2022 AMC 10B Problems/Problem 13
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
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 | |
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