Difference between revisions of "2022 AMC 10B Problems/Problem 13"
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Let the two primes be <math>a</math> and <math>b</math>, with <math>a</math> being the smaller prime. We have <math>a - b = 2</math>, and <math>a^3 - b^3 = 31106</math>. Using difference of cubes, we obtain <math>a^2 + ab + b^2 = 15553</math>. Now, we use the equation <math>a - b = 2</math> to obtain <math>a^2 - 2ab + b^2 = 4</math>. Hence, | Let the two primes be <math>a</math> and <math>b</math>, with <math>a</math> being the smaller prime. We have <math>a - b = 2</math>, and <math>a^3 - b^3 = 31106</math>. Using difference of cubes, we obtain <math>a^2 + ab + b^2 = 15553</math>. Now, we use the equation <math>a - b = 2</math> to obtain <math>a^2 - 2ab + b^2 = 4</math>. Hence, | ||
<cmath> | <cmath> | ||
− | a^2 + ab + b^2 - (a^2 - 2ab + b^2) = 3ab = 15553 - 4 = 15549 | + | a^2 + ab + b^2 - (a^2 - 2ab + b^2) = 3ab = 15553 - 4 = 15549 |
− | ab = 5183. | + | </cmath> |
+ | <cmath> | ||
+ | ab = 5183. | ||
</cmath> | </cmath> | ||
Because we have <math>b = a-2</math>, <math>ab = (a+1)^2 - (1)^2</math>. Thus, <math>(a+1)^2 = 5183 + 1 = 5184</math>, so <math>a+1 = 72</math>. This implies <math>a = 71</math>, <math>b = 73</math>, and thus the next biggest prime is <math>79</math>, so our answer is <math>7 + 9 = \boxed{\textbf{(E) }16}</math> | Because we have <math>b = a-2</math>, <math>ab = (a+1)^2 - (1)^2</math>. Thus, <math>(a+1)^2 = 5183 + 1 = 5184</math>, so <math>a+1 = 72</math>. This implies <math>a = 71</math>, <math>b = 73</math>, and thus the next biggest prime is <math>79</math>, so our answer is <math>7 + 9 = \boxed{\textbf{(E) }16}</math> |
Revision as of 13:14, 18 November 2022
Contents
[hide]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
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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