2021 AMC 12A Problems/Problem 3

The following problem is from both the 2021 AMC 10A #3 and 2021 AMC 12A #3, so both problems redirect to this page.

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 1 (Algebra)

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 $9(1582) = \boxed{\textbf{(D)} ~14{,}238}$ .

~abhinavg0627

Solution 2 (Arithmetic)

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 $\boxed{\textbf{(D)} ~14{,}238}$ .

Another quick solution is to realize that the sum is represents a number $n$ added to $10n$ . The difference is $9n$ , which is $\frac{9}{11}$ of the given sum.

~CoolJupiter 2021

Solution 3 (Vertical Addition and Logic)

Let the bigger of the numbers be $\underline{AB{,}CD0}.$ It follows that the smaller number is $\underline{A{,}BCD}.$ Adding vertically, we have $\begin{array}{cccccc} & A & B & C & D & 0 \\ +\quad & & A & B & C & D \\ \hline & & & & & \\ [-2.5ex] & 1 & 7 & 4 & 0 & 2 \\ \end{array}$ Working from left to right, we get $D=2\implies C=8 \implies B=5 \implies A=1.$ The larger number is $15{,}820$ and the smaller number is $1{,}582.$ Their difference is $15{,}820-1{,}582=\boxed{\textbf{(D)} ~14{,}238}.$

~MRENTHUSIASM

Solution 4 (Logic)

We know that the larger number has a units digit of $0$ since it is divisible by 10. If $D$ is the ten's digit of the larger number, then we have $D$ as the units digit of the smaller number. Since the sum of the natural numbers has a units digit of $2$ , $D=2$ .

The units digit of the larger number is $0$ and the units digit of the smaller number is $2$ , so if we find the positive difference of the numbers, we get 8. There is only one answer choice which has this units digit, and that is \boxed{\textbf{(D)} ~14{,}238}.$

(Similar to MRENTHUSIASM's solution)

~AARUSHGORADIA18

Video Solutions

Video Solution (Simple)

https://youtu.be/SEp9flDYm2c

~ Education, the study Of Everything

-pi_is_3.14

Video Solution by WhyMath

https://youtu.be/VpYmQEKcBpA

~savannahsolver

Video Solution by TheBeautyofMath

https://youtu.be/50CThrk3RcM?t=107 (for AMC 10A)

Video Solution by IceMatrix

https://youtu.be/rEWS75W0Q54?t=198 (for AMC 12A)

~IceMatrix

Video Solution (Problems 1-3)

https://youtu.be/CupJpUzKPB0

~MathWithPi

Video Solution by The Learning Royal

https://youtu.be/slVBYmcDMOI

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AMC 10 Problems and Solutions

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

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2021 AMC 12A Problems/Problem 3

Contents

Problem

Solution 1 (Algebra)

Solution 2 (Arithmetic)

Solution 3 (Vertical Addition and Logic)

Solution 4 (Logic)

Video Solutions

Video Solution (Simple)

Video Solution by North America Math Contest Go Go Go

Video Solution by Aaron He

Video Solution by Punxsutawney Phil

Video Solution by Hawk Math

Video Solution (Using Algebra and Meta-solving)

Video Solution by WhyMath

Video Solution by TheBeautyofMath

Video Solution by IceMatrix

Video Solution (Problems 1-3)

Video Solution by The Learning Royal

See also