Difference between revisions of "2015 AIME I Problems/Problem 8"
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==Solution== | ==Solution== | ||
+ | You know whatever <math>n</math> is, it has to have 3 digits, because if it had only two, the maximum of <math>s(n)</math> is 18. | ||
+ | |||
+ | Now let <math>n=100a_2+10a_1+a_0</math> | ||
+ | |||
+ | So first we know, <math>a_2+a_1+a_0=20</math>. Okay now we have to split into cases based on which digit gets carried. This meaning, when you add a 3 digit number to 864, we have to know when to carry the digits. Note that if you don't understand any of the steps I take, just try adding any 3-digit number to 864 regularly (using the old-fashioned "put one number over the other" method, not mental calculation), and observe what you do at each step. | ||
+ | |||
+ | (1) None of the digit gets carried over to the next space: | ||
+ | So this means <math>a_2<2, a_1<4</math> and <math>a_0<6</math>. So | ||
+ | |||
+ | <math>s(864+n)=(8+a_2)+(6+a_1)+(4+a_0)=\text{Way more than 20}</math> | ||
+ | So it doesn't work. Now: | ||
+ | |||
+ | (2) <math>a_2+8</math> is the only one that carries over | ||
+ | So this means <math>a_2>1, a_1<4</math> and <math>a_0<6</math>. So | ||
+ | |||
+ | <math>s(864+n)=1+(8+a_2-10)+(6+a_1)+(a_0+4)=29</math> | ||
+ | |||
+ | (3)<math>a_0+4</math> is the only one that carries over. | ||
+ | So | ||
+ | |||
+ | <math>s(864+n)=(8+a_2)+(6+a_1+1)+(4+a_0-10)=29</math> | ||
+ | |||
+ | (4)The first and second digit carry over (but not the third) | ||
+ | |||
+ | <math>s(864+n)=1+(8+a_2-10+1)+(6+a_1-10)+(4+a_0)=20</math> | ||
+ | |||
+ | Aha! This case works but we still have to make sure it's possible for <math>a_2+a_1+a_0=20</math> (We assumed this is true, so we have to find a number that works.) Since only the second and first digit carry over, <math>a_2>0, a_1>3</math> and <math>a_0<6</math>. The smallest value we can get with this is 695. Let's see if we can find a smaller one: | ||
+ | |||
+ | (5)The first and third digit carry over (but not the second) | ||
+ | |||
+ | <math>s(864+n)=1+(8+a_2-10)+(7+a_1)+(4+a_0-10)=20</math> | ||
+ | |||
+ | The largest value for the middle digit is 2, so the other digits have to be both 9's. So the smallest possible value is 929 | ||
+ | |||
+ | (6) All the digits carry over | ||
+ | |||
+ | <math>s(864+n)=1+(9+a_2-10)+(7+a_1-10)+(4+a_0-10)=\text{Way less than 20}</math> | ||
+ | |||
+ | |||
+ | So the answer is <math>\box{695}</math> which after a quick test, does indeed work. | ||
== See also == | == See also == |
Revision as of 12:59, 20 March 2015
Problem
For positive integer , let denote the sum of the digits of . Find the smallest positive integer satisfying .
Solution
You know whatever is, it has to have 3 digits, because if it had only two, the maximum of is 18.
Now let
So first we know, . Okay now we have to split into cases based on which digit gets carried. This meaning, when you add a 3 digit number to 864, we have to know when to carry the digits. Note that if you don't understand any of the steps I take, just try adding any 3-digit number to 864 regularly (using the old-fashioned "put one number over the other" method, not mental calculation), and observe what you do at each step.
(1) None of the digit gets carried over to the next space: So this means and . So
So it doesn't work. Now:
(2) is the only one that carries over So this means and . So
(3) is the only one that carries over. So
(4)The first and second digit carry over (but not the third)
Aha! This case works but we still have to make sure it's possible for (We assumed this is true, so we have to find a number that works.) Since only the second and first digit carry over, and . The smallest value we can get with this is 695. Let's see if we can find a smaller one:
(5)The first and third digit carry over (but not the second)
The largest value for the middle digit is 2, so the other digits have to be both 9's. So the smallest possible value is 929
(6) All the digits carry over
So the answer is $\box{695}$ (Error compiling LaTeX. ! Missing number, treated as zero.) which after a quick test, does indeed work.
See also
2015 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.