Difference between revisions of "2017 AMC 10B Problems/Problem 23"
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~Arcticturn | ~Arcticturn | ||
+ | ==Solution 4== | ||
+ | Since <math>N</math> ends with the number <math>4</math>, <math>N\equiv 4\pmod 5</math>. Also, the sum of the digits of <math>N</math> are divisible by <math>9</math>, so <math>N</math> must be divisible by <math>9</math>. Therefore, we have the system of equations: | ||
+ | <math>N\equiv 4\pmod 5</math> | ||
+ | <math>N\equiv 0\pmod 9</math> | ||
+ | According to the second equation, <math>N\equiv \{0, 9, 18, 27, 36\} \pmod {45}</math>. The only one of these solutions that is <math>4\pmod 5</math> is <math>\boxed{\textbf{(C) } 9}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Sid2012 sid2012] | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=B|num-b=22|num-a=24}} | {{AMC10 box|year=2017|ab=B|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:25, 4 November 2024
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 |
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