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>: | ||
| − | <math>N = 1(10)^{78} + 2(10)^{77} + \cdots + 9(10)^{70} + 10(10)^{68} + 11(10)^{66} + \cdots 43(10)^{2} + 44 \equiv 10(1 + 2 + \cdots + 43) + 44 \\ \\ | + | <math>N = 1(10)^{78} + 2(10)^{77} + \cdots + 9(10)^{70} + 10(10)^{68} + 11(10)^{66} + \cdots + 43(10)^{2} + 44 \equiv 10(1 + 2 + \cdots + 43) + 44 \\ \\ |
\equiv 10(\frac{43 \cdot 44}2) + 44 \equiv 10(\frac{-2 \cdot -1}2) - 1 \equiv \boxed{\textbf{(C) } 9} \pmod{45}</math> | \equiv 10(\frac{43 \cdot 44}2) + 44 \equiv 10(\frac{-2 \cdot -1}2) - 1 \equiv \boxed{\textbf{(C) } 9} \pmod{45}</math> | ||
Revision as of 19:06, 23 July 2020
Contents
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
The same way, you can get N==4(Mod 5) and 0(Mod 9). 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
