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

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Therefore, <math>k^2+2^k \equiv 0+6 \equiv 6 \pmod{10}</math>. So the units digit is <math>6 \Rightarrow \boxed{D}</math>.
 
Therefore, <math>k^2+2^k \equiv 0+6 \equiv 6 \pmod{10}</math>. So the units digit is <math>6 \Rightarrow \boxed{D}</math>.
 +
 +
===Note===
 +
Another way to get <math>k \equiv 0 \pmod{10}</math> is to find the cycles of the last digit.
 +
 +
For <math>2008^2</math>, we need only be concerned with the last digit <math>8</math> since the other digits do not affect the last digit. Since <math>8^{2} = 64</math>, the last digit of <math>2008^2</math> is <math>4</math>.
 +
 +
For <math>2^{2008}</math>, note that the last digit cycles through the pattern <math>{2, 4, 8, 6}</math>. (You can try to see this by calculating the first powers of <math>2</math>.)
 +
 +
Since <math>2008</math> is a multiple of <math>4</math>, the last digit of <math>2^{2008}</math> is evidently <math>6.</math>
 +
 +
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 <math>k</math> for an integer <math>n</math> implies <math>n \equiv k \pmod{10}</math>
 +
 +
Let us take this step by step. First, we consider <math>k^2.</math>
 +
 +
Note that <math>k^2 = \left(2008^2 + 2^{2008}\right)^2 = 2008^4 + 2 \cdot 2008^2 \cdot 2^{5016}.</math> Now we calculate <math>k^2 \pmod{10}</math>
 +
 +
Before continuing, though, we must take note of the following:
 +
 +
<cmath>
 +
\begin{align*}
 +
2^1 &\equiv 2 \pmod {10} \\
 +
2^2 &\equiv 4 \pmod {10} \\
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2^3 &\equiv 8 \pmod {10} \\
 +
2^4 &\equiv 6 \pmod {10}
 +
\end{align*}
 +
</cmath>
 +
 +
Now, we continue with the calculation.
 +
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
We do the same with <math>2^k.</math> However, we just need to find <math>k \pmod 4</math> in order to do this calculation since we have the table of <math>2^k \pmod 10.</math>
 +
 +
<cmath>
 +
\begin{align*}
 +
2008^2 + 2^{2008} &\equiv 8^2 \pmod{4} \\
 +
&\equiv 64 \pmod 4\\
 +
&\equiv 0 \pmod 4
 +
\end{align*}
 +
</cmath>
 +
 +
This implies that
 +
<cmath>
 +
\begin{align*}
 +
2^k &\equiv 2^{4} \pmod{10} \\
 +
&\equiv 6 \pmod{10}
 +
\end{align*}
 +
</cmath>
 +
 +
Thus,
 +
<cmath>
 +
\begin{align*}
 +
k^2 + 2^k &\equiv 6+0 \pmod{10}\\
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&\equiv \boxed{\textbf{(D) }6} \pmod{10}
 +
\end{align*}
 +
</cmath>
 +
 +
~mathboy282
 +
 +
 +
 +
==Solution 3 (when you are limited on time)==
 +
By unit digit arithmetic, the unit digit of <math>k^2+2^k</math> need to be either <math>4</math> or <math>6.</math> Hence, we can guess one of them for a <math>50</math>% 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)==
 
==Solution 2 (Video solution)==

Latest revision as of 03:19, 30 September 2023

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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