Difference between revisions of "2003 AMC 10A Problems/Problem 25"
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− | Since q is a quotient and r is a remainder when n is divided by 100. So we have <math>n=100q+r</math>. Since we are counting choices where q+r is divisible by 11, we have <math>n=99q+q+r=99q+ | + | Since q is a quotient and r is a remainder when n is divided by 100. So we have <math>n=100q+r</math>. Since we are counting choices where q+r is divisible by 11, we have <math>n=99q+q+r=99q+11k</math> for some k. This means that n is the sum of two multiples of 11 and would thus itself be a divisor of 11. Then we can count all the four digit divisors of 11 as in Solution 2. (This solution is essentially the same as solution 2, but it does not necessarily involve mods and so could potentially be faster.) |
==== Notes ==== | ==== Notes ==== |
Revision as of 19:41, 2 September 2009
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 equal , where through are digits. Therefore,
We now take :
The divisor trick for 11 is as follows:
"Let be an digit integer. If is divisible by , then is also divisible by 11."
Therefore, the five digit number is divisible by 11. The 5-digit multiples of 11 range from to . There are divisors of 11 between those inclusive.
Solution 3
Since q is a quotient and r is a remainder when n is divided by 100. So we have . Since we are counting choices where q+r is divisible by 11, we have for some k. This means that n is the sum of two multiples of 11 and would thus itself be a divisor of 11. Then we can count all the four digit divisors of 11 as in Solution 2. (This solution is essentially the same as solution 2, but it does not necessarily involve mods and so could potentially be faster.)
Notes
The part labeled "divisor trick" actually follows from the same observation we made in the previous step: , therefore and for all . For a digit number we get , as claimed.
Also note that in the "divisor trick" we actually want to assign the signs backwards - if we make sure that the last sign is a , the result will have the same remainder modulo as the original number.
See Also
2003 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Final Question | |
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 |