Difference between revisions of "2022 AMC 10B Problems/Problem 13"
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Through the givens, we can see that <math>p \approx q</math>. | Through the givens, we can see that <math>p \approx q</math>. | ||
− | Thus, <math>31106=2(p^{2}+pq+q^{2})\approx 6p^{2}\\p^2\approx \tfrac{31106}{6}\approx 5200 | + | Thus, <math>31106=2(p^{2}+pq+q^{2})\approx 6p^{2}\\p^2\approx \tfrac{31106}{6}\approx 5200 |
− | |||
− | The least prime greater than these two primes is <math>79</math> 7 + 9 <math>\implies \boxed{\textbf{(E) }16}</math> | + | Recall that </math>70^2=4900<math> and </math>80^2=6400<math>. It follows that our primes must be only marginally larger than </math>70<math>, where we conveniently find </math>p=73, q=71<math> |
+ | |||
+ | The least prime greater than these two primes is </math>79<math> </math>\implies 7 + 9 <math>\implies \boxed{\textbf{(E) }16}</math> | ||
~BrandonZhang202415 | ~BrandonZhang202415 |
Revision as of 00:08, 15 June 2023
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, $31106=2(p^{2}+pq+q^{2})\approx 6p^{2}\\p^2\approx \tfrac{31106}{6}\approx 5200
Recall that$ (Error compiling LaTeX. Unknown error_msg)70^2=490080^2=6400
70
p=73, q=71
79$$ (Error compiling LaTeX. Unknown error_msg)\implies 7 + 9
~BrandonZhang202415 ~SwordOfJustice (small edits)
Video Solution 1
~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.