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

Latest revision as of 19:24, 24 November 2024

Problem

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

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

Video Solution by OmegaLearn

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 $(-1)^{19} + (-1)^{99} = -1 + -1 \equiv \boxed{(\textbf{D}) \ 8} \pmod{10}$

-ryjs

Solution 3

Experimentation gives $\text{any number ending with }9^{\text{something even}} = \text{has units digit }1$

$\text{any number ending with }9^{\text{something odd}} = \text{has units digit }9$

Using this we have $19^{19} + 99^{99}$ $9^{19} + 9^{99}$

Both $19$ and $99$ are odd, so we are left with $9+9=18,$ which has units digit $\boxed{(\textbf{D}) \ 8}.$ -ryjs

Solution 4

$19^{19}+99^{99}=$ 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

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.

@@ Line 1: / Line 1: @@
-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 D.
+==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}}

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