← A remainder picks the digit; a remainder of zero means the last one · Repeating Decimal Period

A remainder picks the digit; a remainder of zero means the last one · 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: 15th place: 5, 20th place: 8

For the division below, find the digit in the 1515th decimal place and the digit in the 2020th decimal place of the quotient.

14÷2.714 \div 2.7

Show solution
1 · Understandwhat's really being asked

14 divided by 2.7 never ends. We want the digits sitting in the 15th and 20th decimal places.

Givens
  • The division is 14 over 2.7.
  • The digits wanted are in the 15th and 20th decimal places.
Unknowns
  • The 15th and 20th digits after the decimal point.
Constraints
  • The quotient does not terminate, so it cannot simply be written out.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #8 Analyze the Units#9 Solve an Easier Related Problem

Clear the decimal point out of the divisor first -- that turns it into a whole-number division, where the repeating block is easy to see. Then a second division, the place by the block's length, picks the digit.

3 · Execute4 carry out the plan

1Clear the divisor's decimal point

#8 Analyze the Units 6.NS.B.3
Multiplying both numbers by 10 leaves the quotient unchanged.
14÷2.7=140÷2714 \div 2.7 = 140 \div 27
Now both numbers are whole.

2Divide until it repeats

#5 Look for a Pattern 7.NS.A.2
The digits after the point run 185, and the block 185 comes round again.
5.185…5.185\ldots
A block of 3 digits repeats forever.

3Place the 15th digit

#9 Solve an Easier Related Problem 7.NS.A.2
15 divided by 3 leaves 0, so it is digit 3 of the block -- a remainder of 0 would mean the last digit, not the first.
15÷3→remainder 015 \div 3 \rightarrow \text{remainder }0
Digit 3 of 185 is 5.

4Place the 20th digit

#9 Solve an Easier Related Problem 7.NS.A.2
20 divided by 3 leaves 2, so it is digit 2 of the block.
20÷3→remainder 220 \div 3 \rightarrow \text{remainder }2
Digit 2 of 185 is 8.
Answer: 15th place: 5, 20th place: 8
4 · Reviewdoes it hold up?

Both places fall inside the same repeating 185, so however far the division ran, those two digits could only be members of that block.

Another way: Carrying the long division out to 20 places gives the same two digits, at the cost of 20 steps instead of 5.

Standardsmin grade 7
  • 7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading the quotient as a repeating decimal and using its period.
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Clearing the decimal point out of the divisor.
💡Takeaway. When digits repeat, the place number divided by the block length tells you which digit you have landed on.
Variant 2 easy answer: 12th place: 1, 23th place: 6

For the division below, find the digit in the 1212th decimal place and the digit in the 2323th decimal place of the quotient.

14÷11.114 \div 11.1

Show solution
1 · Understandwhat's really being asked

14 divided by 11.1 never ends. We want the digits sitting in the 12th and 23th decimal places.

Givens
  • The division is 14 over 11.1.
  • The digits wanted are in the 12th and 23th decimal places.
Unknowns
  • The 12th and 23th digits after the decimal point.
Constraints
  • The quotient does not terminate, so it cannot simply be written out.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #8 Analyze the Units#9 Solve an Easier Related Problem

Clear the decimal point out of the divisor first -- that turns it into a whole-number division, where the repeating block is easy to see. Then a second division, the place by the block's length, picks the digit.

3 · Execute4 carry out the plan

1Clear the divisor's decimal point

#8 Analyze the Units 6.NS.B.3
Multiplying both numbers by 10 leaves the quotient unchanged.
14÷11.1=140÷11114 \div 11.1 = 140 \div 111
Now both numbers are whole.

2Divide until it repeats

#5 Look for a Pattern 7.NS.A.2
The digits after the point run 261, and the block 261 comes round again.
1.261…1.261\ldots
A block of 3 digits repeats forever.

3Place the 12th digit

#9 Solve an Easier Related Problem 7.NS.A.2
12 divided by 3 leaves 0, so it is digit 3 of the block -- a remainder of 0 would mean the last digit, not the first.
12÷3→remainder 012 \div 3 \rightarrow \text{remainder }0
Digit 3 of 261 is 1.

