Difference between revisions of "2021 AMC 12A Problems/Problem 5"
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==Problem== | ==Problem== | ||
When a student multiplied the number <math>66</math> by the repeating decimal <cmath>\underline{1}.\underline{a}\underline{b}\underline{a}\underline{b}...=\underline{1}.\overline{\underline{a}\underline{b}},</cmath> where <math>a</math> and <math>b</math> are digits, he did not notice the notation and just multiplied <math>66</math> times <math>\underline{1}.\underline{a}\underline{b}</math>. Later he found that his answer is <math>0.5</math> less than the correct answer. What is the <math>2</math>-digit number <math>\underline{a}\underline{b}?</math> | When a student multiplied the number <math>66</math> by the repeating decimal <cmath>\underline{1}.\underline{a}\underline{b}\underline{a}\underline{b}...=\underline{1}.\overline{\underline{a}\underline{b}},</cmath> where <math>a</math> and <math>b</math> are digits, he did not notice the notation and just multiplied <math>66</math> times <math>\underline{1}.\underline{a}\underline{b}</math>. Later he found that his answer is <math>0.5</math> less than the correct answer. What is the <math>2</math>-digit number <math>\underline{a}\underline{b}?</math> |
Revision as of 17:37, 12 February 2021
- The following problem is from both the 2021 AMC 10A #8 and 2021 AMC 12A #5, so both problems redirect to this page.
Contents
[hide]Problem
When a student multiplied the number by the repeating decimal where and are digits, he did not notice the notation and just multiplied times . Later he found that his answer is less than the correct answer. What is the -digit number
Solution
It is known that and . Let . We have that . Solving gives that so . ~aop2014
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=MUHja8TpKGw&t=359s
Video Solution by Hawk Math
https://www.youtube.com/watch?v=P5al76DxyHY
Video Solution (Using repeating decimal properties)
~ pi_is_3.14
See also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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 10 Problems and Solutions |
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 4 |
Followed by Problem 6 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.