To find one such number, you can find in decimal, and use the repeating block of digits.

Understand the Question Dividing a number by yields a non-terminating, recurring decimal expansion. By performing long division of by , we record each digit in the quotient until the remainder becomes again. When the remainder reappears, the entire sequence of quotient digits starts repeating, giving us the repeating block (period) of . Step 1 · Perform Long Division of by Perform long division of by : , remainder , remainder , remainder , remainder , remainder , remainder , remainder , remainder , remainder , remainder , remainder , remainder , remainder , remainder , remainder , remainder Since the remainder repeats the initial dividend, the decimal expansion repeats: Step 2 · Identify the Repeating Block of Digits The repeating block consists of the digits after the decimal point before the cycle repeats: Answer > Common Mistakes Missing the Leading Zero: Forgetting that , which places a directly after the decimal point before dividing by . Stopping Prematurely: Stopping the long division before getting the initial remainder back, leading to an incomplete repeating period (the maximum period for a divisor is digits).

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Question 15

To find one such number, you can find $1 \div 17$ in decimal, and use the repeating block of digits.

Solution

Understand the Question

Dividing a number $1$ by $17$ yields a non-terminating, recurring decimal expansion.
By performing long division of $1.000\dots$ by $17$ , we record each digit in the quotient until the remainder becomes $1$ again.
When the remainder $1$ reappears, the entire sequence of quotient digits starts repeating, giving us the repeating block (period) of $1 \div 17$ .

Step 1 · Perform Long Division of $1$ by $17$

Perform long division of $1.000\dots$ by $17$ : Diagram 1

$1 \times 10 = 10 \implies 10 \div 17 = 0$ , remainder $= 10$
$10 \times 10 = 100 \implies 100 \div 17 = 5$ , remainder $= 15$
$15 \times 10 = 150 \implies 150 \div 17 = 8$ , remainder $= 14$
$14 \times 10 = 140 \implies 140 \div 17 = 8$ , remainder $= 4$
$4 \times 10 = 40 \implies 40 \div 17 = 2$ , remainder $= 6$
$6 \times 10 = 60 \implies 60 \div 17 = 3$ , remainder $= 9$
$9 \times 10 = 90 \implies 90 \div 17 = 5$ , remainder $= 5$
$5 \times 10 = 50 \implies 50 \div 17 = 2$ , remainder $= 16$
$16 \times 10 = 160 \implies 160 \div 17 = 9$ , remainder $= 7$
$7 \times 10 = 70 \implies 70 \div 17 = 4$ , remainder $= 2$
$2 \times 10 = 20 \implies 20 \div 17 = 1$ , remainder $= 3$
$3 \times 10 = 30 \implies 30 \div 17 = 1$ , remainder $= 13$
$13 \times 10 = 130 \implies 130 \div 17 = 7$ , remainder $= 11$
$11 \times 10 = 110 \implies 110 \div 17 = 6$ , remainder $= 8$
$8 \times 10 = 80 \implies 80 \div 17 = 4$ , remainder $= 12$
$12 \times 10 = 120 \implies 120 \div 17 = 7$ , remainder $= 1$

Since the remainder $1$ repeats the initial dividend, the decimal expansion repeats:

$1 \div 17 = 0.\overline{0588235294117647}$

Step 2 · Identify the Repeating Block of Digits

The repeating block consists of the $16$ digits after the decimal point before the cycle repeats:

$\text{Repeating block} = 0588235294117647$

Answer

$0588235294117647$

Common Mistakes

Missing the Leading Zero: Forgetting that $10 < 17$ , which places a $0$ directly after the decimal point before dividing $100$ by $17$ .
Stopping Prematurely: Stopping the long division before getting the initial remainder $1$ back, leading to an incomplete repeating period (the maximum period for a divisor $17$ is $17 - 1 = 16$ digits).

Question 15

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