 7+9=16 so the answer is .
Solution 1
Let the two primes be and such that and
By the difference of cubes formula,
Plugging in and ,
Through the givens, we can see that .
Thus,
Checking prime pairs near , we find that
The least prime greater than these two primes is
~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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