Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (→‎Solution 1a)
 
(14 intermediate revisions by 7 users not shown)
Line 1: Line 1:
 
==Problem ==
 
==Problem ==
 
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==
+
==Solution 1==
−
We claim <math>m = 132</math>. Note <math>132^2 - 132 + 11 = 11^3 \cdot 13</math>.
+
<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>.
  
−
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>.
+
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:
  
−
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  
+
===Solution 1a===
 +
 
 +
Suppose <math>p=11</math>; then <math>m^2-m+11=11qrs</math>. Reducing modulo 11, we get <math>m\equiv  1,0 \pmod{11}</math> so <math>k(11k\pm 1)+1 = qrs</math>.
 +
 
 +
Suppose <math>q=11</math>. Then we must have <math>11k^2\pm k + 1 = 11rs</math>, which leads to <math>k\equiv \mp 1 \pmod{11}</math>, i.e., <math>k\in \{1,10,12,21,23,\ldots\}</math>.
 +
 
 +
<math>k=1</math> leads to <math>rs=1</math> (impossible)! Then <math>k=10</math> leads to <math>rs=101</math>, a prime (impossible). Finally, for <math>k=12</math> we get <math>rs=143=11\cdot 13</math>.
 +
 
 +
Thus our answer is <math>m=11k= \boxed{132}</math>.
 +
 
 +
===Solution 1b===
 +
 
 +
Let <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>
 
<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>.
+
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 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.
 +
 
 +
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>.
 +
 
 +
-AlexLikeMath
 +
 
 +
==Video Solution==
 +
https://youtu.be/KRleD8iDRhI
 +
 
 +
~MathProblemSolvingSkills.com
 +
 
 +
 
  
 
==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}}

Latest revision as of 17:33, 3 September 2024

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 $k\equiv \mp 1 \pmod{11}$, 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 $m^2 - m + 11 \not |$ $2,3,5,7$. This can be done by just testing all residue classes.

For example, we can test $m \equiv 0 \mod 2$ or $m \equiv 1 \mod 2$ to show that $m^2 - m + 11$ is not divisible by 2.

Case 1: m = 2k 

      $m^2 - m + 11  \equiv 2(2 \cdot k^2 - k + 5) +1 \equiv 1 \mod 2$

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


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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_I_Problems/Problem_12&oldid=226340"