Difference between revisions of "2008 AMC 12A Problems/Problem 15"

Revision as of 21:12, 7 November 2022

The following problem is from both the 2008 AMC 12A #15 and 2008 AMC 10A #24, so both problems redirect to this page.

Problem

Let $k={2008}^{2}+{2}^{2008}$ . What is the units digit of $k^2+2^k$ ?

$\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 8$

Solution

$k \equiv 2008^2 + 2^{2008} \equiv 8^2 + 2^4 \equiv 4+6 \equiv 0 \pmod{10}$ .

So, $k^2 \equiv 0 \pmod{10}$ . Since $k = 2008^2+2^{2008}$ is a multiple of four and the units digit of powers of two repeat in cycles of four, $2^k \equiv 2^4 \equiv 6 \pmod{10}$ .

Therefore, $k^2+2^k \equiv 0+6 \equiv 6 \pmod{10}$ . So the units digit is $6 \Rightarrow \boxed{D}$ .

Note

Another way to get $k \equiv 0 \pmod{10}$ is to find the cycles of the last digit.

For $2008^2$ , we need only be concerned with the last digit $8$ since the other digits do not affect the last digit. Since $8^{2} = 64$ , the last digit of $2008^2$ is $4$ .

For $2^{2008}$ , note that the last digit cycles through the pattern ${2, 4, 8, 6}$ . (You can try to see this by calculating the first powers of $2$ .)

Since $2008$ is a multiple of $4$ , the last digit of $2^{2008}$ is evidently $6.$

Continue as follows.

~mathboy282

Mathboy282, That is actually what solution 1 is explaining in the first sentence but I think yours is a more detailed and easier to comprehend explanation.

@graceandmymommath, I will offer my own solution below. Thanks for the comment.

Solution 2

I am going to share another approach to this problem.

A units digit $k$ for an integer $n$ implies $n \equiv k \pmod{10}$

Let us take this step by step. First, we consider $k^2.$

Note that $k^2 = \left(2008^2 + 2^{2008}\right)^2 = 2008^4 + 2 \cdot 2008^2 \cdot 2^{5016}.$ Now we calculate $k^2 \pmod{10}$

Before continuing, though, we must take note of the following:

$\begin{align*} 2^1 &\equiv 2 \pmod {10} \\ 2^2 &\equiv 4 \pmod {10} \\ 2^3 &\equiv 8 \pmod {10} \\ 2^4 &\equiv 6 \pmod {10} \end{align*}$

Now, we continue with the calculation.

$\begin{align*} 2008^4 + 2 \cdot 2008^2 \cdot 2^{5016} &\equiv 8^4 + 2 \cdot 8^2 \cdot 2^{2008} + 2^{5016} \pmod{10} \\ &\equiv 8^4 + 2^1 \cdot 2^6 \cdot 2^{2008} + 2^{5016} \pmod{10} \\ &\equiv 8^4 + 2^{1+6+2008} + 2^{5016} \pmod{10} \\ &\equiv 6 + 2^{2015} + 6 \pmod{10} \\ &\equiv 6 + 2^{3} + 6 \pmod{10} \\ &\equiv 6 + 8 + 6 \pmod{10} \\ &\equiv 20 \pmod{10} \\ &\equiv 0 \pmod{10} \\ \end{align*}$

We do the same with $2^k.$ However, we just need to find $k \pmod 4$ in order to do this calculation since we have the table of $2^k \pmod 10.$

$\begin{align*} 2008^2 + 2^{2008} &\equiv 8^2 \pmod{4} \\ &\equiv 64 \pmod 4\\ &\equiv 0 \pmod 4 \end{align*}$

This implies that $\begin{align*} 2^k &\equiv 2^{4} \pmod{10} \\ &\equiv 6 \pmod{10} \end{align*}$

Thus, $\begin{align*} k^2 + 2^k &\equiv 6+0 \pmod{10}\\ &\equiv \boxed{\textbf{(D) }6} \pmod{10} \end{align*}$

~mathboy282

Solution 3 (when you are limited on time)

By unit digit arithmetic, the unit digit of $k^2+2^k$ need to be either $4$ or $6.$ Hence, we can guess one of them for a $50$ % of getting it right. This should only take 20 seconds or less. ~peelybonehead

Note: this solution is not recommended and is only advised when you have <5 minutes left.

Video Solution by OmegaLearn

https://youtu.be/-H4n-QplQew?t=36

~ pi_is_3.14

Solution 2 (Video solution)

Video: https://youtu.be/Ib-onAecb1I

Another Video

Video: https://www.youtube.com/watch?v=PzqQSOaCcnw

@@ Line 108: / Line 108: @@
 ==Solution 2 (Video solution)==
 Video: https://youtu.be/Ib-onAecb1I
+==Another Video==
+Video: https://www.youtube.com/watch?v=PzqQSOaCcnw
 ==See Also==

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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

2008 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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

Page

Toolbox

Search