Difference between revisions of "2017 AMC 10B Problems/Problem 23"
(→Solution) |
|||
| 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 1== | ==Solution 1== | ||
| Line 16: | Line 12: | ||
==Alternative Ending to Solution 1== | ==Alternative Ending to Solution 1== | ||
Once we find our 2 modular congruences, we can narrow our options down to <math>{C}</math> and <math>{D}</math> because the remainder when <math>N</math> is divided by <math>45</math> should be a multiple of 9 by our modular congruence that states <math>N</math> has a remainder of <math>0</math> when divided by <math>9</math>. Also, our other modular congruence states that the remainder when divided by <math>45</math> should have a remainder of <math>4</math> when divided by <math>5</math>. Out of options <math>C</math> and <math>D</math>, only <math>\boxed{\textbf{(C) } 9}</math> satisfies that the remainder when <math>N</math> is divided by 45 <math>\equiv 4 \text{ (mod 5)}</math>. | Once we find our 2 modular congruences, we can narrow our options down to <math>{C}</math> and <math>{D}</math> because the remainder when <math>N</math> is divided by <math>45</math> should be a multiple of 9 by our modular congruence that states <math>N</math> has a remainder of <math>0</math> when divided by <math>9</math>. Also, our other modular congruence states that the remainder when divided by <math>45</math> should have a remainder of <math>4</math> when divided by <math>5</math>. Out of options <math>C</math> and <math>D</math>, only <math>\boxed{\textbf{(C) } 9}</math> satisfies that the remainder when <math>N</math> is divided by 45 <math>\equiv 4 \text{ (mod 5)}</math>. | ||
| − | + | ==Solution2== | |
| + | The same way, you can get N=4(Mod 5) and 0(Mod 9). By The Chinese remainder Theorem, the answer come out to be 9-(C) | ||
==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 21:36, 15 April 2020
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 ge the remainder to be (mod ). 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 
.
Solution2
The same way, you can get N=4(Mod 5) and 0(Mod 9). By The Chinese remainder Theorem, the answer come out to be 9-(C)
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. 