4Place the 23th digit

#9 Solve an Easier Related Problem 7.NS.A.2
23 divided by 3 leaves 2, so it is digit 2 of the block.
23÷3→remainder 223 \div 3 \rightarrow \text{remainder }2
Digit 2 of 261 is 6.
Answer: 12th place: 1, 23th place: 6
4 · Reviewdoes it hold up?

Both places fall inside the same repeating 261, so however far the division ran, those two digits could only be members of that block.

Another way: Carrying the long division out to 23 places gives the same two digits, at the cost of 23 steps instead of 5.

Standardsmin grade 7
  • 7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading the quotient as a repeating decimal and using its period.
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Clearing the decimal point out of the divisor.
💡Takeaway. When digits repeat, the place number divided by the block length tells you which digit you have landed on.
Variant 3 easy answer: 13th place: 7, 24th place: 2

For the division below, find the digit in the 1313th decimal place and the digit in the 2424th decimal place of the quotient.

25÷1.125 \div 1.1

Show solution
1 · Understandwhat's really being asked

25 divided by 1.1 never ends. We want the digits sitting in the 13th and 24th decimal places.

Givens
  • The division is 25 over 1.1.
  • The digits wanted are in the 13th and 24th decimal places.
Unknowns
  • The 13th and 24th digits after the decimal point.
Constraints
  • The quotient does not terminate, so it cannot simply be written out.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #8 Analyze the Units#9 Solve an Easier Related Problem

Clear the decimal point out of the divisor first -- that turns it into a whole-number division, where the repeating block is easy to see. Then a second division, the place by the block's length, picks the digit.

3 · Execute4 carry out the plan

1Clear the divisor's decimal point

#8 Analyze the Units 6.NS.B.3
Multiplying both numbers by 10 leaves the quotient unchanged.
25÷1.1=250÷1125 \div 1.1 = 250 \div 11
Now both numbers are whole.

2Divide until it repeats

#5 Look for a Pattern 7.NS.A.2
The digits after the point run 72, and the block 72 comes round again.
22.72…22.72\ldots
A block of 2 digits repeats forever.

3Place the 13th digit

#9 Solve an Easier Related Problem 7.NS.A.2
13 divided by 2 leaves 1, so it is digit 1 of the block -- a remainder of 0 would mean the last digit, not the first.
13÷2→remainder 113 \div 2 \rightarrow \text{remainder }1
Digit 1 of 72 is 7.

4Place the 24th digit

#9 Solve an Easier Related Problem 7.NS.A.2
24 divided by 2 leaves 0, so it is digit 2 of the block.
24÷2→remainder 024 \div 2 \rightarrow \text{remainder }0
Digit 2 of 72 is 2.
Answer: 13th place: 7, 24th place: 2
4 · Reviewdoes it hold up?

Both places fall inside the same repeating 72, so however far the division ran, those two digits could only be members of that block.

Another way: Carrying the long division out to 24 places gives the same two digits, 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 the quotient as a repeating decimal and using its period.
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Clearing the decimal point out of the divisor.
💡Takeaway. When digits repeat, the place number divided by the block length tells you which digit you have landed on.
Variant 4 easy answer: 21th place: 1, 26th place: 3

For the division below, find the digit in the 2121th decimal place and the digit in the 2626th decimal place of the quotient.

17÷11.117 \div 11.1

Show solution
1 · Understandwhat's really being asked

17 divided by 11.1 never ends. We want the digits sitting in the 21th and 26th decimal places.

Givens
  • The division is 17 over 11.1.
  • The digits wanted are in the 21th and 26th decimal places.
Unknowns
  • The 21th and 26th digits after the decimal point.
Constraints
  • The quotient does not terminate, so it cannot simply be written out.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #8 Analyze the Units#9 Solve an Easier Related Problem

Clear the decimal point out of the divisor first -- that turns it into a whole-number division, where the repeating block is easy to see. Then a second division, the place by the block's length, picks the digit.

3 · Execute4 carry out the plan

1Clear the divisor's decimal point

#8 Analyze the Units 6.NS.B.3
Multiplying both numbers by 10 leaves the quotient unchanged.
17÷11.1=170÷11117 \div 11.1 = 170 \div 111
Now both numbers are whole.

