2021 AMC 12A Problems/Problem 3

Revision as of 22:23, 11 February 2021 by Cooljupiter (talk | contribs) (Solution)

Problem

The sum of two natural numbers is $17{,}402$. One of the two numbers is divisible by $10$. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?

$\textbf{(A)} ~10{,}272\qquad\textbf{(B)} ~11{,}700\qquad\textbf{(C)} ~13{,}362\qquad\textbf{(D)} ~14{,}238\qquad\textbf{(E)} ~15{,}426$

Solution

The units digit of a multiple of $10$ will always be $0$. We add a $0$ whenever we multiply by $10$. So, removing the units digit is equal to dividing by $10$.

Let the smaller number (the one we get after removing the units digit) be $a$. This means the bigger number would be $10a$.

We know the sum is $10a+a = 11a$ so $11a=17402$. So $a=1582$. The difference is $10a-a = 9a$. So, the answer is $9(1582) = 14238 = \boxed{\textbf{(D)}}$.


--abhinavg0627

Solution 2(Lazy Speed)

Since the ones place of a multiple of $10$ is $0$, this implies the other integer has to end with a $2$ since both integers sum up to a number that ends with a $2$. Thus, the ones place of the difference has to be $10-2=8$, and the only answer choice that ends with an $8$ is $\boxed{\textbf{(D)~14238}}$

~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)

https://youtu.be/d2musztzDjw

-pi_is_3.14

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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All AMC 12 Problems and Solutions

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