← A far-off digit is decided by a remainder, not by dividing further · Repeating Decimal Period

A far-off digit is decided by a remainder, not by dividing further · 12 practice problems

7.NS.A.26.NS.B.3

Generated variants — 12

Freshly produced from the archetype’s parameters — problem, figure, and solution derived together.

Variant 1 easy answer: 1

The numerator of an improper fraction was divided by its denominator, $7$ , giving a quotient of $1$ and a remainder of $1$ . When this fraction is written as a decimal, what is the digit in the $13$ th decimal place?

Show solution

1 · Understandwhat's really being asked

An improper fraction over 7 leaves quotient 1 and remainder 1. Written as a decimal, we want the digit in the 13th place.

Givens

The denominator is 7.
Dividing gives a quotient of 1 and a remainder of 1.
The digit wanted is in the 13th decimal place.

Unknowns

The 13th digit after the decimal point.

Constraints

The fraction is improper, so the numerator is bigger than the denominator.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #11 Work Backwards#9 Solve an Easier Related Problem

Dividing 13 places by hand is not the plan. Divide far enough to see the block that repeats, then use a second division -- the place number by the block's length -- to pick the digit.

3 · Execute4 carry out the plan

1Rebuild the fraction

#11 Work Backwards 6.NS.B.3

Quotient 1 remainder 1 over 7 means the numerator is 7 plus 1.

7 \times 1 + 1 = 8

The fraction is 8 over 7.

2Divide until it repeats

#5 Look for a Pattern 7.NS.A.2

The digits after the point start 142857, and the block 142857 comes round again.

8 \div 7 = 1.142857\ldots

A block of 6 digits repeats forever.

3Divide the place by the block's length

#9 Solve an Easier Related Problem 7.NS.A.2

13 over 6 leaves a remainder of 1, and the remainder picks the digit -- a remainder of 0 means the last digit of the block, not the first.

13 \div 6 \rightarrow \text{remainder }1

It is digit number 1 of the block.

4Read that digit off

#5 Look for a Pattern 7.NS.A.2

Digit 1 of 142857 is 1.

1

The 13th decimal place holds 1.

Answer: 1

4 · Reviewdoes it hold up?

Counting out the first 12 digits by hand and stepping through the repeating block lands on the same 1, and the block never changes however far it runs.

Another way: Long division carried out 13 places would also give 1, at the cost of 13 steps instead of 7.

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading a fraction as a repeating decimal and using its period.
6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Rebuilding the numerator from the quotient and remainder.

💡Takeaway. When digits repeat, you never have to go all the way. Find the block, then find where your place lands inside it.

Variant 2 easy answer: 0

The numerator of an improper fraction was divided by its denominator, $11$ , giving a quotient of $1$ and a remainder of $10$ . When this fraction is written as a decimal, what is the digit in the $26$ th decimal place?

Show solution

1 · Understandwhat's really being asked

An improper fraction over 11 leaves quotient 1 and remainder 10. Written as a decimal, we want the digit in the 26th place.

Givens

The denominator is 11.
Dividing gives a quotient of 1 and a remainder of 10.
The digit wanted is in the 26th decimal place.

Unknowns

The 26th digit after the decimal point.

Constraints

The fraction is improper, so the numerator is bigger than the denominator.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #11 Work Backwards#9 Solve an Easier Related Problem

Dividing 26 places by hand is not the plan. Divide far enough to see the block that repeats, then use a second division -- the place number by the block's length -- to pick the digit.

3 · Execute4 carry out the plan

1Rebuild the fraction

#11 Work Backwards 6.NS.B.3

Quotient 1 remainder 10 over 11 means the numerator is 11 plus 10.

11 \times 1 + 10 = 21

The fraction is 21 over 11.

2Divide until it repeats

#5 Look for a Pattern 7.NS.A.2

The digits after the point start 90, and the block 90 comes round again.

21 \div 11 = 1.90\ldots

A block of 2 digits repeats forever.

