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2021 AMC 10A Problem 8

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2021 AMC 10A Problem 8AMC 10 Step by Step
Name:                             Date: September 13, 2026
Fractions & Decimals · difficulty 2 of 5 · about 3 min · position P1-10

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}?

  1. A)

    1515

  2. B)

    3030

  3. C)

    4545

  4. D)

    6060

  5. E)

    7575

Answer

E

Key insight

The error is the tail 0.00ababab..., which equals ab/9900; so 66 * ab/9900 = 0.5 gives ab = 75.

Solution

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}.

Why this works

Comparing the wrong product to the right one isolates the piece of the number the student ignored, so you never need aa and bb separately. A two-digit repeating block is N99\frac{N}{99}, and each place it is shifted right divides by 1010; here two places give N9900\frac{N}{9900}.

Alternative approach

Test choices: the difference is 66N9900=N150\frac{66N}{9900} = \frac{N}{150}. Only N=75N = 75 gives 0.50.5; for instance N=60N = 60 gives 0.40.4 and N=45N = 45 gives 0.30.3.

The trap

Writing the difference as ab/99 (forgetting the two-place shift), which gives ab = 0.75 and no valid digits.

Common mistakes

  • Writing the difference as ab/99 (forgetting the two-place shift), which gives ab = 0.75 and no valid digits.
  • Trying to compute 66⋅1.ab‾66 \cdot 1.\overline{ab} and 66⋅1.ab66 \cdot 1.ab separately with unknown digits instead of subtracting first.

Techniques

Set up the equation/formula and compute; no special trick needed