2017 AMC 10B Problems/Problem 23
Contents
[hide]Problem
Let be the -digit number that is formed by writing the integers from to in order, one after the other. What is the remainder when is divided by ?
Solution 1
We only need to find the remainders of N when divided by 5 and 9 to determine the answer. By inspection, . The remainder when is divided by is , but since , we can also write this as , which has a remainder of 0 mod 9. Solving these modular congruence using the Chinese Remainder Theorem we get the remainder to be . Therefore, the answer is .
Alternative Ending to Solution 1
Once we find our 2 modular congruences, we can narrow our options down to and because the remainder when is divided by should be a multiple of 9 by our modular congruence that states has a remainder of when divided by . Also, our other modular congruence states that the remainder when divided by should have a remainder of when divided by . Out of options and , only satisfies that the remainder when is divided by .
Solution 2
Realize that for all positive integers .
Apply this on the expanded form of :
Solution 3 (Clever way using divisibility rules)
We know that , so we can apply our restrictions to that. We know that the units digit must be or , and the digits must add up to a multiple of . . We can quickly see this is a multiple of because . We know is not a multiple of because the units digit isn't or . We can just subtract by 9 until we get a number whose units digit is 5 or 0.
We have is divisible by , so we can subtract by to get and we know that this is divisible by 5. So our answer is
~Arcticturn
Solution 4
Since ends with the number , . Also, the sum of the digits of are divisible by , so must be divisible by . Therefore, we have the system of equations:
According to the second equation, . The only one of these solutions that is is
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.