Difference between revisions of "2017 AMC 10B Problems/Problem 23"
Stanleyh0608 (talk | contribs) m (→Solution2) |
|||
Line 12: | 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>. | ||
− | == | + | ==Solution 2== |
− | The same way, you can get N=4(Mod 5) and 0(Mod 9). By The Chinese remainder Theorem, the answer | + | 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 <math>\boxed{C}</math> |
+ | |||
==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 15:13, 5 June 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 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 .
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
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.