Difference between revisions of "2000 AMC 8 Problems/Problem 14"
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<math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9</math> | <math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9</math> | ||
| + | |||
| + | ==Video Solution by OmegaLearn== | ||
| + | https://youtu.be/7an5wU9Q5hk?t=1552 | ||
==Solution== | ==Solution== | ||
| Line 19: | Line 22: | ||
We have | We have | ||
| − | <cmath>-1^{19} + -1^{99} = -1 + -1 \equiv \boxed{(\textbf{D}) \ 8} \pmod{10}</cmath> | + | <cmath>(-1)^{19} + (-1)^{99} = -1 + -1 \equiv \boxed{(\textbf{D}) \ 8} \pmod{10}</cmath> |
| + | |||
| + | -ryjs | ||
| + | |||
| + | ==Solution 3== | ||
| + | Experimentation gives | ||
| + | <cmath>\text{any number ending with }9^{\text{something even}} = \text{has units digit }1</cmath> | ||
| + | |||
| + | <cmath>\text{any number ending with }9^{\text{something odd}} = \text{has units digit }9</cmath> | ||
| + | |||
| + | Using this we have | ||
| + | <cmath>19^{19} + 99^{99}</cmath> | ||
| + | <cmath>9^{19} + 9^{99}</cmath> | ||
| + | |||
| + | Both <math>19</math> and <math>99</math> are odd, so we are left with | ||
| + | <cmath>9+9=18,</cmath> which has units digit <math>\boxed{(\textbf{D}) \ 8}.</math> | ||
| + | -ryjs | ||
| + | |||
| + | ==Solution 4== | ||
| + | <math>19^{19}+99^{99}=</math>36972963764972677265718790562880544059566876428174110243025997242355257045527752342141065001012823272794097888 | ||
| + | 9548326540119429996769494359451621570193644014418071060667659303363419435659472789623878 which ends in 8. :-) If you are really stupid and have time (like about 3 hours), this is a real brainless bash. | ||
| + | |||
| + | --hefei417 | ||
==See Also== | ==See Also== | ||
Latest revision as of 04:23, 31 December 2022
Contents
Problem
What is the units digit of 
?
Video Solution by OmegaLearn
https://youtu.be/7an5wU9Q5hk?t=1552
Solution
Finding a pattern for each half of the sum, even powers of 
have a units digit of 
, and odd powers of have a units digit of 
. So, 
has a units digit of .
Powers of 
have the exact same property, so 
also has a units digit of . 
which has a units digit of 
, so the answer is 
.
Solution 2
Using modular arithmetic:
![\[99 \equiv 9 \equiv -1 \pmod{10}\]](http://latex.artofproblemsolving.com/7/5/e/75ef9dad0de69599575109d040e7639b9fc39fa3.png)
Similarly,
![\[19 \equiv 9 \equiv -1 \pmod{10}\]](http://latex.artofproblemsolving.com/1/5/c/15c3fc39f1bc9a64c1f573eb133b3078538c5951.png)
We have
![\[(-1)^{19} + (-1)^{99} = -1 + -1 \equiv \boxed{(\textbf{D}) \ 8} \pmod{10}\]](http://latex.artofproblemsolving.com/e/c/9/ec91d9d5444d75ac00161eb77f2d743d167f99d9.png)
-ryjs
Solution 3
Experimentation gives
![\[\text{any number ending with }9^{\text{something even}} = \text{has units digit }1\]](http://latex.artofproblemsolving.com/9/c/1/9c129f4219d96b4b10cbcca3beeae338ca91108b.png)
![\[\text{any number ending with }9^{\text{something odd}} = \text{has units digit }9\]](http://latex.artofproblemsolving.com/7/3/4/734c9872ac0a9506801bb56c341c75ee06743bc7.png)
Using this we have
![\[19^{19} + 99^{99}\]](http://latex.artofproblemsolving.com/4/a/8/4a8c39504c5a4b8990eefee52a93c968bf42a6b8.png)
![\[9^{19} + 9^{99}\]](http://latex.artofproblemsolving.com/8/6/8/868a2cc7eb6f1a7e280d6578b8f519761df7dfb7.png)
Both and are odd, so we are left with
![\[9+9=18,\]](http://latex.artofproblemsolving.com/e/5/9/e59c1c010c314809684f48e73b1bb48502d78b0e.png)
which has units digit 
-ryjs
Solution 4
36972963764972677265718790562880544059566876428174110243025997242355257045527752342141065001012823272794097888
9548326540119429996769494359451621570193644014418071060667659303363419435659472789623878 which ends in 8. :-) If you are really stupid and have time (like about 3 hours), this is a real brainless bash.
--hefei417
See Also
| 2000 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
