Difference between revisions of "2016 AIME I Problems/Problem 8"
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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>. | ||
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− | + | ==Video Solutions== | |
https://www.youtube.com/watch?v=WBtMUzgqfwI | https://www.youtube.com/watch?v=WBtMUzgqfwI | ||
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https://www.youtube.com/watch?v=QBHakfd2gnQ | https://www.youtube.com/watch?v=QBHakfd2gnQ | ||
+ | == 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}} |
Latest revision as of 19:17, 3 May 2021
Contents
[hide]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 contributes more to the final sum than the ones digit, and we are looking for the minimum value of , 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. .
Video Solutions
https://www.youtube.com/watch?v=WBtMUzgqfwI
https://www.youtube.com/watch?v=QBHakfd2gnQ
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.