2Divide until it repeats

#5 Look for a Pattern 7.NS.A.2
The digits after the point run 531, and the block 531 comes round again.
1.531…1.531\ldots
A block of 3 digits repeats forever.

3Place the 21th digit

#9 Solve an Easier Related Problem 7.NS.A.2
21 divided by 3 leaves 0, so it is digit 3 of the block -- a remainder of 0 would mean the last digit, not the first.
21÷3→remainder 021 \div 3 \rightarrow \text{remainder }0
Digit 3 of 531 is 1.

4Place the 26th digit

#9 Solve an Easier Related Problem 7.NS.A.2
26 divided by 3 leaves 2, so it is digit 2 of the block.
26÷3→remainder 226 \div 3 \rightarrow \text{remainder }2
Digit 2 of 531 is 3.
Answer: 21th place: 1, 26th place: 3
4 · Reviewdoes it hold up?

Both places fall inside the same repeating 531, so however far the division ran, those two digits could only be members of that block.

Another way: Carrying the long division out to 26 places gives the same two digits, at the cost of 26 steps instead of 5.

Standardsmin grade 7
  • 7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading the quotient as a repeating decimal and using its period.
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Clearing the decimal point out of the divisor.
💡Takeaway. When digits repeat, the place number divided by the block length tells you which digit you have landed on.
Variant 5 medium answer: 27th place: 8, 29th place: 6

For the division below, find the digit in the 2727th decimal place and the digit in the 2929th decimal place of the quotient.

28÷1.328 \div 1.3

Show solution
1 · Understandwhat's really being asked

28 divided by 1.3 never ends. We want the digits sitting in the 27th and 29th decimal places.

Givens
  • The division is 28 over 1.3.
  • The digits wanted are in the 27th and 29th decimal places.
Unknowns
  • The 27th and 29th digits after the decimal point.
Constraints
  • The quotient does not terminate, so it cannot simply be written out.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #8 Analyze the Units#9 Solve an Easier Related Problem

Clear the decimal point out of the divisor first -- that turns it into a whole-number division, where the repeating block is easy to see. Then a second division, the place by the block's length, picks the digit.

3 · Execute4 carry out the plan

1Clear the divisor's decimal point

#8 Analyze the Units 6.NS.B.3
Multiplying both numbers by 10 leaves the quotient unchanged.
28÷1.3=280÷1328 \div 1.3 = 280 \div 13
Now both numbers are whole.

2Divide until it repeats

#5 Look for a Pattern 7.NS.A.2
The digits after the point run 538461, and the block 538461 comes round again.
21.538461…21.538461\ldots
A block of 6 digits repeats forever.

3Place the 27th digit

#9 Solve an Easier Related Problem 7.NS.A.2
27 divided by 6 leaves 3, so it is digit 3 of the block -- a remainder of 0 would mean the last digit, not the first.
27÷6→remainder 327 \div 6 \rightarrow \text{remainder }3
Digit 3 of 538461 is 8.

4Place the 29th digit

#9 Solve an Easier Related Problem 7.NS.A.2
29 divided by 6 leaves 5, so it is digit 5 of the block.
29÷6→remainder 529 \div 6 \rightarrow \text{remainder }5
Digit 5 of 538461 is 6.
Answer: 27th place: 8, 29th place: 6
4 · Reviewdoes it hold up?

Both places fall inside the same repeating 538461, so however far the division ran, those two digits could only be members of that block.

Another way: Carrying the long division out to 29 places gives the same two digits, at the cost of 29 steps instead of 8.

Standardsmin grade 7
  • 7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading the quotient as a repeating decimal and using its period.
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Clearing the decimal point out of the divisor.
💡Takeaway. When digits repeat, the place number divided by the block length tells you which digit you have landed on.
Variant 6 medium answer: 19th place: 9, 29th place: 7

For the division below, find the digit in the 1919th decimal place and the digit in the 2929th decimal place of the quotient.

9÷1.39 \div 1.3

Show solution
1 · Understandwhat's really being asked

9 divided by 1.3 never ends. We want the digits sitting in the 19th and 29th decimal places.

