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Difference between revisions of "2010 AMC 8 Problems/Problem 22"

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The result must hold for any three-digit number with hundreds digit being <math>2</math> more than the units digit. <math>301</math> is such a number. Evaluating, we get <math>301-103=198</math>. Thus, the units digit in the final result is <math>\boxed{\textbf{(E)}\ 8}</math>
 
The result must hold for any three-digit number with hundreds digit being <math>2</math> more than the units digit. <math>301</math> is such a number. Evaluating, we get <math>301-103=198</math>. Thus, the units digit in the final result is <math>\boxed{\textbf{(E)}\ 8}</math>
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==Solution 3==
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Set the units digit of the original number as <math> x </math>. Thus, its hundreds digit is <math> x+2 </math>. After the digits are reversed, the hundreds digit of the original number will be the units digit of the new number. Since <math> x-(x+2) = -2 </math>, we can do regrouping and "borrow" <math> 1 </math> from the tens digit and bring it to the units digit as a <math> 10 </math>. Therefore, the units digit will end up as <math> -2 + 10 = \boxed{\textbf{(E)}\ 8}</math>
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~[[User:Bloggish|Bloggish]]
  
 
==Video by MathTalks==
 
==Video by MathTalks==

Revision as of 04:19, 27 January 2023

Problem

The hundreds digit of a three-digit number is $2$ more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?

$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8$

Solution 1

Let the hundreds, tens, and units digits of the original three-digit number be $a$, $b$, and $c$, respectively. We are given that $a=c+2$. The original three-digit number is equal to $100a+10b+c = 100(c+2)+10b+c = 101c+10b+200$. The hundreds, tens, and units digits of the reversed three-digit number are $c$, $b$, and $a$, respectively. This number is equal to $100c+10b+a = 100c+10b+(c+2) = 101c+10b+2$. Subtracting this expression from the expression for the original number, we get $(101c+10b+200) - (101c+10b+2) = 198$. Thus, the units digit in the final result is $\boxed{\textbf{(E)}\ 8}$

Solution 2

The result must hold for any three-digit number with hundreds digit being $2$ more than the units digit. $301$ is such a number. Evaluating, we get $301-103=198$. Thus, the units digit in the final result is $\boxed{\textbf{(E)}\ 8}$

Solution 3

Set the units digit of the original number as $x$. Thus, its hundreds digit is $x+2$. After the digits are reversed, the hundreds digit of the original number will be the units digit of the new number. Since $x-(x+2) = -2$, we can do regrouping and "borrow" $1$ from the tens digit and bring it to the units digit as a $10$. Therefore, the units digit will end up as $-2 + 10 = \boxed{\textbf{(E)}\ 8}$

~Bloggish

Video by MathTalks

https://youtu.be/mSCQzmfdX-g





See Also

2010 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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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