Difference between revisions of "2007 AMC 10B Problems/Problem 24"
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==Problem== | ==Problem== | ||
| − | Let <math>n</math> denote the smallest positive integer that is divisible by both <math>4</math> and <math>9,</math> and whose base-<math>10</math> representation consists of only <math>4</math>'s and <math>9</math>'s, with at least one of each. What are the last four digits of <math>n?</math> | + | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Let <math>n</math> denote the smallest positive integer that is divisible by both <math>4</math> and <math>9,</math> and whose base-<math>10</math> representation consists of only <math>4</math>'s and <math>9</math>'s, with at least one of each. What are the last four digits of <math>n?</math><!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
<math>\textbf{(A) } 4444 \qquad\textbf{(B) } 4494 \qquad\textbf{(C) } 4944 \qquad\textbf{(D) } 9444 \qquad\textbf{(E) } 9944</math> | <math>\textbf{(A) } 4444 \qquad\textbf{(B) } 4494 \qquad\textbf{(C) } 4944 \qquad\textbf{(D) } 9444 \qquad\textbf{(E) } 9944</math> | ||
Revision as of 18:11, 11 November 2015
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
