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

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==Solution 1==
 
==Solution 1==
 
<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>.
 
<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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 +
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:
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===Solution 1a===
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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>.
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Suppose <math>q=11</math>. Then we must have <math>11k^2\pm k + 1 = 11rs</math>, i.e., <math>k\equiv \mp 1 \pmod{11}</math>. Smallest value for <math>k</math> in this range is <math>k=1</math> but that leads to <math>rs=1</math> (impossible)! The next smallest is <math>k=10</math> which leads to <math>rs=103</math>, a prime, (again not possible). Checking <math>k=12</math> we get <math>rs=143=11\cdot 13</math>. This works; so our answer is <math>m=11k= \boxed{132}</math>.
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===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  
 
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  

Revision as of 12:20, 23 November 2021

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 greater than or equal to $11$.

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 $11k^2\pm k + 1 = 11rs$, i.e., $k\equiv \mp 1 \pmod{11}$. Smallest value for $k$ in this range is $k=1$ but that leads to $rs=1$ (impossible)! The next smallest is $k=10$ which leads to $rs=103$, a prime, (again not possible). Checking $k=12$ we get $rs=143=11\cdot 13$. This works; so 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

Let $m=11k$, then $s=m^2-m+11=11(11k^2-k+1)$. We can see $k = 1 \mod 11$ for $s$ to have a second factor of 11. Let $k=12$, we get $11k^2-k+1=11*11*13$, so $m=11*12=132$. -Mathdummy

Solution 3

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

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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