Difference between revisions of "2016 AIME I Problems/Problem 12"
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Find the least positive integer <math>m</math> such that <math>m^2 - m + 11</math> is a product of at least four not necessarily distinct primes. | Find the least positive integer <math>m</math> such that <math>m^2 - m + 11</math> is a product of at least four not necessarily distinct primes. | ||
==Solution 1== | ==Solution 1== | ||
− | <math>m(m-1)</math> is the product of two consecutive integers, so it is always even. Thus <math>m(m-1)+11</math> is odd and never divisible by <math>2</math>. Thus any prime <math>p</math> that divides <math>m^2-m+11</math> must divide <math>4m^2-4m+44=(2m-1)^2+43</math>. We see that <math>(2m-1)^2\equiv -43\pmod{p}</math>. We can verify that <math>-43</math> is not a perfect square mod <math>p</math> for each of <math>p=3,5,7</math>. Therefore, all prime factors of <math>m^2-m+11</math> are | + | <math>m(m-1)</math> is the product of two consecutive integers, so it is always even. Thus <math>m(m-1)+11</math> is odd and never divisible by <math>2</math>. Thus any prime <math>p</math> that divides <math>m^2-m+11</math> must divide <math>4m^2-4m+44=(2m-1)^2+43</math>. We see that <math>(2m-1)^2\equiv -43\pmod{p}</math>. We can verify that <math>-43</math> is not a perfect square mod <math>p</math> for each of <math>p=3,5,7</math>. Therefore, all prime factors of <math>m^2-m+11</math> are <math>\ge 11</math>. |
Let <math>m^2 - m + 11 = pqrs</math> for primes <math>11\le p \le q \le r \le s</math>. From here, we could go a few different ways: | Let <math>m^2 - m + 11 = pqrs</math> for primes <math>11\le p \le q \le r \le s</math>. From here, we could go a few different ways: |
Revision as of 02:29, 9 December 2021
Contents
[hide]Problem
Find the least positive integer such that is a product of at least four not necessarily distinct primes.
Solution 1
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 .
Let for primes . From here, we could go a few different ways:
Solution 1a
Suppose ; then . Reducing modulo 11, we get so .
Suppose . Then we must have , which leads to , i.e., .
leads to (impossible)! Then leads to , a prime (impossible). Finally, for we get .
Thus our answer is .
Solution 1b
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.
Case 1: m = 2k
Case 2: m = 2k+1
Now, we can test , which fails, so we test , and we get m = .
-AlexLikeMath
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.