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Difference between revisions of "2023 AMC 10B Problems/Problem 8"

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<math>2022^{2023} + 2023^{2022} \equiv 2^3 + 3^2 \equiv 17 \equiv 7</math> (mod 10). <math>\boxed{\text{A}}</math>
 
<math>2022^{2023} + 2023^{2022} \equiv 2^3 + 3^2 \equiv 17 \equiv 7</math> (mod 10). <math>\boxed{\text{A}}</math>
 
~andliu766
 
~andliu766
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==Solution 2 (Digit Cycles)==
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Note that the units digit will be the same regardless of the tens, hundreds, and thousands digits, so we can simplify this problem to finding the last digit of <math>2^{2023} + 3^{2022}</math>. We can find the units digit of <math>2^{2023}</math>, by listing the units digits of the first few powers of two, and trying to find a pattern.
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<math>2^1=2</math>
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<math>2^2=4</math>
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<math>2^3=8</math>
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<math>2^4=6+10</math>
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<math>2^5=2+30</math>
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<math>2^6=4+60</math>
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As we can see the units digits of powers of two repeat after every four iterations. Now we know the units digit of <math>2^{2020}</math> is <math>6</math> and the units digit of <math>2^{2023} \Rightarrow 2^3\cdot 2^{2020} \Rightarrow 6\cdot 8 \Rightarrow 8</math>. Similarly we can find the last digits of powers of three repeat after every four so the units digit of <math>3^{2022}</math> to be <math>1\cdot 3^2 = 9</math>. Adding these together, the ones digit is the same as the ones digit of <math>9+8</math> which is <math>7</math>. <math>\boxed{\text{(A) 7}}</math>
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~vsinghminhas

Revision as of 16:23, 15 November 2023

What is the units digit of $2022^{2023} + 2023^{2022}$?

$\text{(A)}\ 7 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 3$

Solution

$2022^{2023} + 2023^{2022} \equiv 2^3 + 3^2 \equiv 17 \equiv 7$ (mod 10). $\boxed{\text{A}}$ ~andliu766

Solution 2 (Digit Cycles)

Note that the units digit will be the same regardless of the tens, hundreds, and thousands digits, so we can simplify this problem to finding the last digit of $2^{2023} + 3^{2022}$. We can find the units digit of $2^{2023}$, by listing the units digits of the first few powers of two, and trying to find a pattern.

$2^1=2$

$2^2=4$

$2^3=8$

$2^4=6+10$

$2^5=2+30$

$2^6=4+60$

As we can see the units digits of powers of two repeat after every four iterations. Now we know the units digit of $2^{2020}$ is $6$ and the units digit of $2^{2023} \Rightarrow 2^3\cdot 2^{2020} \Rightarrow 6\cdot 8 \Rightarrow 8$. Similarly we can find the last digits of powers of three repeat after every four so the units digit of $3^{2022}$ to be $1\cdot 3^2 = 9$. Adding these together, the ones digit is the same as the ones digit of $9+8$ which is $7$. $\boxed{\text{(A) 7}}$

~vsinghminhas

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