Difference between revisions of "2023 AMC 10B Problems/Problem 8"
(→Video Solution) |
|||
(5 intermediate revisions by 4 users not shown) | |||
Line 4: | Line 4: | ||
<math>\text{(A)}\ 7 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 3</math> | <math>\text{(A)}\ 7 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 3</math> | ||
− | ==Solution== | + | ==Solution 1== |
<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 | ||
− | |||
− | |||
− | + | ==Solution 2== | |
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
When looking at the units digit patterns of the powers of <math>2</math>, we see that | When looking at the units digit patterns of the powers of <math>2</math>, we see that | ||
Line 48: | Line 30: | ||
We only need to find the units digit in the end, so we just add those <math>2</math> already found units digits, to get a new units digit of <math>7</math>. Therefore the answer is <math>\boxed{\text{(A) 7}}</math> | We only need to find the units digit in the end, so we just add those <math>2</math> already found units digits, to get a new units digit of <math>7</math>. Therefore the answer is <math>\boxed{\text{(A) 7}}</math> | ||
− | ~ | + | ~mk00 |
− | |||
− | |||
− | == | + | ==Solution 3 (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 <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. | |
+ | |||
+ | <math>2^1=2</math> | ||
+ | |||
+ | <math>2^2=4</math> | ||
+ | |||
+ | <math>2^3=8</math> | ||
+ | |||
+ | <math>2^4=6+10</math> | ||
+ | |||
+ | <math>2^5=2+30</math> | ||
+ | |||
+ | <math>2^6=4+60</math> | ||
+ | |||
+ | 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> is <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> | ||
− | ~ | + | ~vsinghminhas |
− | == | + | == Solution 4 (Bashy) == |
− | + | <math>2022^{2023}\equiv 2^{2023}=128^{289}\equiv 8^{289}=2^{289\times3}=131072^{51}\equiv 2^{51}=131072^3\equiv 2^3=8</math> <math>(\text{mod}</math> <math>10)</math> | |
+ | <math>2023^{2022}\equiv 3^{2022}=9^{1011}\equiv (-1)^{1011}=-1\equiv 9</math> <math>(\text{mod}</math> <math>10)</math> | ||
− | + | <math>8+9=17\equiv 7</math> <math>(\text{mod}</math> <math>10)</math> | |
− | + | ~ [https://artofproblemsolving.com/wiki/index.php?title=User:Yrock Yrock] | |
− | https:// | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2023|ab=B|num-b=7|num-a=9}} | {{AMC10 box|year=2023|ab=B|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:53, 9 December 2023
Contents
Problem
What is the units digit of ?
Solution 1
(mod 10).
~andliu766
Solution 2
When looking at the units digit patterns of the powers of , we see that
, units digit
, units digit
, units digit
, units digit
, units digit
And the pattern repeats. This pattern will apply for the powers of as well, since the units digit of
is
. We can find the pattern for the powers of
too. The pattern follows with units digits,
,
,
,
,
,
, ...
Similarly, the units digit of
will follow the same pattern as the powers of
.
Both of these powers cycle in groups of . When diving
by
, we get
remainder
, meaning
complete cycles; or the power being a multiple of
,
times, and
extra. So the units digit of
is
.
divided by
is
reminder
, which means
complete cycles, or the power being a multiple of
,
times, and
extra. So the units digit of
is
.
We only need to find the units digit in the end, so we just add those already found units digits, to get a new units digit of
. Therefore the answer is
~mk00
Solution 3 (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 . We can find the units digit of
, by listing the units digits of the first few powers of two, and trying to find a pattern.
As we can see the units digits of powers of two repeat after every four iterations. Now we know the units digit of is
and the units digit of
. Similarly we can find the last digits of powers of three repeat after every four, so the units digit of
is
. Adding these together, the ones digit is the same as the ones digit of
which is
.
~vsinghminhas
Solution 4 (Bashy)
~ Yrock
See also
2023 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.