2024 AMC 8 Problems/Problem 1: Difference between revisions
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We only care about the unit's digits. | We only care about the unit's digits. | ||
Thus, <math>2-2</math> ends in <math>0</math>, <math>0-2</math> ends in <math>8</math>, <math>8-2</math> ends in <math>6</math>, <math>6-2</math> ends in <math>4</math>, and <math>4-2</math> ends in <math>\boxed{\textbf{(B) } 2}</math> | Thus, <math>2-2</math> ends in <math>0</math>, <math>0-2</math> ends in <math>8</math>, <math>8-2</math> ends in <math>6</math>, <math>6-2</math> ends in <math>4</math>, and <math>4-2</math> ends in <math>\boxed{\textbf{(B) } 2}</math>. | ||
~MrThinker | ~MrThinker | ||
Revision as of 17:32, 25 January 2024
Problem
What is the ones digit of
Solution 1
We can rewrite the expression as
.
We note that the units digit of the addition is
because all the units digits of the five numbers are
and
, which has a units digit of
.
Now, we have something with a units digit of
subtracted from
. The units digit of this expression is obviously
, and we get
as our answer.
~ Dreamer1297
Solution 2
This means the ones digit is
~ nikhil
~ CXP
Solution 3
We only care about the unit's digits.
Thus,
ends in
,
ends in
,
ends in
,
ends in
, and
ends in
.
~MrThinker
Video Solution 1 (easy to digest) by Power Solve
https://www.youtube.com/watch?v=HE7JjZQ6xCk
See Also
| 2024 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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 | ||
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