Difference between revisions of "2016 AIME I Problems/Problem 12"
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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>. <math>132^2 - 132 + 11 = 11^3 \cdot 13</math>, so <math>m = \boxed{132}</math> and we are done. | 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>. <math>132^2 - 132 + 11 = 11^3 \cdot 13</math>, so <math>m = \boxed{132}</math> and we are done. | ||
− | ==Solution | + | ==Solution 2== |
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. | 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. | ||
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-AlexLikeMath | -AlexLikeMath | ||
− | |||
==Video Solution== | ==Video Solution== |
Revision as of 19:32, 22 November 2022
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
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
Video Solution
~MathProblemSolvingSkills.com
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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