Difference between revisions of "2016 AIME I Problems/Problem 12"
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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. | ||
− | 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. | + | 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. |
+ | |||
+ | Case 1: m = 2k | ||
+ | <math>m^2 - m + 11 \equiv 2(2 \cdot k^2 - k + 5) +1 \equiv 1 \mod 2 </math> | ||
+ | Case 2: m = 2k+1 | ||
+ | <math>m^2 - m + 11 \equiv 2(2 \cdot k^2 + k + 5) +1 \equiv 1 \mod 2 </math> | ||
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>. | 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>. |
Revision as of 17:52, 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.
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.