Difference between revisions of "2016 AIME I Problems/Problem 12"
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==Solution 2== | ==Solution 2== | ||
Let <math>m=11k</math>, then <math>s=m^2-m+11=11(11k^2-k+1)</math>. We can see <math>k = 1 \mod 11</math> for <math>s</math> to have a second factor of 11. Let <math>k=12</math>, we get <math>11k^2-k+1=11*11*13</math>, so <math>m=11*12=132</math>. -Mathdummy | Let <math>m=11k</math>, then <math>s=m^2-m+11=11(11k^2-k+1)</math>. We can see <math>k = 1 \mod 11</math> for <math>s</math> to have a second factor of 11. Let <math>k=12</math>, we get <math>11k^2-k+1=11*11*13</math>, so <math>m=11*12=132</math>. -Mathdummy | ||
+ | ==Solution 3== | ||
+ | First, we can show that <math>m^2 - m + 11 \not |</math> <math> 2,3,5,7</math>. This can be done by just testing all residue classes. | ||
+ | |||
+ | For example, we can test <math>m \equiv 0 \mod 2</math> or <math>m \equiv 1 \mod 2</math> to show that <math>m^2 - m + 11</math> is not divisible by 2. | ||
+ | |||
+ | Now, we can test <math>m^2 - m + 11 = 11^4</math>, which fails, so we test <math>m^2 - m + 11 = 11^3 \cdot 13</math>, and we get m = <math>132</math>. | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2016|n=I|num-b=11|num-a=13}} | {{AIME box|year=2016|n=I|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:07, 17 February 2020
Contents
[hide]Problem
Find the least positive integer such that is a product of at least four not necessarily distinct primes.
Solution
is the product of two consecutive integers, so it is always even. Thus is odd and never divisible by . Thus any prime that divides must divide . We see that . We can verify that is not a perfect square mod for each of . Therefore, all prime factors of are greater than or equal to .
Let for primes . If , then . We can multiply this by and complete the square to find . But hence we have pinned a perfect square strictly between two consecutive perfect squares, a contradiction. Hence . Thus , or . From the inequality, we see that . , so and we are done.
Solution 2
Let , then . We can see for to have a second factor of 11. Let , we get , so . -Mathdummy
Solution 3
First, we can show that . This can be done by just testing all residue classes.
For example, we can test or to show that is not divisible by 2.
Now, we can test , which fails, so we test , and we get m = .
See Also
2016 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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