Difference between revisions of "2016 AIME I Problems/Problem 8"
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==Solution== | ==Solution== | ||
− | 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 | + | 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>. |
− | 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 | + | 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>. |
== See also == | == See also == | ||
{{AIME box|year=2016|n=I|num-b=7|num-a=9}} | {{AIME box|year=2016|n=I|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:06, 21 March 2016
Problem 8
For a permutation of the digits , let denote the sum of the three -digit numbers , , and . Let be the minimum value of subject to the condition that the units digit of is . Let denote the number of permutations with . Find .
Solution
To minimize , the numbers , , and (which sum to ) must be in the hundreds places. For the units digit of to be , the numbers in the ones places must have a sum of either or . However, since the tens digit is more significant that the ones digit, we take the sum's units digit to be . We know that the sum of the numbers in the tens digits is . Therefore, .
To find , realize that there are ways of ordering the numbers in each of the places. Additionally, there are three possibilities for the numbers in the ones place: , , and . Therefore there are ways in total. .
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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