2003 AMC 10A Problems/Problem 25
Contents
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 ?
Solutions
Simple Solution
implies that and therefore , so . Then, can range from to for a total of numbers.
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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 . Another way to think about it is the number of possible values of q when r, the remainder, is . In this case, q itself has to be a multiple of . . Then we'll need to subtract from since we only want multiples of greater than
Therefore, the number of possible values of such that is .
~ Minor Edit by PlainOldNumberTheory
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 ."
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.
Note
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 want to assign the signs backward - if we make sure that the last sign is a , the result will have the same remainder modulo as the original number.
Solution 3
Since is a quotient and is a remainder when is divided by . So we have . Since we are counting choices where is divisible by , we have for some . This means that is the sum of two multiples of and would thus itself be a multiple of . Then we can count all the five-digit multiples of 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.)
Solution 4
Defining and in terms of floor functions and would yield: and . Since , . Simplifying gets us ( is always true since floor function always yields an integer, and 99 is divisible by 11 w/o any remainder). After we come to this conclusion, it becomes easy to solve the rest of the problem (). ~hw21
Video Solution 1
https://youtu.be/OpGHj-B0_hg?t=672
~IceMatrix
Video Solution 2
~savannahsolver
See Also
2003 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last 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 |
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 |
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