Difference between revisions of "2017 AMC 12B Problems/Problem 19"
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==Solution== | ==Solution== | ||
| − | We will consider this number <math>\bmod 5</math> and <math>\bmod 9</math>. By looking at the last digit, it is obvious that the number is <math>\equiv 4\bmod 5</math>. To calculate the number <math>\bmod 9</math>, note that | + | We will consider this number <math>\bmod\ 5</math> and <math>\bmod\ 9</math>. By looking at the last digit, it is obvious that the number is <math>\equiv 4\bmod\ 5</math>. To calculate the number <math>\bmod\ 9</math>, note that |
| − | <cmath>123456\cdots 4344 \equiv 1+2+3+4+5+6+7+8+9+(1+0)+(1+1)+\cdots+(4+3)+(4+4) \equiv 1+2+\cdots+44 \bmod 9,</cmath> | + | <cmath>123456\cdots 4344 \equiv 1+2+3+4+5+6+7+8+9+(1+0)+(1+1)+\cdots+(4+3)+(4+4) \equiv 1+2+\cdots+44 \bmod\ 9,</cmath> |
so it is equivalent to | so it is equivalent to | ||
| − | <cmath>\frac{44\cdot 45}{2} = 22\cdot 45 \equiv 0\ | + | <cmath>\frac{44\cdot 45}{2} = 22\cdot 45 \equiv 0\bmod\ 9.</cmath> |
| − | Thus it is <math>0\bmod 9</math> and <math>4\bmod 5</math>, so it is <math>\textbf{(C)} = 9\bmod 45.</math> | + | Thus it is <math>0\bmod 9</math> and <math>4\bmod 5</math>, so it is <math>\textbf{(C)} = 9\bmod\ 45.</math> |
==See Also== | ==See Also== | ||
Revision as of 16:32, 16 February 2017
Problem
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 will consider this number 
and 
. By looking at the last digit, it is obvious that the number is 
. To calculate the number , note that
![\[123456\cdots 4344 \equiv 1+2+3+4+5+6+7+8+9+(1+0)+(1+1)+\cdots+(4+3)+(4+4) \equiv 1+2+\cdots+44 \bmod\ 9,\]](http://latex.artofproblemsolving.com/8/5/d/85d2840185e459a74b0b33b2f38da35ccf8231f6.png)
so it is equivalent to
![\[\frac{44\cdot 45}{2} = 22\cdot 45 \equiv 0\bmod\ 9.\]](http://latex.artofproblemsolving.com/c/9/3/c93e7ccf6d2d377c1101b951cac39ba47a59cbf7.png)
Thus it is 
and 
, so it is 
See Also
| 2017 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 18 |
Followed by Problem 20 |
| 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 12 Problems and Solutions | |
| 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. 
