# Difference between revisions of "2003 AMC 12A Problems/Problem 18"

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== See Also == | == See Also == | ||

*[[2003 AMC 12A Problems]] | *[[2003 AMC 12A Problems]] | ||

− | *[[2003 AMC 12A/Problem 17|Previous Problem]] | + | *[[2003 AMC 12A Problems/Problem 17|Previous Problem]] |

− | *[[2003 AMC 12A/Problem 19|Next Problem]] | + | *[[2003 AMC 12A Problems/Problem 19|Next Problem]] |

[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] |

## Revision as of 13:19, 1 January 2012

## 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).