3Divide the place by the block's length

#9 Solve an Easier Related Problem 7.NS.A.2

26 over 2 leaves a remainder of 0, and the remainder picks the digit -- a remainder of 0 means the last digit of the block, not the first.

26 \div 2 \rightarrow \text{remainder }0

It is digit number 2 of the block.

4Read that digit off

#5 Look for a Pattern 7.NS.A.2

Digit 2 of 90 is 0.

0

The 26th decimal place holds 0.

Answer: 0

4 · Reviewdoes it hold up?

Counting out the first 12 digits by hand and stepping through the repeating block lands on the same 0, and the block never changes however far it runs.

Another way: Long division carried out 26 places would also give 0, at the cost of 26 steps instead of 3.

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading a fraction as a repeating decimal and using its period.
6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Rebuilding the numerator from the quotient and remainder.

💡Takeaway. When digits repeat, you never have to go all the way. Find the block, then find where your place lands inside it.

Variant 3 easy answer: 1

The numerator of an improper fraction was divided by its denominator, $27$ , giving a quotient of $1$ and a remainder of $13$ . When this fraction is written as a decimal, what is the digit in the $18$ th decimal place?

Show solution

1 · Understandwhat's really being asked

An improper fraction over 27 leaves quotient 1 and remainder 13. Written as a decimal, we want the digit in the 18th place.

Givens

The denominator is 27.
Dividing gives a quotient of 1 and a remainder of 13.
The digit wanted is in the 18th decimal place.

Unknowns

The 18th digit after the decimal point.

Constraints

The fraction is improper, so the numerator is bigger than the denominator.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #11 Work Backwards#9 Solve an Easier Related Problem

Dividing 18 places by hand is not the plan. Divide far enough to see the block that repeats, then use a second division -- the place number by the block's length -- to pick the digit.

3 · Execute4 carry out the plan

1Rebuild the fraction

#11 Work Backwards 6.NS.B.3

Quotient 1 remainder 13 over 27 means the numerator is 27 plus 13.

27 \times 1 + 13 = 40

The fraction is 40 over 27.

2Divide until it repeats

#5 Look for a Pattern 7.NS.A.2

The digits after the point start 481, and the block 481 comes round again.

40 \div 27 = 1.481\ldots

A block of 3 digits repeats forever.

3Divide the place by the block's length

#9 Solve an Easier Related Problem 7.NS.A.2

18 over 3 leaves a remainder of 0, and the remainder picks the digit -- a remainder of 0 means the last digit of the block, not the first.

18 \div 3 \rightarrow \text{remainder }0

It is digit number 3 of the block.

4Read that digit off

#5 Look for a Pattern 7.NS.A.2

Digit 3 of 481 is 1.

1

The 18th decimal place holds 1.

Answer: 1

4 · Reviewdoes it hold up?

Counting out the first 12 digits by hand and stepping through the repeating block lands on the same 1, and the block never changes however far it runs.

Another way: Long division carried out 18 places would also give 1, at the cost of 18 steps instead of 4.

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading a fraction as a repeating decimal and using its period.
6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Rebuilding the numerator from the quotient and remainder.

💡Takeaway. When digits repeat, you never have to go all the way. Find the block, then find where your place lands inside it.

Variant 4 easy answer: 1

The numerator of an improper fraction was divided by its denominator, $27$ , giving a quotient of $1$ and a remainder of $14$ . When this fraction is written as a decimal, what is the digit in the $20$ th decimal place?

Show solution

1 · Understandwhat's really being asked

An improper fraction over 27 leaves quotient 1 and remainder 14. Written as a decimal, we want the digit in the 20th place.

Givens

The denominator is 27.
Dividing gives a quotient of 1 and a remainder of 14.
The digit wanted is in the 20th decimal place.

Unknowns

The 20th digit after the decimal point.

Constraints

The fraction is improper, so the numerator is bigger than the denominator.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #11 Work Backwards#9 Solve an Easier Related Problem

Dividing 20 places by hand is not the plan. Divide far enough to see the block that repeats, then use a second division -- the place number by the block's length -- to pick the digit.

