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

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==Solution 1==
 
==Solution 1==
  
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Let the hundreds, tens, and units digits of the original three-digit number be <math>a</math>, <math>b</math>, and <math>c</math>, respectively. We are given that <math>a=c+2</math>. The original three-digit number is equal to <math>100a+10b+c = 100(c+2)+10b+c = 101c+10b+200</math>. The hundreds, tens, and units digits of the reversed three-digit number are <math>c</math>, <math>b</math>, and <math>a</math>, respectively. This number is equal to <math>100c+10b+a = 100c+10b+(c+2) = 101c+10b+2</math>. Subtracting this expression from the expression for the original number, we get <math>(101c+10b+200) - (101c+10b+2) = 198</math>. Thus, the units digit in the final result is <math>\textbf{(E)}\ 8</math>
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Let the hundreds, tens, and units digits of the original three-digit number be <math>a</math>, <math>b</math>, and <math>c</math>, respectively. We are given that <math>a=c+2</math>. The original three-digit number is equal to <math>100a+10b+c = 100(c+2)+10b+c = 101c+10b+200</math>. The hundreds, tens, and units digits of the reversed three-digit number are <math>c</math>, <math>b</math>, and <math>a</math>, respectively. This number is equal to <math>100c+10b+a = 100c+10b+(c+2) = 101c+10b+2</math>. Subtracting this expression from the expression for the original number, we get <math>(101c+10b+200) - (101c+10b+2) = 198</math>. Thus, the units digit in the final result is <math>\boxed{\textbf{(E)}\ 8}</math>
  
 
==Solution 2==
 
==Solution 2==

Revision as of 18:37, 4 November 2012

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 $\boxed{\textbf{(E)}\ 8}$

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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