Difference between revisions of "2000 AMC 8 Problems/Problem 14"

m (Solution)
(Solution)
Line 10: Line 10:
  
 
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>.
 
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{(\text{D })8} \pmod{10}</cmath>
  
 
==See Also==
 
==See Also==

Revision as of 23:15, 11 April 2020

Problem

What is the units digit of $19^{19} + 99^{99}$?

$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$

Solution

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

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

Solution 2

Using modular arithmetic: \[99 \equiv 9 \equiv -1 \pmod{10}\]

Similarly, \[19 \equiv 9 \equiv -1 \pmod{10}\]

We have \[-1^{19} + -1^{99} = -1 + -1 \equiv \boxed{(\text{D })8} \pmod{10}\]

See Also

2000 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS