Difference between revisions of "2017 AMC 10B Problems/Problem 23"
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Apply this on the expanded form of <math>N</math>: | Apply this on the expanded form of <math>N</math>: | ||
<cmath>N = 1(10)^{78} + 2(10)^{77} + \cdots + 9(10)^{70} + 10(10)^{68} + 11(10)^{66} + \cdots + 43(10)^{2} + 44 \equiv</cmath> | <cmath>N = 1(10)^{78} + 2(10)^{77} + \cdots + 9(10)^{70} + 10(10)^{68} + 11(10)^{66} + \cdots + 43(10)^{2} + 44 \equiv</cmath> | ||
− | <cmath>10(1 + 2 + \cdots + 43) + 44 \equiv 10(\frac{43 \cdot 44}2) + 44 \equiv</cmath> | + | <cmath>10(1 + 2 + \cdots + 43) + 44 \equiv 10 \left (\frac{43 \cdot 44}2 \right ) + 44 \equiv</cmath> |
− | <cmath>10(\frac{-2 \cdot -1}2) - 1 \equiv \boxed{\textbf{(C) } 9} \pmod{45}</cmath> | + | <cmath>10 \left (\frac{-2 \cdot -1}2 \right ) - 1 \equiv \boxed{\textbf{(C) } 9} \pmod{45}</cmath> |
==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}} |
Revision as of 14:13, 7 August 2020
Contents
[hide]Problem 23
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 CRT(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 45 .
Solution 2
In the same way, you can get and . By the Chinese remainder Theorem, the answer comes out to be
Solution 3
Realize that for all positive integers .
Apply this on the expanded form of :
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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