Difference between revisions of "2016 AIME I Problems/Problem 12"

Revision as of 00:57, 4 November 2025

Problem

Find the least positive integer $m$ such that $m^2 - m + 11$ is a product of at least four not necessarily distinct primes.

Solution 1

$m(m-1)$ is the product of two consecutive integers, so it is always even. Thus $m(m-1)+11$ is odd and never divisible by $2$ . Thus any prime $p$ that divides $m^2-m+11$ must divide $4m^2-4m+44=(2m-1)^2+43$ . We see that $(2m-1)^2\equiv -43\pmod{p}$ . We can verify that $-43$ is not a perfect square mod $p$ for each of $p=3,5,7$ . Therefore, all prime factors of $m^2-m+11$ are $\ge 11$ .

Let $m^2 - m + 11 = pqrs$ for primes $11\le p \le q \le r \le s$ . From here, we could go a few different ways:

Solution 1a

Suppose $p=11$ ; then $m^2-m+11=11qrs$ . Reducing modulo 11, we get $m\equiv 1,0 \pmod{11}$ so $k(11k\pm 1)+1 = qrs$ .

Suppose $q=11$ . Then we must have $11k^2\pm k + 1 = 11rs$ , which leads to $k\equiv \pm 1 \pmod{11}$ , i.e., $k\in \{1,10,12,21,23,\ldots\}$ .

$k=1$ leads to $rs=1$ (impossible)! Then $k=10$ leads to $rs=101$ , a prime (impossible). Finally, for $k=12$ we get $rs=143=11\cdot 13$ .

Thus our answer is $m=11k= \boxed{132}$ .

Solution 1b

Let $m^2 - m + 11 = pqrs$ for primes $p, q, r, s\ge11$ . If $p, q, r, s = 11$ , then $m^2-m+11=11^4$ . We can multiply this by $4$ and complete the square to find $(2m-1)^2=4\cdot 11^4-43$ . But $(2\cdot 11^2-1)^2=4\cdot 11^4-4\cdot 11^2+1 <4\cdot 11^4-43<(2\cdot 11^2)^2,$ hence we have pinned a perfect square $(2m-1)^2=4\cdot 11^4-43$ strictly between two consecutive perfect squares, a contradiction. Hence $pqrs \ge 11^3 \cdot 13$ . Thus $m^2-m+11\ge 11^3\cdot 13$ , or $(m-132)(m+131)\ge0$ . From the inequality, we see that $m \ge 132$ . $132^2 - 132 + 11 = 11^3 \cdot 13$ , so $m = \boxed{132}$ and we are done.

Solution 2

First, we can show that $2,3,5,7$ $\not |$ $m^2 - m + 11. This can be done by just testing all residue classes.

For example, we can test$ (Error compiling LaTeX. Unknown error_msg)m \equiv 0 \mod 2 $or$ m \equiv 1 \mod 2 $to show that$ m^2 - m + 11$is not divisible by 2.

Case 1: m = 2k$ (Error compiling LaTeX. Unknown error_msg)m^2 - m + 11 \equiv 2(2 \cdot k^2 - k + 5) +1 \equiv 1 \mod 2 $Case 2: m = 2k+1$ m^2 - m + 11 \equiv 2(2 \cdot k^2 + k + 5) +1 \equiv 1 \mod 2 $Now, we can test$ m^2 - m + 11 = 11^4 $, which fails, so we test$ m^2 - m + 11 = 11^3 \cdot 13 $, and we get m =$ 132$.

-AlexLikeMath

Video Solution

https://youtu.be/KRleD8iDRhI

~MathProblemSolvingSkills.com

@@ Line 23: / Line 23: @@
 ==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>2,3,5,7</math> <math>\not |</math>  <math>m^2 - m + 11. 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>
+        </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>
+        </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$.
 -AlexLikeMath

2016 AIME I (Problems • Answer Key • Resources)
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