Givens
  • The division is 9 over 1.3.
  • The digits wanted are in the 19th and 29th decimal places.
Unknowns
  • The 19th and 29th digits after the decimal point.
Constraints
  • The quotient does not terminate, so it cannot simply be written out.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #8 Analyze the Units#9 Solve an Easier Related Problem

Clear the decimal point out of the divisor first -- that turns it into a whole-number division, where the repeating block is easy to see. Then a second division, the place by the block's length, picks the digit.

3 · Execute4 carry out the plan

1Clear the divisor's decimal point

#8 Analyze the Units 6.NS.B.3
Multiplying both numbers by 10 leaves the quotient unchanged.
9÷1.3=90÷139 \div 1.3 = 90 \div 13
Now both numbers are whole.

2Divide until it repeats

#5 Look for a Pattern 7.NS.A.2
The digits after the point run 923076, and the block 923076 comes round again.
6.923076…6.923076\ldots
A block of 6 digits repeats forever.

3Place the 19th digit

#9 Solve an Easier Related Problem 7.NS.A.2
19 divided by 6 leaves 1, so it is digit 1 of the block -- a remainder of 0 would mean the last digit, not the first.
19÷6→remainder 119 \div 6 \rightarrow \text{remainder }1
Digit 1 of 923076 is 9.

4Place the 29th digit

#9 Solve an Easier Related Problem 7.NS.A.2
29 divided by 6 leaves 5, so it is digit 5 of the block.
29÷6→remainder 529 \div 6 \rightarrow \text{remainder }5
Digit 5 of 923076 is 7.
Answer: 19th place: 9, 29th place: 7
4 · Reviewdoes it hold up?

Both places fall inside the same repeating 923076, so however far the division ran, those two digits could only be members of that block.

Another way: Carrying the long division out to 29 places gives the same two digits, at the cost of 29 steps instead of 8.

Standardsmin grade 7
  • 7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading the quotient as a repeating decimal and using its period.
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Clearing the decimal point out of the divisor.
💡Takeaway. When digits repeat, the place number divided by the block length tells you which digit you have landed on.
Variant 7 medium answer: 21th place: 3, 29th place: 0

For the division below, find the digit in the 2121th decimal place and the digit in the 2929th decimal place of the quotient.

20÷2.120 \div 2.1

Show solution
1 · Understandwhat's really being asked

20 divided by 2.1 never ends. We want the digits sitting in the 21th and 29th decimal places.

Givens
  • The division is 20 over 2.1.
  • The digits wanted are in the 21th and 29th decimal places.
Unknowns
  • The 21th and 29th digits after the decimal point.
Constraints
  • The quotient does not terminate, so it cannot simply be written out.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #8 Analyze the Units#9 Solve an Easier Related Problem

Clear the decimal point out of the divisor first -- that turns it into a whole-number division, where the repeating block is easy to see. Then a second division, the place by the block's length, picks the digit.

3 · Execute4 carry out the plan

1Clear the divisor's decimal point

#8 Analyze the Units 6.NS.B.3
Multiplying both numbers by 10 leaves the quotient unchanged.
20÷2.1=200÷2120 \div 2.1 = 200 \div 21
Now both numbers are whole.

2Divide until it repeats

#5 Look for a Pattern 7.NS.A.2
The digits after the point run 523809, and the block 523809 comes round again.
9.523809…9.523809\ldots
A block of 6 digits repeats forever.

3Place the 21th digit

#9 Solve an Easier Related Problem 7.NS.A.2
21 divided by 6 leaves 3, so it is digit 3 of the block -- a remainder of 0 would mean the last digit, not the first.
21÷6→remainder 321 \div 6 \rightarrow \text{remainder }3
Digit 3 of 523809 is 3.

4Place the 29th digit

#9 Solve an Easier Related Problem 7.NS.A.2
29 divided by 6 leaves 5, so it is digit 5 of the block.
29÷6→remainder 529 \div 6 \rightarrow \text{remainder }5
Digit 5 of 523809 is 0.
Answer: 21th place: 3, 29th place: 0
4 · Reviewdoes it hold up?

Both places fall inside the same repeating 523809, so however far the division ran, those two digits could only be members of that block.