3 · Execute4 carry out the plan

1Rebuild the fraction

#11 Work Backwards 6.NS.B.3

Quotient 1 remainder 14 over 27 means the numerator is 27 plus 14.

27 \times 1 + 14 = 41

The fraction is 41 over 27.

2Divide until it repeats

#5 Look for a Pattern 7.NS.A.2

The digits after the point start 518, and the block 518 comes round again.

41 \div 27 = 1.518\ldots

A block of 3 digits repeats forever.

3Divide the place by the block's length

#9 Solve an Easier Related Problem 7.NS.A.2

20 over 3 leaves a remainder of 2, and the remainder picks the digit -- a remainder of 0 means the last digit of the block, not the first.

20 \div 3 \rightarrow \text{remainder }2

It is digit number 2 of the block.

4Read that digit off

#5 Look for a Pattern 7.NS.A.2

Digit 2 of 518 is 1.

1

The 20th decimal place holds 1.

Answer: 1

4 · Reviewdoes it hold up?

Counting out the first 12 digits by hand and stepping through the repeating block lands on the same 1, and the block never changes however far it runs.

Another way: Long division carried out 20 places would also give 1, at the cost of 20 steps instead of 4.

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading a fraction as a repeating decimal and using its period.
6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Rebuilding the numerator from the quotient and remainder.

💡Takeaway. When digits repeat, you never have to go all the way. Find the block, then find where your place lands inside it.

Variant 5 medium answer: 7

The numerator of an improper fraction was divided by its denominator, $7$ , giving a quotient of $1$ and a remainder of $6$ . When this fraction is written as a decimal, what is the digit in the $27$ th decimal place?

Show solution

1 · Understandwhat's really being asked

An improper fraction over 7 leaves quotient 1 and remainder 6. Written as a decimal, we want the digit in the 27th place.

Givens

The denominator is 7.
Dividing gives a quotient of 1 and a remainder of 6.
The digit wanted is in the 27th decimal place.

Unknowns

The 27th digit after the decimal point.

Constraints

The fraction is improper, so the numerator is bigger than the denominator.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #11 Work Backwards#9 Solve an Easier Related Problem

Dividing 27 places by hand is not the plan. Divide far enough to see the block that repeats, then use a second division -- the place number by the block's length -- to pick the digit.

3 · Execute4 carry out the plan

1Rebuild the fraction

#11 Work Backwards 6.NS.B.3

Quotient 1 remainder 6 over 7 means the numerator is 7 plus 6.

7 \times 1 + 6 = 13

The fraction is 13 over 7.

2Divide until it repeats

#5 Look for a Pattern 7.NS.A.2

The digits after the point start 857142, and the block 857142 comes round again.

13 \div 7 = 1.857142\ldots

A block of 6 digits repeats forever.

3Divide the place by the block's length

#9 Solve an Easier Related Problem 7.NS.A.2

27 over 6 leaves a remainder of 3, and the remainder picks the digit -- a remainder of 0 means the last digit of the block, not the first.

27 \div 6 \rightarrow \text{remainder }3

It is digit number 3 of the block.

4Read that digit off

#5 Look for a Pattern 7.NS.A.2

Digit 3 of 857142 is 7.

7

The 27th decimal place holds 7.

Answer: 7

4 · Reviewdoes it hold up?

Counting out the first 12 digits by hand and stepping through the repeating block lands on the same 7, and the block never changes however far it runs.

Another way: Long division carried out 27 places would also give 7, at the cost of 27 steps instead of 7.

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading a fraction as a repeating decimal and using its period.
6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Rebuilding the numerator from the quotient and remainder.

💡Takeaway. When digits repeat, you never have to go all the way. Find the block, then find where your place lands inside it.

Variant 6 medium answer: 0

The numerator of an improper fraction was divided by its denominator, $13$ , giving a quotient of $1$ and a remainder of $3$ . When this fraction is written as a decimal, what is the digit in the $33$ th decimal place?

