Practice · A remainder picks the digit; a remainder of zero means the last one · Sensim Math

Variant 1 easy

For the division below, find the digit in the $15$ th decimal place and the digit in the $20$ th decimal place of the quotient.

$14 \div 2.7$

Your answer (check it yourself)

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 \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\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 \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 \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

For the division below, find the digit in the $12$ th decimal place and the digit in the $23$ th decimal place of the quotient.

$14 \div 11.1$

Your answer (check it yourself)

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 \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\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 \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 \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

For the division below, find the digit in the $13$ th decimal place and the digit in the $24$ th decimal place of the quotient.

$25 \div 1.1$

Your answer (check it yourself)

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 \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\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 \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 \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

For the division below, find the digit in the $21$ th decimal place and the digit in the $26$ th decimal place of the quotient.

$17 \div 11.1$

Your answer (check it yourself)

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 \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\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 \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 \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

For the division below, find the digit in the $27$ th decimal place and the digit in the $29$ th decimal place of the quotient.

$28 \div 1.3$

Your answer (check it yourself)

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 \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\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 \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 \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

For the division below, find the digit in the $19$ th decimal place and the digit in the $29$ th decimal place of the quotient.

$9 \div 1.3$

Your answer (check it yourself)

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 \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\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 \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 \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

For the division below, find the digit in the $21$ th decimal place and the digit in the $29$ th decimal place of the quotient.

$20 \div 2.1$

Your answer (check it yourself)

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 \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\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 \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 \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

For the division below, find the digit in the $19$ th decimal place and the digit in the $30$ th decimal place of the quotient.

$26 \div 6.3$

Your answer (check it yourself)

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 \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\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 \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 \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

For the division below, find the digit in the $27$ th decimal place and the digit in the $31$ th decimal place of the quotient.

$3 \div 4.1$

Your answer (check it yourself)

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 \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\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 \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 \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

For the division below, find the digit in the $26$ th decimal place and the digit in the $33$ th decimal place of the quotient.

$22 \div 3.7$

Your answer (check it yourself)

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 \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\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 \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 \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

For the division below, find the digit in the $23$ th decimal place and the digit in the $34$ th decimal place of the quotient.

$24 \div 1.1$

Your answer (check it yourself)

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 \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\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 \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 \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

For the division below, find the digit in the $27$ th decimal place and the digit in the $39$ th decimal place of the quotient.

$19 \div 4.1$

Your answer (check it yourself)

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 \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\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 \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 \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.

A remainder picks the digit; a remainder of zero means the last one · 12 practice problems

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