Difference between revisions of "2017 AMC 10B Problems/Problem 23"
m (→Problem 23) |
(→Problem 23) |
||
Line 4: | Line 4: | ||
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44</math> | <math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44</math> | ||
− | + | ==Solution== | |
− | + | We only need to find the remainders of N when divided by 5 and 9 to determine the answer. | |
+ | By inspection, <math>N \equiv 4 \text{ (mod 5)}</math>. | ||
+ | The remainder when divided by 9 is <math>1+2+3+4 \cdot 1+0+1+1 \cdot 4+3+4+4</math>, but since <math>10 \equiv 1 \text{ (mod 9)}</math>, we can also write this as <math>1+2+3 \cdot 10+11+12 \cdot 43 + 44 = \frac{44 \cdot 45}2 = 22 \cdot 45</math>, which clearly has a remainder of 0 mod 9. Therefore, by CRT, the answer is <math>\boxed{\textbf{(C) } 9}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=B|num-b=24|after=Last Problem}} | {{AMC10 box|year=2017|ab=B|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 09:43, 16 February 2017
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
We only need to find the remainders of N when divided by 5 and 9 to determine the answer. By inspection, . The remainder when divided by 9 is , but since , we can also write this as , which clearly has a remainder of 0 mod 9. Therefore, by CRT, the answer is .
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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.