Another way: Carrying the long division out to 29 places gives the same two digits, at the cost of 29 steps instead of 8.

Standardsmin grade 7
  • 7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading the quotient as a repeating decimal and using its period.
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Clearing the decimal point out of the divisor.
💡Takeaway. When digits repeat, the place number divided by the block length tells you which digit you have landed on.
Variant 8 medium answer: 19th place: 1, 30th place: 4

For the division below, find the digit in the 1919th decimal place and the digit in the 3030th decimal place of the quotient.

26÷6.326 \div 6.3

Show solution
1 · Understandwhat's really being asked

26 divided by 6.3 never ends. We want the digits sitting in the 19th and 30th decimal places.

Givens
  • The division is 26 over 6.3.
  • The digits wanted are in the 19th and 30th decimal places.
Unknowns
  • The 19th and 30th digits after the decimal point.
Constraints
  • The quotient does not terminate, so it cannot simply be written out.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #8 Analyze the Units#9 Solve an Easier Related Problem

Clear the decimal point out of the divisor first -- that turns it into a whole-number division, where the repeating block is easy to see. Then a second division, the place by the block's length, picks the digit.

3 · Execute4 carry out the plan

1Clear the divisor's decimal point

#8 Analyze the Units 6.NS.B.3
Multiplying both numbers by 10 leaves the quotient unchanged.
26÷6.3=260÷6326 \div 6.3 = 260 \div 63
Now both numbers are whole.

2Divide until it repeats

#5 Look for a Pattern 7.NS.A.2
The digits after the point run 126984, and the block 126984 comes round again.
4.126984…4.126984\ldots
A block of 6 digits repeats forever.

3Place the 19th digit

#9 Solve an Easier Related Problem 7.NS.A.2
19 divided by 6 leaves 1, so it is digit 1 of the block -- a remainder of 0 would mean the last digit, not the first.
19÷6→remainder 119 \div 6 \rightarrow \text{remainder }1
Digit 1 of 126984 is 1.

4Place the 30th digit

#9 Solve an Easier Related Problem 7.NS.A.2
30 divided by 6 leaves 0, so it is digit 6 of the block.
30÷6→remainder 030 \div 6 \rightarrow \text{remainder }0
Digit 6 of 126984 is 4.
Answer: 19th place: 1, 30th place: 4
4 · Reviewdoes it hold up?

Both places fall inside the same repeating 126984, so however far the division ran, those two digits could only be members of that block.

Another way: Carrying the long division out to 30 places gives the same two digits, at the cost of 30 steps instead of 8.

Standardsmin grade 7
  • 7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading the quotient as a repeating decimal and using its period.
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Clearing the decimal point out of the divisor.
💡Takeaway. When digits repeat, the place number divided by the block length tells you which digit you have landed on.
Variant 9 hard answer: 27th place: 3, 31th place: 7

For the division below, find the digit in the 2727th decimal place and the digit in the 3131th decimal place of the quotient.

3÷4.13 \div 4.1

Show solution
1 · Understandwhat's really being asked

3 divided by 4.1 never ends. We want the digits sitting in the 27th and 31th decimal places.

Givens
  • The division is 3 over 4.1.
  • The digits wanted are in the 27th and 31th decimal places.
Unknowns
  • The 27th and 31th digits after the decimal point.
Constraints
  • The quotient does not terminate, so it cannot simply be written out.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #8 Analyze the Units#9 Solve an Easier Related Problem

Clear the decimal point out of the divisor first -- that turns it into a whole-number division, where the repeating block is easy to see. Then a second division, the place by the block's length, picks the digit.

3 · Execute4 carry out the plan

1Clear the divisor's decimal point

#8 Analyze the Units 6.NS.B.3
Multiplying both numbers by 10 leaves the quotient unchanged.
3÷4.1=30÷413 \div 4.1 = 30 \div 41
Now both numbers are whole.

2Divide until it repeats

#5 Look for a Pattern 7.NS.A.2
The digits after the point run 73170, and the block 73170 comes round again.
0.73170…0.73170\ldots
A block of 5 digits repeats forever.

