Difference between revisions of "2003 AMC 12A Problems/Problem 18"
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Therefore, the number of possible values of <math>n</math> such that <math>q+r \equiv 0\pmod{11}</math> is <math>900\cdot9+81\cdot1=8181 \Rightarrow\boxed{(B)} </math>. | Therefore, the number of possible values of <math>n</math> such that <math>q+r \equiv 0\pmod{11}</math> is <math>900\cdot9+81\cdot1=8181 \Rightarrow\boxed{(B)} </math>. | ||
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==See Also== | ==See Also== | ||
{{AMC12 box|year=2003|ab=A|num-b=17|num-a=19}} | {{AMC12 box|year=2003|ab=A|num-b=17|num-a=19}} | ||
Revision as of 03:54, 24 May 2013
Problem
Let 
be a 
-digit number, and let 
and 
be the quotient and the remainder, respectively, when is divided by 
. For how many values of is 
divisible by 
?
Solution
When a -digit number is divided by , the first 
digits become the quotient, , and the last 
digits become the remainder, .
Therefore, can be any integer from to 
inclusive, and can be any integer from 
to 
inclusive.
For each of the 
possible values of , there are at least 
possible values of such that 
.
Since there is 
"extra" possible value of that is congruent to 
, each of the 
values of that are congruent to have more possible value of such that .
Therefore, the number of possible values of such that is 
.
See Also
| 2003 AMC 12A (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 | |
