Difference between revisions of "2010 AMC 8 Problems/Problem 22"
(→See Also) |
|||
Line 11: | Line 11: | ||
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> | ||
+ | |||
+ | ==Video by MathTalks== | ||
+ | |||
+ | https://youtu.be/mSCQzmfdX-g | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2010|num-b=21|num-a=23}} | {{AMC8 box|year=2010|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:49, 22 January 2023
Problem
The hundreds digit of a three-digit number is 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?
Solution 1
Let the hundreds, tens, and units digits of the original three-digit number be , , and , respectively. We are given that . The original three-digit number is equal to . The hundreds, tens, and units digits of the reversed three-digit number are , , and , respectively. This number is equal to . Subtracting this expression from the expression for the original number, we get . Thus, the units digit in the final result is
Solution 2
The result must hold for any three-digit number with hundreds digit being more than the units digit. is such a number. Evaluating, we get . Thus, the units digit in the final result is
Video by MathTalks
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.