Show solution

1 · Understandwhat's really being asked

An improper fraction over 13 leaves quotient 1 and remainder 3. Written as a decimal, we want the digit in the 33th place.

Givens

The denominator is 13.
Dividing gives a quotient of 1 and a remainder of 3.
The digit wanted is in the 33th decimal place.

Unknowns

The 33th digit after the decimal point.

Constraints

The fraction is improper, so the numerator is bigger than the denominator.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #11 Work Backwards#9 Solve an Easier Related Problem

Dividing 33 places by hand is not the plan. Divide far enough to see the block that repeats, then use a second division -- the place number by the block's length -- to pick the digit.

3 · Execute4 carry out the plan

1Rebuild the fraction

#11 Work Backwards 6.NS.B.3

Quotient 1 remainder 3 over 13 means the numerator is 13 plus 3.

13 \times 1 + 3 = 16

The fraction is 16 over 13.

2Divide until it repeats

#5 Look for a Pattern 7.NS.A.2

The digits after the point start 230769, and the block 230769 comes round again.

16 \div 13 = 1.230769\ldots

A block of 6 digits repeats forever.

3Divide the place by the block's length

#9 Solve an Easier Related Problem 7.NS.A.2

33 over 6 leaves a remainder of 3, and the remainder picks the digit -- a remainder of 0 means the last digit of the block, not the first.

33 \div 6 \rightarrow \text{remainder }3

It is digit number 3 of the block.

4Read that digit off

#5 Look for a Pattern 7.NS.A.2

Digit 3 of 230769 is 0.

0

The 33th decimal place holds 0.

Answer: 0

4 · Reviewdoes it hold up?

Counting out the first 12 digits by hand and stepping through the repeating block lands on the same 0, and the block never changes however far it runs.

Another way: Long division carried out 33 places would also give 0, at the cost of 33 steps instead of 7.

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading a fraction as a repeating decimal and using its period.
6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Rebuilding the numerator from the quotient and remainder.

💡Takeaway. When digits repeat, you never have to go all the way. Find the block, then find where your place lands inside it.

Variant 7 medium answer: 7

The numerator of an improper fraction was divided by its denominator, $11$ , giving a quotient of $1$ and a remainder of $8$ . When this fraction is written as a decimal, what is the digit in the $37$ th decimal place?

Show solution

1 · Understandwhat's really being asked

An improper fraction over 11 leaves quotient 1 and remainder 8. Written as a decimal, we want the digit in the 37th place.

Givens

The denominator is 11.
Dividing gives a quotient of 1 and a remainder of 8.
The digit wanted is in the 37th decimal place.

Unknowns

The 37th digit after the decimal point.

Constraints

The fraction is improper, so the numerator is bigger than the denominator.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #11 Work Backwards#9 Solve an Easier Related Problem

Dividing 37 places by hand is not the plan. Divide far enough to see the block that repeats, then use a second division -- the place number by the block's length -- to pick the digit.

3 · Execute4 carry out the plan

1Rebuild the fraction

#11 Work Backwards 6.NS.B.3

Quotient 1 remainder 8 over 11 means the numerator is 11 plus 8.

11 \times 1 + 8 = 19

The fraction is 19 over 11.

2Divide until it repeats

#5 Look for a Pattern 7.NS.A.2

The digits after the point start 72, and the block 72 comes round again.

19 \div 11 = 1.72\ldots

A block of 2 digits repeats forever.

3Divide the place by the block's length

#9 Solve an Easier Related Problem 7.NS.A.2

37 over 2 leaves a remainder of 1, and the remainder picks the digit -- a remainder of 0 means the last digit of the block, not the first.

37 \div 2 \rightarrow \text{remainder }1

It is digit number 1 of the block.

4Read that digit off

#5 Look for a Pattern 7.NS.A.2

Digit 1 of 72 is 7.

7

The 37th decimal place holds 7.

