View source for 2000 AMC 8 Problems/Problem 14

==Problem==

What is the units digit of <math>19^{19} + 99^{99}</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==

Finding a pattern for each half of the sum, even powers of <math>19</math> have a units digit of <math>1</math>, and odd powers of <math>19</math> have a units digit of <math>9</math>. So, <math>19^{19}</math> has a units digit of <math>9</math>. 

Powers of <math>99</math> have the exact same property, so <math>99^{99}</math> also has a units digit of <math>9</math>. <math>9+9=18</math> which has a units digit of <math>8</math>, so the answer is <math>\boxed{D}</math>.

==Solution 2==
Using modular arithmetic:
<cmath>99 \equiv 9 \equiv -1 \pmod{10}</cmath>

Similarly,
<cmath>19 \equiv 9 \equiv -1 \pmod{10}</cmath>

We have 
<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==

{{AMC8 box|year=2000|num-b=13|num-a=15}}
{{MAA Notice}}