Difference between revisions of "2003 AMC 12A Problems/Problem 18"
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== Solution == | == Solution == | ||
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| + | Solution 1: | ||
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When a <math>5</math>-digit number is divided by <math>100</math>, the first <math>3</math> digits become the quotient, <math>q</math>, and the last <math>2</math> digits become the remainder, <math>r</math>. | When a <math>5</math>-digit number is divided by <math>100</math>, the first <math>3</math> digits become the quotient, <math>q</math>, and the last <math>2</math> digits become the remainder, <math>r</math>. | ||
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Since there is <math>1</math> "extra" possible value of <math>r</math> that is congruent to <math>0\pmod{11}</math>, each of the <math>\lfloor \frac{900}{11} \rfloor = 81</math> values of <math>q</math> that are congruent to <math>0\pmod{11}</math> have <math>1</math> more possible value of <math>r</math> such that <math>q+r \equiv 0\pmod{11}</math>. | Since there is <math>1</math> "extra" possible value of <math>r</math> that is congruent to <math>0\pmod{11}</math>, each of the <math>\lfloor \frac{900}{11} \rfloor = 81</math> values of <math>q</math> that are congruent to <math>0\pmod{11}</math> have <math>1</math> more possible value of <math>r</math> such that <math>q+r \equiv 0\pmod{11}</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 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 B</math>. |
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| + | Solution 2: | ||
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| + | Let the 5 digit number be abcde, where a, b, c, d, and e are digits. Then q = abc, and r = de. Then the digits of q +r theoretically are a, then b+d, then c + e. If this is to be divisible by 11, we can use the divisibility check for 11 to say that a + c + e - b -d is divisible by 11. If we revert this again, but we want all the digits to be single digit, then we get abcde again. So if q + r is to be divisible by 11, then the number we started with, abcde, must also be divisible by 11. There are 90000 total possible values of abcde, of which 8181 of them are divisible by 11. So for those numbers q + r is divisible by 11, so we have 8181 (B). | ||
== See Also == | == See Also == | ||
Revision as of 17:56, 21 February 2011
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
Solution 1:
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 
.
Solution 2:
Let the 5 digit number be abcde, where a, b, c, d, and e are digits. Then q = abc, and r = de. Then the digits of q +r theoretically are a, then b+d, then c + e. If this is to be divisible by 11, we can use the divisibility check for 11 to say that a + c + e - b -d is divisible by 11. If we revert this again, but we want all the digits to be single digit, then we get abcde again. So if q + r is to be divisible by 11, then the number we started with, abcde, must also be divisible by 11. There are 90000 total possible values of abcde, of which 8181 of them are divisible by 11. So for those numbers q + r is divisible by 11, so we have 8181 (B).