Answer: 7

4 · Reviewdoes it hold up?

Counting out the first 12 digits by hand and stepping through the repeating block lands on the same 7, and the block never changes however far it runs.

Another way: Long division carried out 37 places would also give 7, at the cost of 37 steps instead of 3.

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading a fraction as a repeating decimal and using its period.
6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Rebuilding the numerator from the quotient and remainder.

💡Takeaway. When digits repeat, you never have to go all the way. Find the block, then find where your place lands inside it.

Variant 8 medium answer: 1

The numerator of an improper fraction was divided by its denominator, $37$ , giving a quotient of $1$ and a remainder of $3$ . When this fraction is written as a decimal, what is the digit in the $24$ th decimal place?

Show solution

1 · Understandwhat's really being asked

An improper fraction over 37 leaves quotient 1 and remainder 3. Written as a decimal, we want the digit in the 24th place.

Givens

The denominator is 37.
Dividing gives a quotient of 1 and a remainder of 3.
The digit wanted is in the 24th decimal place.

Unknowns

The 24th digit after the decimal point.

Constraints

The fraction is improper, so the numerator is bigger than the denominator.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #11 Work Backwards#9 Solve an Easier Related Problem

Dividing 24 places by hand is not the plan. Divide far enough to see the block that repeats, then use a second division -- the place number by the block's length -- to pick the digit.

3 · Execute4 carry out the plan

1Rebuild the fraction

#11 Work Backwards 6.NS.B.3

Quotient 1 remainder 3 over 37 means the numerator is 37 plus 3.

37 \times 1 + 3 = 40

The fraction is 40 over 37.

2Divide until it repeats

#5 Look for a Pattern 7.NS.A.2

The digits after the point start 081, and the block 081 comes round again.

40 \div 37 = 1.081\ldots

A block of 3 digits repeats forever.

3Divide the place by the block's length

#9 Solve an Easier Related Problem 7.NS.A.2

24 over 3 leaves a remainder of 0, and the remainder picks the digit -- a remainder of 0 means the last digit of the block, not the first.

24 \div 3 \rightarrow \text{remainder }0

It is digit number 3 of the block.

4Read that digit off

#5 Look for a Pattern 7.NS.A.2

Digit 3 of 081 is 1.

1

The 24th decimal place holds 1.

Answer: 1

4 · Reviewdoes it hold up?

Counting out the first 12 digits by hand and stepping through the repeating block lands on the same 1, and the block never changes however far it runs.

Another way: Long division carried out 24 places would also give 1, at the cost of 24 steps instead of 4.

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading a fraction as a repeating decimal and using its period.
6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Rebuilding the numerator from the quotient and remainder.

💡Takeaway. When digits repeat, you never have to go all the way. Find the block, then find where your place lands inside it.

Variant 9 hard answer: 8

The numerator of an improper fraction was divided by its denominator, $27$ , giving a quotient of $1$ and a remainder of $5$ . When this fraction is written as a decimal, what is the digit in the $38$ th decimal place?

Show solution

1 · Understandwhat's really being asked

An improper fraction over 27 leaves quotient 1 and remainder 5. Written as a decimal, we want the digit in the 38th place.

Givens

The denominator is 27.
Dividing gives a quotient of 1 and a remainder of 5.
The digit wanted is in the 38th decimal place.

Unknowns

The 38th digit after the decimal point.

Constraints

The fraction is improper, so the numerator is bigger than the denominator.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #11 Work Backwards#9 Solve an Easier Related Problem

Dividing 38 places by hand is not the plan. Divide far enough to see the block that repeats, then use a second division -- the place number by the block's length -- to pick the digit.

3 · Execute4 carry out the plan

1Rebuild the fraction

#11 Work Backwards 6.NS.B.3

Quotient 1 remainder 5 over 27 means the numerator is 27 plus 5.

27 \times 1 + 5 = 32

The fraction is 32 over 27.

2Divide until it repeats

#5 Look for a Pattern 7.NS.A.2

The digits after the point start 185, and the block 185 comes round again.

