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What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 
(1)/(17) ? Perform the division to check your answer.

What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 117 \frac{1}{17} ? Perform the division to check your answer.

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Q. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 117 \frac{1}{17} ? Perform the division to check your answer.
  1. Perform Long Division: To determine the maximum number of digits in the repeating block of the decimal expansion of 117\frac{1}{17}, we need to perform the long division of 11 by 1717.
  2. Determine Fits into 100100: We start the division by determining how many times 1717 fits into 100100 (since 1717 does not fit into 11, we add a decimal point and a zero to 11, making it 10.010.0, and then another zero, making it 100100).\newline1717 goes into 100100 a total of 55 times (10010000), leaving a remainder of 10010011.
  3. Calculate Remainder 1515: We bring down another zero to the remainder, making it 150150, and divide again by 1717. 1717 goes into 150150 a total of 88 times (17×8=13617 \times 8 = 136), leaving a remainder of 1414.
  4. Divide by 1717 Again: We bring down another zero to the remainder, making it 140140, and divide again by 1717. 1717 goes into 140140 a total of 88 times (17×8=13617 \times 8 = 136), leaving a remainder of 44.
  5. Repeat Remainders: We bring down another zero to the remainder, making it 4040, and divide again by 1717. 1717 goes into 4040 a total of 22 times (17×2=3417 \times 2 = 34), leaving a remainder of 66.
  6. Identify Repeating Block: We bring down another zero to the remainder, making it 6060, and divide again by 1717. 1717 goes into 6060 a total of 33 times (17×3=5117 \times 3 = 51), leaving a remainder of 99.
  7. Identify Repeating Block: We bring down another zero to the remainder, making it 6060, and divide again by 1717. 1717 goes into 6060 a total of 33 times (17×3=5117 \times 3 = 51), leaving a remainder of 99.We bring down another zero to the remainder, making it 9090, and divide again by 1717. 1717 goes into 9090 a total of 171711 times (171722), leaving a remainder of 171711.
  8. Identify Repeating Block: We bring down another zero to the remainder, making it 6060, and divide again by 1717. 1717 goes into 6060 a total of 33 times (17×3=5117 \times 3 = 51), leaving a remainder of 99.We bring down another zero to the remainder, making it 9090, and divide again by 1717. 1717 goes into 9090 a total of 171711 times (171722), leaving a remainder of 171711.We notice that we have seen the remainder 171711 before, which means the digits will start to repeat from this point. The remainders we have seen so far are 171755, 171766, 171777, 171788, and 99, which correspond to the digits 171711, 171711, 171711, 171733, and 33 in the decimal expansion.
  9. Identify Repeating Block: We bring down another zero to the remainder, making it 6060, and divide again by 1717. 1717 goes into 6060 a total of 33 times (17×3=5117 \times 3 = 51), leaving a remainder of 99.We bring down another zero to the remainder, making it 9090, and divide again by 1717. 1717 goes into 9090 a total of 171711 times (171722), leaving a remainder of 171711.We notice that we have seen the remainder 171711 before, which means the digits will start to repeat from this point. The remainders we have seen so far are 171755, 171766, 171777, 171788, and 99, which correspond to the digits 171711, 171711, 171711, 171733, and 33 in the decimal expansion.Since the remainders are starting to repeat, we have found the repeating block of the decimal expansion of 171755. The repeating block is 171766, which has 171777 digits.

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