2022 AMC 10B Problems/Problem 13
Contents
Problem
The positive difference between a pair of primes is equal to , and the positive difference between the cubes of the two primes is . What is the sum of the digits of the least prime that is greater than those two primes?
Solution 1
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 is too small to be a valid number, the two primes must be odd, therefore is the number in the middle of them. Conveniently enough, so the two numbers are and . The next prime number is , and so the answer is .
~Trex226
Solution 2
Let the two primes be and , with being the smaller prime. We have , and . Using difference of cubes, we obtain . Now, we use the equation to obtain . Hence, Because we have , . Thus, , so . This implies , , and thus the next biggest prime is , so our answer is
~mathboy100
Solution 3 (Estimation)
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,
Recall that and . It follows that our primes must be only marginally larger than , where we conveniently find
The least prime greater than these two primes is
~BrandonZhang202415 ~SwordOfJustice (small edits)
Video Solution (⚡️Lightning Fast⚡️)
~Education, the Study of Everything
Video Solution by Interstigation
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.