Difference between revisions of "2007 AMC 10B Problems/Problem 24"
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For a number to be divisible by <math>4,</math> the last two digits have to be divisible by <math>4.</math> That means the last two digits of this integer must be <math>4.</math> | For a number to be divisible by <math>4,</math> the last two digits have to be divisible by <math>4.</math> That means the last two digits of this integer must be <math>4.</math> | ||
| − | For a number to be divisible by <math>9,</math> the sum of all the digits must be divisible by | + | For a number to be divisible by <math>9,</math> the sum of all the digits must be divisible by <math>9.</math> The only way to make this happen is with 9 <math>4</math>'s. However, we also need one <math>9.</math> |
The smallest integer that meets all these conditions is <math>4444444944</math>. The last four digits are <math>\boxed{\mathrm{(C) \ } 4944}</math> | The smallest integer that meets all these conditions is <math>4444444944</math>. The last four digits are <math>\boxed{\mathrm{(C) \ } 4944}</math> | ||
Revision as of 18:25, 1 August 2012
Problem
Let 
denote the smallest positive integer that is divisible by both 
and 
and whose base-
representation consists of only 's and 
's, with at least one of each. What are the last four digits of 
Solution
For a number to be divisible by 
the last two digits have to be divisible by 
That means the last two digits of this integer must be
For a number to be divisible by the sum of all the digits must be divisible by 
The only way to make this happen is with 9 's. However, we also need one
The smallest integer that meets all these conditions is 
. The last four digits are 
See Also
| 2007 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
