Difference between revisions of "2003 AMC 12B Problems/Problem 18"
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<math>10000 \leq n \leq 99999</math>, so there are <math>\left\lfloor\frac{99999}{11}\right\rfloor-\left\lceil\frac{10000}{11}\right\rceil+1 = \boxed{8181}</math> values of <math>q+r</math> that are divisible by <math>11 \Rightarrow {B}</math>. | <math>10000 \leq n \leq 99999</math>, so there are <math>\left\lfloor\frac{99999}{11}\right\rfloor-\left\lceil\frac{10000}{11}\right\rceil+1 = \boxed{8181}</math> values of <math>q+r</math> that are divisible by <math>11 \Rightarrow {B}</math>. | ||
| + | |||
| + | == See Also== | ||
| + | {{AMC12 box|year=2003|ab=B|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 00:09, 5 January 2014
Problem
Let 
be a 5-digit number, and let 
and 
be the quotient and remainder, respectively, when is divided by 
. For how many values of is 
divisible by 
?
Solution
Suppose 
Since 
and 
, 
, so there are 
values of that are divisible by 
.
See Also
| 2003 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 17 |
Followed by Problem 19 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