32 \div 27 = 1.185\ldots

A block of 3 digits repeats forever.

3Divide the place by the block's length

#9 Solve an Easier Related Problem 7.NS.A.2

38 over 3 leaves a remainder of 2, and the remainder picks the digit -- a remainder of 0 means the last digit of the block, not the first.

38 \div 3 \rightarrow \text{remainder }2

It is digit number 2 of the block.

4Read that digit off

#5 Look for a Pattern 7.NS.A.2

Digit 2 of 185 is 8.

8

The 38th decimal place holds 8.

Answer: 8

4 · Reviewdoes it hold up?

Counting out the first 12 digits by hand and stepping through the repeating block lands on the same 8, and the block never changes however far it runs.

Another way: Long division carried out 38 places would also give 8, at the cost of 38 steps instead of 4.

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading a fraction as a repeating decimal and using its period.
6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Rebuilding the numerator from the quotient and remainder.

💡Takeaway. When digits repeat, you never have to go all the way. Find the block, then find where your place lands inside it.

Variant 10 hard answer: 3

The numerator of an improper fraction was divided by its denominator, $13$ , giving a quotient of $1$ and a remainder of $9$ . When this fraction is written as a decimal, what is the digit in the $40$ th decimal place?

Show solution

1 · Understandwhat's really being asked

An improper fraction over 13 leaves quotient 1 and remainder 9. Written as a decimal, we want the digit in the 40th place.

Givens

The denominator is 13.
Dividing gives a quotient of 1 and a remainder of 9.
The digit wanted is in the 40th decimal place.

Unknowns

The 40th digit after the decimal point.

Constraints

The fraction is improper, so the numerator is bigger than the denominator.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #11 Work Backwards#9 Solve an Easier Related Problem

Dividing 40 places by hand is not the plan. Divide far enough to see the block that repeats, then use a second division -- the place number by the block's length -- to pick the digit.

3 · Execute4 carry out the plan

1Rebuild the fraction

#11 Work Backwards 6.NS.B.3

Quotient 1 remainder 9 over 13 means the numerator is 13 plus 9.

13 \times 1 + 9 = 22

The fraction is 22 over 13.

2Divide until it repeats

#5 Look for a Pattern 7.NS.A.2

The digits after the point start 692307, and the block 692307 comes round again.

22 \div 13 = 1.692307\ldots

A block of 6 digits repeats forever.

3Divide the place by the block's length

#9 Solve an Easier Related Problem 7.NS.A.2

40 over 6 leaves a remainder of 4, and the remainder picks the digit -- a remainder of 0 means the last digit of the block, not the first.

40 \div 6 \rightarrow \text{remainder }4

It is digit number 4 of the block.

4Read that digit off

#5 Look for a Pattern 7.NS.A.2

Digit 4 of 692307 is 3.

3

The 40th decimal place holds 3.

Answer: 3

4 · Reviewdoes it hold up?

Counting out the first 12 digits by hand and stepping through the repeating block lands on the same 3, and the block never changes however far it runs.

Another way: Long division carried out 40 places would also give 3, at the cost of 40 steps instead of 7.

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading a fraction as a repeating decimal and using its period.
6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Rebuilding the numerator from the quotient and remainder.

💡Takeaway. When digits repeat, you never have to go all the way. Find the block, then find where your place lands inside it.

Variant 11 hard answer: 9

The numerator of an improper fraction was divided by its denominator, $41$ , giving a quotient of $1$ and a remainder of $5$ . When this fraction is written as a decimal, what is the digit in the $19$ th decimal place?

Show solution

1 · Understandwhat's really being asked

An improper fraction over 41 leaves quotient 1 and remainder 5. Written as a decimal, we want the digit in the 19th place.

Givens

The denominator is 41.
Dividing gives a quotient of 1 and a remainder of 5.
The digit wanted is in the 19th decimal place.

Unknowns

The 19th digit after the decimal point.

