When expanded as a decimal, the fraction 1/97 has a repetend (the repeating part of the decimal) that begins right after the decimal point and is 96 digits long. If the last three digits of the repetend are A67

, compute the digit A

To determine the digit \(A\) in the repetend of the decimal expansion of \(\frac{1}{97}\), let's analyze the repeating sequence.

We know that the decimal expansion of \(\frac{1}{97}\) is a repeating decimal with a period of 96 digits. The repetend can be represented as:
\[
0.\overline{d_1 d_2 \ldots d_{96}}
\]

Given that the last three digits of the repetend are \(A67\), we want to find the value of \(A\). This implies that the last three digits of the repetend must satisfy the relation:
\[
d_{94} = A, \quad d_{95} = 6, \quad d_{96} = 7
\]

We can solve this problem by examining the properties of the repetend in the decimal expansion of \(\frac{1}{97}\). We know:
\[
\frac{1}{97} = 0.\overline{d_1 d_2 \ldots d_{96}}
\]

Since the decimal expansion of \(\frac{1}{97}\) is repeating with a period of 96, the repetend satisfies:
\[
10^{96} \equiv 1 \pmod{97}
\]

To determine the last three digits of the repetend, we consider the equation:
\[
10^3 \equiv 1 \pmod{97}
\]
This means:
\[
10^3 = 1000 \equiv 29 \pmod{97}
\]

Thus, \(1000k \equiv 29k \pmod{97}\) for any integer \(k\). Since we are interested in the repetend, we need to find:
\[
10^{95} \times 10^2 \equiv 1 \pmod{97}
\]

Let's compute the position where the last three digits \(A67\) fit. It is easier to use modular arithmetic to identify the digits. By long division or using modular arithmetic properties, we can systematically determine the digit positions.

Continuing with the calculations, the decimal expansion of \(\frac{1}{97}\) proceeds as:
\[
1 / 97 = 0.010309278350515463917525773195876288659793814432989690721649484536082474226804123711340206185567
\]

Reading the last three digits from the repetend gives us the digits: \(567\), which confirms:
\[
d_{94} = 5, \quad d_{95} = 6, \quad d_{96} = 7
\]

Thus, the digit \(A\) is:
\[
A = 5
\]

Therefore, the value of \(A\) is:
\[
\boxed{5}
\]

Learn how to find the digit A in the repeating decimal expansion of 1/97 where the last three digits are A67. Explore the properties of modular arithmetic and decimal expansion to solve this advanced high school level problem.

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