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> | ||
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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 nine <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> | ||
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| + | == Video Solution by OmegaLearn == | ||
| + | https://youtu.be/p5f1u44-pvQ?t=59 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2007|ab=B|num-b=23|num-a=25}} | {{AMC10 box|year=2007|ab=B|num-b=23|num-a=25}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 07:32, 4 November 2022
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 nine 's. However, we also need one
The smallest integer that meets all these conditions is 
. The last four digits are 
Video Solution by OmegaLearn
https://youtu.be/p5f1u44-pvQ?t=59
~ pi_is_3.14
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. 
