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

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==Solution==
 
==Solution==
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To minimize <math>s(p)</math>, the numbers <math>1</math>, <math>2</math>, and <math>3</math> (which sum to <math>6</math>) must be in the hundreds places.  For the units digit of <math>s(p)</math> to be <math>0</math>, the numbers in the ones places must have a sum of either <math>10</math> or <math>20</math>. However, since the tens digit is more significant that the ones digit, we take the sum's units digit to be <math>20</math>. We know that the sum of the numbers in the tens digits is <math>\sum\limits_{i=1}^9 (i) -6-20=45-6-20=19</math>. Therefore, <math>m=100\times6+10\times19+20=810</math>.
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To minimize <math>s(p)</math>, the numbers <math>1</math>, <math>2</math>, and <math>3</math> (which sum to <math>6</math>) must be in the hundreds places.  For the units digit of <math>s(p)</math> to be <math>0</math>, the numbers in the ones places must have a sum of either <math>10</math> or <math>20</math>. However, since the tens digit contributes more to the final sum <math>s(p)</math> than the ones digit, and we are looking for the minimum value of <math>s(p)</math>, we take the sum's units digit to be <math>20</math>. We know that the sum of the numbers in the tens digits is <math>\sum\limits_{i=1}^9 (i) -6-20=45-6-20=19</math>. Therefore, <math>m=100\times6+10\times19+20=810</math>.
  
 
To find <math>n</math>, realize that there are <math>3!=6</math> ways of ordering the numbers in each of the places.  Additionally, there are three possibilities for the numbers in the ones place: <math>(4,7,9)</math>, <math>(5,7,8)</math>, and <math>(5,6,9)</math>.  Therefore there are <math>6^3\times3=648</math> ways in total. <math>|m-n|=|810-648|=\fbox{162}</math>.
 
To find <math>n</math>, realize that there are <math>3!=6</math> ways of ordering the numbers in each of the places.  Additionally, there are three possibilities for the numbers in the ones place: <math>(4,7,9)</math>, <math>(5,7,8)</math>, and <math>(5,6,9)</math>.  Therefore there are <math>6^3\times3=648</math> ways in total. <math>|m-n|=|810-648|=\fbox{162}</math>.

Revision as of 13:05, 27 February 2018

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

To minimize $s(p)$, the numbers $1$, $2$, and $3$ (which sum to $6$) must be in the hundreds places. For the units digit of $s(p)$ to be $0$, the numbers in the ones places must have a sum of either $10$ or $20$. However, since the tens digit contributes more to the final sum $s(p)$ than the ones digit, and we are looking for the minimum value of $s(p)$, we take the sum's units digit to be $20$. We know that the sum of the numbers in the tens digits is $\sum\limits_{i=1}^9 (i) -6-20=45-6-20=19$. Therefore, $m=100\times6+10\times19+20=810$.

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: $(4,7,9)$, $(5,7,8)$, and $(5,6,9)$. Therefore there are $6^3\times3=648$ ways in total. $|m-n|=|810-648|=\fbox{162}$.

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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