Troubling Repetend Problem

Here are two "Repetends" of 1/97=

1/97=0.0103092783 5051546391 7525773195 8762886597 9381443298 9690721649 4845360824 7422680412 3711340206 185567.......starts repeating here. 0103 0927835051 5463917525 7731958762 8865979381 4432989690 7216494845 3608247422 6804123711 3402061855 67.....etc.

I don't know if they expect you to " calculate" manually, or some other way.

Suppose that k = the 96 digit repetend, (as a whole number), then

1/97 = k.10^(-96) + k.10^(192) + k.10^(-288) + ...

        = k.10^(-96)(1 + 10^(-96) + 10^(-192) + ...)

        = k.10(-96).1/(1 - 10^(-96)),     (summing the GP),

        = k/(10^96 - 1)

        = k/ 9999....99999,

so

97k = 9999...99999 .

Carry out the long multiplication on the lhs, (matching up with the 9's on the rhs).

The last digit of k is given as a 7, so 97* 7 = 679 which gives us a 9 in the final column and a carry of 67.

The next to last digit of k is a 6 so 97*6 + 'the carry' = 582 + 67 = 649, which gives us the next 9 with a carry of 64.

The third digit from the end of k, denoted by A in the question, has the requirement that the result of 97*A + 64 should end with a 9. That means that 97*A has to end which a 5 which in turn means that A = 5.

- Bertie

\(\begin{array}{|rrcrrll|} \hline 1 &:97 &=& 0. &\text{ remainder }& 1 \\ 1\cdot 10 &:97 &=& 0 &\text{ remainder }& 10 \\ 10\cdot 10 &:97 &=& 1 &\text{ remainder }& 3 \\ 3\cdot 10 &:97 &=& 0 &\text{ remainder }& 30 \\ 30\cdot 10 &:97 &=& 3 &\text{ remainder }& 9 \\ 9\cdot 10 &:97 &=& 0 &\text{ remainder }& 90 \\ 90\cdot 10 &:97 &=& 9 &\text{ remainder }& 27 \\ 27\cdot 10 &:97 &=& 2 &\text{ remainder }& 76 \\ \dots\\ z\cdot 10 &:97 &=& A &\text{ remainder }& y \\ y\cdot 10 &:97 &=& 6 & \text{ remainder }& x \\ x\cdot 10 &:97 &=& 7 & \text{ remainder }& 1 & \qquad \text{ end of repeating part remainder } = 1\\ \hline \end{array} \)

We solve A:

We have:

\(\begin{array}{|rcl|} \hline \mathbf{x\cdot 10 }&\mathbf{:97} &\mathbf{=}& \mathbf{7} & \mathbf{\text{ remainder }}& \mathbf{1} \\ \text{ or }\\ x\cdot 10&&=& 7\cdot 97 & +& 1\\\\ x &=& \frac{ 7\cdot 97+1}{10} \\ x &=& \frac{ 680}{10} \\ \mathbf{x} & \mathbf{=}& \mathbf{68}\\ \hline \end{array}\)

...and we have:

\(\begin{array}{|rcl|} \hline \mathbf{y\cdot 10 }&\mathbf{:97} &\mathbf{=}& \mathbf{6} & \mathbf{\text{ remainder }}& \mathbf{x} \\ \text{ or }\\ y\cdot 10&&=& 6\cdot 97 & +& 68\\\\ y &=& \frac{ 6\cdot 97+68}{10} \\ y &=& \frac{ 650}{10} \\ \mathbf{y} & \mathbf{=}& \mathbf{65}\\ \hline \end{array}\)

...and...

\(\begin{array}{|rcl|} \hline \mathbf{z\cdot 10 }&\mathbf{:97} &\mathbf{=}& \mathbf{A} & \mathbf{\text{ remainder }}& \mathbf{y} \\ \text{ or }\\ z\cdot 10&&=& A\cdot 97 & +& 65\\\\ \hline \end{array}\)

We have to solve the Diophantine equation  \(10z -97A = 65.\)

We can do it with a variation of the Euclidean algorithm, or we can do it with Euler's Method etc.

I will do it with Euler's theorem:

\(\begin{array}{l} \text{In number theory, }\\ \text{Euler's theorem (also known as the Fermat–Euler theorem or Euler's totient theorem)}\\ \text{states that if } \mathbf{m} \text{ and } \mathbf{a} \text{ are coprime positive integers, then }\\ \hline a^{\varphi (m)} \equiv 1 \pmod{m} \qquad \text{, if }~ gcd(a,m)=1 \\\\ \text{or} \\ \\ a^{\varphi (m)} \pmod{m} \equiv 1 \qquad | \qquad \cdot \frac{1}{a}\\ \frac{ a^{\varphi (m)} } {a} \pmod{m} \equiv \frac{1}{a} = a^{-1}\\ a^{\varphi (m)-1} \pmod{m} \equiv \frac{1}{a} = a^{-1}\\ \begin{array}{|rcl|} \hline a^{-1} = \frac{1}{a} \equiv a^{\varphi (m)-1} \pmod{m} \qquad \text{, if }~ gcd(a,m)=1\\ \hline \end{array} \end{array} \)

We need this Formula: \( \begin{array}{|rcl|} \hline a^{-1} = \frac{1}{a} \equiv a^{\varphi (m)-1} \pmod{m} \qquad \text{, if }~ gcd(a,m)=1\\ \hline \end{array}\)

an we need the Euler's Totient Function: \(\varphi(10) = 10\cdot (1- \frac{1}{2} )\cdot (1- \frac{1}{5} ) = 4, \text{ because } ~ 10 = 2\cdot 5\)

so we have:

\(\begin{array}{rcl} 10z -97A &=& 65\\\\ 97A &=& 10\cdot z - 65 \qquad | \qquad \pmod{10} \\ 97A &\equiv& -65 \pmod{10} \\ 97A &\equiv& -65 +7\cdot 10 \pmod{10} \\ \mathbf{97A} & \mathbf{\equiv}& \mathbf{5 \pmod{10}}\\\\ 97\cdot A\cdot \frac{1}{97} & \equiv & \frac{5}{97} \pmod{10}\\ A & \equiv & 5\cdot \frac{1}{97} \pmod{10}\\\\ \frac{1}{97} \pmod{10} &\equiv& 97^{\varphi(10)-1} \pmod{10} \qquad \text{ see Formula above,}\\ &&\qquad \text{we can do it because, } ~gcd(97,10)=1\\ &\equiv& 97^{4-1} \pmod{10}\\ &\equiv& 97^3 \pmod{10}\\ &\equiv& 912673 \pmod{10}\\ &\equiv& 3 \pmod{10}\\\\ A & \equiv & 5\cdot \frac{1}{97} \pmod{10}\\\\ A & \equiv & 5\cdot 3 \pmod{10}\\ A & \equiv & 15 \pmod{10}\\ A & \equiv & 5 \pmod{10}\\ \mathbf{A} & \mathbf{=} & \mathbf{5} \end{array}\)