Difference between revisions of "2021 AMC 12A Problems/Problem 3"

(Solution)
(Solution)
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Let the smaller number (the one we get after removing the units digit) be <math>a</math>. This means the bigger number would be <math>10a</math>.
 
Let the smaller number (the one we get after removing the units digit) be <math>a</math>. This means the bigger number would be <math>10a</math>.
  
We know the sum is <math>10a+a = 11a</math> so <math>11a=17402</math>. So <math>a=1582</math>. The difference is <math>10a-a = 9a</math>. So, the answer is <math>9(1582) = 14238</math>.
+
We know the sum is <math>10a+a = 11a</math> so <math>11a=17402</math>. So <math>a=1582</math>. The difference is <math>10a-a = 9a</math>. So, the answer is <math>9(1582) = 14238 = \boxed{D}</math>.
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--abhinavg0627
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2021|ab=A|num-b=2|num-a=4}}
 
{{AMC12 box|year=2021|ab=A|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:39, 11 February 2021

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 $9(1582) = 14238 = \boxed{D}$.

--abhinavg0627

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
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

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