Difference between revisions of "2021 AMC 12A Problems/Problem 3"
Pi is 3.14 (talk | contribs) (→Video Solution) |
Cooljupiter (talk | contribs) (→Solution) |
||
Line 14: | Line 14: | ||
--abhinavg0627 | --abhinavg0627 | ||
+ | |||
+ | ==Solution 2(Lazy Speed)== | ||
+ | |||
+ | Since the ones place of a multiple of <math>10</math> is <math>0</math>, this implies the other integer has to end with a <math>2</math> since both integers sum up to a number that ends with a <math>2</math>. Thus, the ones place of the difference has to be <math>10-2=8</math>, and the only answer choice that ends with an <math>8</math> is <math>\boxed{\textbf{(D)~14238}}</math> | ||
+ | |||
+ | ~CoolJupiter 2021 | ||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== |
Revision as of 22:23, 11 February 2021
Contents
Problem
The sum of two natural numbers is . One of the two numbers is divisible by . If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?
Solution
The units digit of a multiple of will always be . We add a whenever we multiply by . So, removing the units digit is equal to dividing by .
Let the smaller number (the one we get after removing the units digit) be . This means the bigger number would be .
We know the sum is so . So . The difference is . So, the answer is .
--abhinavg0627
Solution 2(Lazy Speed)
Since the ones place of a multiple of is , this implies the other integer has to end with a since both integers sum up to a number that ends with a . Thus, the ones place of the difference has to be , and the only answer choice that ends with an is
~CoolJupiter 2021
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=MUHja8TpKGw&t=143s
Video Solution by Hawk Math
https://www.youtube.com/watch?v=P5al76DxyHY
Video Solution (Using Algebra and Meta-solving)
-pi_is_3.14
See also
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 2 |
Followed by Problem 4 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.