Constraints

The fraction is improper, so the numerator is bigger than the denominator.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #11 Work Backwards#9 Solve an Easier Related Problem

Dividing 19 places by hand is not the plan. Divide far enough to see the block that repeats, then use a second division -- the place number by the block's length -- to pick the digit.

3 · Execute4 carry out the plan

1Rebuild the fraction

#11 Work Backwards 6.NS.B.3

Quotient 1 remainder 5 over 41 means the numerator is 41 plus 5.

41 \times 1 + 5 = 46

The fraction is 46 over 41.

2Divide until it repeats

#5 Look for a Pattern 7.NS.A.2

The digits after the point start 12195, and the block 12195 comes round again.

46 \div 41 = 1.12195\ldots

A block of 5 digits repeats forever.

3Divide the place by the block's length

#9 Solve an Easier Related Problem 7.NS.A.2

19 over 5 leaves a remainder of 4, and the remainder picks the digit -- a remainder of 0 means the last digit of the block, not the first.

19 \div 5 \rightarrow \text{remainder }4

It is digit number 4 of the block.

4Read that digit off

#5 Look for a Pattern 7.NS.A.2

Digit 4 of 12195 is 9.

9

The 19th decimal place holds 9.

Answer: 9

4 · Reviewdoes it hold up?

Counting out the first 12 digits by hand and stepping through the repeating block lands on the same 9, and the block never changes however far it runs.

Another way: Long division carried out 19 places would also give 9, at the cost of 19 steps instead of 6.

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading a fraction as a repeating decimal and using its period.
6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Rebuilding the numerator from the quotient and remainder.

💡Takeaway. When digits repeat, you never have to go all the way. Find the block, then find where your place lands inside it.

Variant 12 hard answer: 2

The numerator of an improper fraction was divided by its denominator, $101$ , giving a quotient of $1$ and a remainder of $63$ . When this fraction is written as a decimal, what is the digit in the $38$ th decimal place?

Show solution

1 · Understandwhat's really being asked

An improper fraction over 101 leaves quotient 1 and remainder 63. Written as a decimal, we want the digit in the 38th place.

Givens

The denominator is 101.
Dividing gives a quotient of 1 and a remainder of 63.
The digit wanted is in the 38th decimal place.

Unknowns

The 38th digit after the decimal point.

Constraints

The fraction is improper, so the numerator is bigger than the denominator.

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #11 Work Backwards#9 Solve an Easier Related Problem

Dividing 38 places by hand is not the plan. Divide far enough to see the block that repeats, then use a second division -- the place number by the block's length -- to pick the digit.

3 · Execute4 carry out the plan

1Rebuild the fraction

#11 Work Backwards 6.NS.B.3

Quotient 1 remainder 63 over 101 means the numerator is 101 plus 63.

101 \times 1 + 63 = 164

The fraction is 164 over 101.

2Divide until it repeats

#5 Look for a Pattern 7.NS.A.2

The digits after the point start 6237, and the block 6237 comes round again.

164 \div 101 = 1.6237\ldots

A block of 4 digits repeats forever.

3Divide the place by the block's length

#9 Solve an Easier Related Problem 7.NS.A.2

38 over 4 leaves a remainder of 2, and the remainder picks the digit -- a remainder of 0 means the last digit of the block, not the first.

38 \div 4 \rightarrow \text{remainder }2

It is digit number 2 of the block.

4Read that digit off

#5 Look for a Pattern 7.NS.A.2

Digit 2 of 6237 is 2.

2

The 38th decimal place holds 2.

Answer: 2

4 · Reviewdoes it hold up?

Counting out the first 12 digits by hand and stepping through the repeating block lands on the same 2, and the block never changes however far it runs.

Another way: Long division carried out 38 places would also give 2, at the cost of 38 steps instead of 5.

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading a fraction as a repeating decimal and using its period.
6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Rebuilding the numerator from the quotient and remainder.

💡Takeaway. When digits repeat, you never have to go all the way. Find the block, then find where your place lands inside it.