2021 AMC 12A Problems/Problem 5

The following problem is from both the 2021 AMC 10A #8 and 2021 AMC 12A #5, so both problems redirect to this page.

Problem

When a student multiplied the number $66$ by the repeating decimal, $\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}},$ 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 1

We are given that $66\Bigl(\underline{1}.\overline{\underline{a} \ \underline{b}}\Bigr)-0.5=66\Bigl(\underline{1}.\underline{a} \ \underline{b}\Bigr),$ from which $\begin{align*} 66\Bigl(\underline{1}.\overline{\underline{a} \ \underline{b}}\Bigr)-66\Bigl(\underline{1}.\underline{a} \ \underline{b}\Bigr)&=0.5 \\ 66\Bigl(\underline{1}.\overline{\underline{a} \ \underline{b}} - \underline{1}.\underline{a} \ \underline{b}\Bigr)&=0.5 \\ 66\Bigl(\underline{0}.\underline{0} \ \underline{0} \ \overline{\underline{a} \ \underline{b}}\Bigr)&=0.5 \\ 66\left(\frac{1}{100}\cdot\underline{0}.\overline{\underline{a} \ \underline{b}}\right)&=\frac12 \\ \underline{0}.\overline{\underline{a} \ \underline{b}}&=\frac{25}{33} \\ \underline{0}.\overline{\underline{a} \ \underline{b}}&=0.\overline{75} \\ \underline{a} \ \underline{b}&=\boxed{\textbf{(E) }75}. \end{align*}$ ~MRENTHUSIASM

Solution 2

It is known that $\underline{0}.\overline{\underline{a} \ \underline{b}}=\frac{\underline{a} \ \underline{b}}{99}$ and $\underline{0}.\underline{a} \ \underline{b}=\frac{\underline{a} \ \underline{b}}{100}.$

Let $x=\underline{a} \ \underline{b}.$ We have $66\biggl(1+\frac{x}{99}\biggr)-66\biggl(1+\frac{x}{100}\biggr)=0.5.$ Expanding and simplifying give $\frac{x}{150}=0.5,$ so $x=\boxed{\textbf{(E) }75}.$

~aop2014 ~BakedPotato66 ~MRENTHUSIASM

Solution 3 (Similar to Solution 2)

We have $66 \cdot \left(1 + \frac{10a+b}{100}\right) + \frac{1}{2} = 66 \cdot \left(1+ \frac{10a+b}{99}\right).$ Expanding both sides, we have $66 + \frac{33(10a+b)}{50} + \frac{1}{2} = 66 + \frac{2(10a+b)}{3}.$ Subtracting $66$ from both sides, we have $\frac{33(10a+b)}{50} + \frac{1}{2} = \frac{2(10a+b)}{3}.$ Multiplying both sides by $50 \cdot 3 = 150,$ we have $99(10a+b) + 75 = 100(10a+b).$ Thus, the answer is $10a+b = \boxed{\textbf{(E) }75}.$

By letting $x=\underline{a} \ \underline{b}=10a+b,$ this solution is similar to Solution 2. In this solution, we solve for $10a+b$ as a whole.

-mathboy282 (Solution)

~MRENTHUSIASM (Minor Revision)

Solution 4 (Answer Choices & Modular)

Let $\underline{a} \ \underline{b}$ represent the two-digit number $ab$ , not the product of the digits $a$ and $b$ . We can construct fractions for the values $1.\underline{a} \ \underline{b}$ , and $1.\overline{\underline{a} \ \underline{b}}$ , which are $\frac{1\underline{a} \ \underline{b}}{100}$ and $\frac{1\underline{a} \ \underline{b}}{99}$ respectively. Multiplying by $66$ on both sides and adding $1/2$ to $\frac{1\underline{a} \ \underline{b}}{100}$ and simplifying, we get this: $\frac{33 * \underline{a}\underline{b} + 25}{50} = \frac{2\underline{a}\underline{b}}{3}.$

Looking at the answer choices, we notice that all of them are divisible by $3$ . This means that since the right-hand side will result in an integer, the left-hand side needs to as well. This means that the numerator of the left-hand side fraction has to be divisible by $50$ . So, we get this expression:

$33 * \underline{a}\underline{b}\ + 25 \equiv 0\pmod{50}.$

$33 * \underline{a}\underline{b}\ \equiv -25\pmod{50}.$

$33 * \underline{a}\underline{b}\ \equiv 25\pmod{50}.$

This means that the product of $33$ and $\underline{a}\underline{b}$ must have a remainder of $25$ when divided by $50$ . Since it must have a remainder of $25$ , the product should have a units digit of $5$ , which eliminates $\textbf{(B) }$ and $\textbf{(D) }$ . Multiplying $33$ to the rest of the answer choices, the only one which fills this requirement is $75$ , which is $\boxed{\textbf{(E) }75}.$

~neeyakkid23

Video Solution (Simple & Quick)

https://youtu.be/9HI79V-vtCU

~ Education, the Study of Everything

Video Solution by Aaron He

https://www.youtube.com/watch?v=xTGDKBthWsw&t=4m12s

Video Solution (Use of Properties of Repeating Decimals)

https://www.youtube.com/watch?v=zS1u-ohUDzQ&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=6\

~North America Math Contest Go Go Go

Video Solution by Punxsutawney Phil

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

Note: This video is unavailable

Video Solution by Hawk Math

https://www.youtube.com/watch?v=P5al76DxyHY

Video Solution by OmegaLearn (Using Repeating Decimal Properties)

https://youtu.be/vQZ13WiL4WU

~ pi_is_3.14

Video Solution

https://youtu.be/DOF3FYUsXsU

~savannahsolver

Video Solution by TheBeautyofMath

https://youtu.be/s6E4E06XhPU?t=360 (AMC 10A)

https://youtu.be/rEWS75W0Q54?t=511 (AMC 12A)

~IceMatrix

Video Solution by The Learning Royal

https://youtu.be/AWjOeBFyeb4

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

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