3Place the 27th digit

#9 Solve an Easier Related Problem 7.NS.A.2
27 divided by 5 leaves 2, so it is digit 2 of the block -- a remainder of 0 would mean the last digit, not the first.
27÷5→remainder 227 \div 5 \rightarrow \text{remainder }2
Digit 2 of 73170 is 3.

4Place the 31th digit

#9 Solve an Easier Related Problem 7.NS.A.2
31 divided by 5 leaves 1, so it is digit 1 of the block.
31÷5→remainder 131 \div 5 \rightarrow \text{remainder }1
Digit 1 of 73170 is 7.
Answer: 27th place: 3, 31th place: 7
4 · Reviewdoes it hold up?

Both places fall inside the same repeating 73170, so however far the division ran, those two digits could only be members of that block.

Another way: Carrying the long division out to 31 places gives the same two digits, at the cost of 31 steps instead of 7.

Standardsmin grade 7
  • 7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading the quotient as a repeating decimal and using its period.
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Clearing the decimal point out of the divisor.
💡Takeaway. When digits repeat, the place number divided by the block length tells you which digit you have landed on.
Variant 10 hard answer: 26th place: 4, 33th place: 5

For the division below, find the digit in the 2626th decimal place and the digit in the 3333th decimal place of the quotient.

22÷3.722 \div 3.7

Show solution
1 · Understandwhat's really being asked

22 divided by 3.7 never ends. We want the digits sitting in the 26th and 33th decimal places.

Givens
  • The division is 22 over 3.7.
  • The digits wanted are in the 26th and 33th decimal places.
Unknowns
  • The 26th and 33th digits after the decimal point.
Constraints
  • The quotient does not terminate, so it cannot simply be written out.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #8 Analyze the Units#9 Solve an Easier Related Problem

Clear the decimal point out of the divisor first -- that turns it into a whole-number division, where the repeating block is easy to see. Then a second division, the place by the block's length, picks the digit.

3 · Execute4 carry out the plan

1Clear the divisor's decimal point

#8 Analyze the Units 6.NS.B.3
Multiplying both numbers by 10 leaves the quotient unchanged.
22÷3.7=220÷3722 \div 3.7 = 220 \div 37
Now both numbers are whole.

2Divide until it repeats

#5 Look for a Pattern 7.NS.A.2
The digits after the point run 945, and the block 945 comes round again.
5.945…5.945\ldots
A block of 3 digits repeats forever.

3Place the 26th digit

#9 Solve an Easier Related Problem 7.NS.A.2
26 divided by 3 leaves 2, so it is digit 2 of the block -- a remainder of 0 would mean the last digit, not the first.
26÷3→remainder 226 \div 3 \rightarrow \text{remainder }2
Digit 2 of 945 is 4.

4Place the 33th digit

#9 Solve an Easier Related Problem 7.NS.A.2
33 divided by 3 leaves 0, so it is digit 3 of the block.
33÷3→remainder 033 \div 3 \rightarrow \text{remainder }0
Digit 3 of 945 is 5.
Answer: 26th place: 4, 33th place: 5
4 · Reviewdoes it hold up?

Both places fall inside the same repeating 945, so however far the division ran, those two digits could only be members of that block.

Another way: Carrying the long division out to 33 places gives the same two digits, at the cost of 33 steps instead of 5.

Standardsmin grade 7
  • 7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading the quotient as a repeating decimal and using its period.
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Clearing the decimal point out of the divisor.
💡Takeaway. When digits repeat, the place number divided by the block length tells you which digit you have landed on.
Variant 11 hard answer: 23th place: 8, 34th place: 1

For the division below, find the digit in the 2323th decimal place and the digit in the 3434th decimal place of the quotient.

24÷1.124 \div 1.1

Show solution
1 · Understandwhat's really being asked

24 divided by 1.1 never ends. We want the digits sitting in the 23th and 34th decimal places.

Givens
  • The division is 24 over 1.1.
  • The digits wanted are in the 23th and 34th decimal places.
Unknowns
  • The 23th and 34th digits after the decimal point.
Constraints
  • The quotient does not terminate, so it cannot simply be written out.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #8 Analyze the Units#9 Solve an Easier Related Problem

Clear the decimal point out of the divisor first -- that turns it into a whole-number division, where the repeating block is easy to see. Then a second division, the place by the block's length, picks the digit.

