Difference between revisions of "2021 AMC 12A Problems/Problem 5"

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==Problem==
 
==Problem==
Coming out very soon!
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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>
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<math>\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75</math>
  
 
==Solution==
 
==Solution==

Revision as of 15:00, 11 February 2021

Problem

When a student multiplied the number $66$ by the repeating decimal \[\underline{1}.\underline{a}\underline{b}\underline{a}\underline{b}...=\underline{1}.\overline{\underline{a}\underline{b}},\] where $a$ and $b$ are digits, he did not notice the notation and just multiplied $66$ times $\underline{1}.\underline{a}\underline{b}$. Later he found that his answer is $0.5$ less than the correct answer. What is the $2$-digit number $\underline{a}\underline{b}?$

$\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75$

Solution

It is known that $0.\overline{ab}=\frac{ab}{99}$ and $0.ab=\frac{ab}{100}$. Let $\overline {ab} = x$. We have that $66(1+\frac{x}{100})+0.5=66(1+\frac{x}{99})$. Solving gives that $100x-75=99x$ so $x=75$. ~aop2014

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=MUHja8TpKGw&t=359s

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
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

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