AMC 10 Step by Step

2021 AMC 10A Problem 8

P1-10~3 minPrint

When a student multiplied the number 6666 by the repeating decimal, 1‾.a‾ b‾ a‾ b‾…=1‾.a‾ b‾‾,\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}}, where aa and bb are digits, he did not notice the notation and just multiplied 6666 times 1‾.a‾ b‾.\underline{1}.\underline{a} \ \underline{b}. Later he found that his answer is 0.50.5 less than the correct answer. What is the 22 -digit number a‾ b‾?\underline{a} \ \underline{b}?

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Key insight

The student only lost the tail 0.00ababab... Multiplying that lost part by 66 must give exactly 0.5.

Solution

The student's answer came out too small because he used a smaller number. So look at exactly what he left out.

1. The correct number is 1.ababab…1.ababab\ldots and he used 1.ab1.ab. The difference is the tail he dropped: 0.00ababab…0.00ababab\ldots
For example, if abab were 1212: 1.121212…−1.12=0.001212…1.121212\ldots - 1.12 = 0.001212\ldots
2. Let NN be the two-digit number a‾ b‾\underline{a}\,\underline{b}. A known fact about repeating decimals: 0.ababab…=N990.ababab\ldots = \frac{N}{99}. (For example, 0.121212…=12990.121212\ldots = \frac{12}{99}.)
3. The dropped tail 0.00abab…0.00abab\ldots is that same decimal moved two places to the right, which divides it by 100100. So the tail is N9900\frac{N}{9900}.
4. Multiplying by 6666, his answer was short by
66×N9900=N150.66 \times \frac{N}{9900} = \frac{N}{150}.
5. This shortfall is 0.50.5, so N150=12\frac{N}{150} = \frac{1}{2} and N=75N = 75.
6. Check: 66×1.75‾=66×(1+7599)=66+50=11666 \times 1.\overline{75} = 66 \times \left(1 + \frac{75}{99}\right) = 66 + 50 = 116, while 66×1.75=115.566 \times 1.75 = 115.5. The difference is 0.50.5.

The answer is (E) 75\boxed{\textbf{(E)}\ 75}.

Why this works

Subtracting the wrong product from the right one leaves only the piece of the number that was ignored, so you never need aa and bb separately. A two-digit repeating block is N99\frac{N}{99}, and every place it is shifted to the right divides it by another 1010.

Alternative approach

Test the choices using the shortfall N150\frac{N}{150} from step 4: N=45N = 45 gives 0.30.3, N=60N = 60 gives 0.40.4, and N=75N = 75 gives 0.50.5. Only 7575 works.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Let N=a‾ b‾N = \underline{a}\,\underline{b} be the two-digit number. The correct factor is 1.ab‾1.\overline{ab} and the student used 1.ab1.ab; their difference is the part he dropped:
1.ab‾−1.ab=0.00ab‾=1100⋅0.ab‾=N9900.1.\overline{ab} - 1.ab = 0.00\overline{ab} = \frac{1}{100}\cdot 0.\overline{ab} = \frac{N}{9900}.

Multiplying by 6666 changes the product by 66⋅N9900=N15066 \cdot \frac{N}{9900} = \frac{N}{150}, and this must equal 0.50.5:
N150=12⟹N=75.\frac{N}{150} = \frac{1}{2} \quad\Longrightarrow\quad N = 75.

The answer is (E) 75\boxed{\textbf{(E)}\ 75}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.