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Difference between revisions of "2016 AIME I Problems/Problem 8"

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To find n, realize that there are 3!=6 ways of ordering the numbers in each of the places.  Additionally, there are three possibilities for the numbers in the ones place-4 7 and 9, 5 7 and 8, and 6 7 and 9.  Therefore there are 6^3 * 3 ways total, which is 648. <math>|m-n|</math>=<math>|810-648|</math>=162.
 
To find n, realize that there are 3!=6 ways of ordering the numbers in each of the places.  Additionally, there are three possibilities for the numbers in the ones place-4 7 and 9, 5 7 and 8, and 6 7 and 9.  Therefore there are 6^3 * 3 ways total, which is 648. <math>|m-n|</math>=<math>|810-648|</math>=162.
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== See also ==
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{{AIME box|year=2016|n=I|num-b=7|num-a=9}}
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{{MAA Notice}}

Revision as of 16:55, 4 March 2016

Problem 8

For a permutation $p = (a_1,a_2,\ldots,a_9)$ of the digits $1,2,\ldots,9$, let $s(p)$ denote the sum of the three $3$-digit numbers $a_1a_2a_3$, $a_4a_5a_6$, and $a_7a_8a_9$. Let $m$ be the minimum value of $s(p)$ subject to the condition that the units digit of $s(p)$ is $0$. Let $n$ denote the number of permutations $p$ with $s(p) = m$. Find $|m - n|$.

Solution

Solution by jonnyboyg: To minimize $s(p)$, the numbers 1, 2, and 3 must be in the hundreds places. The numbers in the ones places must have a sum of 20. This means that the numbers in the tens places will always have the same sum. One way to do this is 154+267+389=810. Therefore m=810.

To find n, realize that there are 3!=6 ways of ordering the numbers in each of the places. Additionally, there are three possibilities for the numbers in the ones place-4 7 and 9, 5 7 and 8, and 6 7 and 9. Therefore there are 6^3 * 3 ways total, which is 648. $|m-n|$=$|810-648|$=162.

See also

2016 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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