Difference between revisions of "2016 AIME I Problems/Problem 12"
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We claim <math>m = 132</math>. Note <math>132^2 - 132 + 11 = 11^3 \cdot 13</math>. | We claim <math>m = 132</math>. Note <math>132^2 - 132 + 11 = 11^3 \cdot 13</math>. | ||
− | + | Now <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 greater than or equal to <math>11</math>. | |
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+ | Now suppose <math>m < 132</math> and <math>m^2 - m + 11 = pqrs</math> for primes <math>p, q, r, s\ge11</math>. If <math>p, q, r, s = 11</math>, then <math>m^2-m+11=11^4</math>. We can multiply this by <math>4</math> and complete the square to find <math>(2m-1)^2=4\cdot 11^4-43</math>. But | ||
+ | <cmath>(2\cdot 11^2-1)^2=4\cdot 11^4-4\cdot 11^2+1 <4\cdot 11^4-43<(2\cdot 11^2)^2,</cmath> | ||
+ | hence we have pinned a perfect square <math>(2m-1)^2=4\cdot 11^4-43</math> strictly between two consecutive perfect squares, a contradiction. Hence <math>pqrs \ge 11^3 \cdot 13</math>. Thus <math>m^2-m+11\ge 11^3\cdot 13</math>, or <math>(m-132)(m+131)\ge0</math>. From the inequality, we see that <math>m \ge 132</math>, hence <math>m = \fbox{132}</math>. | ||
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==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 05:35, 7 March 2016
Problem
Find the least positive integer such that is a product of at least four not necessarily distinct primes.
Solution
We claim . Note .
Now 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 .
Now suppose and 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 , hence .
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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