3 · Execute4 carry out the plan

1Clear the divisor's decimal point

#8 Analyze the Units 6.NS.B.3
Multiplying both numbers by 10 leaves the quotient unchanged.
24÷1.1=240÷1124 \div 1.1 = 240 \div 11
Now both numbers are whole.

2Divide until it repeats

#5 Look for a Pattern 7.NS.A.2
The digits after the point run 81, and the block 81 comes round again.
21.81…21.81\ldots
A block of 2 digits repeats forever.

3Place the 23th digit

#9 Solve an Easier Related Problem 7.NS.A.2
23 divided by 2 leaves 1, so it is digit 1 of the block -- a remainder of 0 would mean the last digit, not the first.
23÷2→remainder 123 \div 2 \rightarrow \text{remainder }1
Digit 1 of 81 is 8.

4Place the 34th digit

#9 Solve an Easier Related Problem 7.NS.A.2
34 divided by 2 leaves 0, so it is digit 2 of the block.
34÷2→remainder 034 \div 2 \rightarrow \text{remainder }0
Digit 2 of 81 is 1.
Answer: 23th place: 8, 34th place: 1
4 · Reviewdoes it hold up?

Both places fall inside the same repeating 81, so however far the division ran, those two digits could only be members of that block.

Another way: Carrying the long division out to 34 places gives the same two digits, at the cost of 34 steps instead of 4.

Standardsmin grade 7
  • 7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading the quotient as a repeating decimal and using its period.
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Clearing the decimal point out of the divisor.
💡Takeaway. When digits repeat, the place number divided by the block length tells you which digit you have landed on.
Variant 12 hard answer: 27th place: 3, 39th place: 1

For the division below, find the digit in the 2727th decimal place and the digit in the 3939th decimal place of the quotient.

19÷4.119 \div 4.1

Show solution
1 · Understandwhat's really being asked

19 divided by 4.1 never ends. We want the digits sitting in the 27th and 39th decimal places.

Givens
  • The division is 19 over 4.1.
  • The digits wanted are in the 27th and 39th decimal places.
Unknowns
  • The 27th and 39th digits after the decimal point.
Constraints
  • The quotient does not terminate, so it cannot simply be written out.
2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #8 Analyze the Units#9 Solve an Easier Related Problem

Clear the decimal point out of the divisor first -- that turns it into a whole-number division, where the repeating block is easy to see. Then a second division, the place by the block's length, picks the digit.

3 · Execute4 carry out the plan

1Clear the divisor's decimal point

#8 Analyze the Units 6.NS.B.3
Multiplying both numbers by 10 leaves the quotient unchanged.
19÷4.1=190÷4119 \div 4.1 = 190 \div 41
Now both numbers are whole.

2Divide until it repeats

#5 Look for a Pattern 7.NS.A.2
The digits after the point run 63414, and the block 63414 comes round again.
4.63414…4.63414\ldots
A block of 5 digits repeats forever.

3Place the 27th digit

#9 Solve an Easier Related Problem 7.NS.A.2
27 divided by 5 leaves 2, so it is digit 2 of the block -- a remainder of 0 would mean the last digit, not the first.
27÷5→remainder 227 \div 5 \rightarrow \text{remainder }2
Digit 2 of 63414 is 3.

4Place the 39th digit

#9 Solve an Easier Related Problem 7.NS.A.2
39 divided by 5 leaves 4, so it is digit 4 of the block.
39÷5→remainder 439 \div 5 \rightarrow \text{remainder }4
Digit 4 of 63414 is 1.
Answer: 27th place: 3, 39th place: 1
4 · Reviewdoes it hold up?

Both places fall inside the same repeating 63414, so however far the division ran, those two digits could only be members of that block.

Another way: Carrying the long division out to 39 places gives the same two digits, at the cost of 39 steps instead of 7.

Standardsmin grade 7
  • 7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Reading the quotient as a repeating decimal and using its period.
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Clearing the decimal point out of the divisor.
💡Takeaway. When digits repeat, the place number divided by the block length tells you which digit you have landed on.