Practice · Count in hundredths and an exact division is a remainder question · Sensim Math

Variant 1 easy answer: 0.04

Dividing $25$ by $13$ does not come out exactly. When $0.\bigcirc\triangle$ is subtracted from $25$ first, the division by $13$ ends exactly at the second decimal place. Find the smallest such number $0.\bigcirc\triangle$ .

Show solution

1 · Understandwhat's really being asked

25 does not divide by 13 exactly. We want the smallest two-digit decimal that can be taken off 25 so that the division stops at the second decimal place.

Givens

The number is 25 and the divisor is 13.
The amount removed is written 0 point two digits.
After removing it, the division ends at the second decimal place.

Unknowns

The smallest such two-digit decimal.

Constraints

Ending at the second decimal place means the quotient is a whole number of hundredths.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #13 Convert to Algebra#11 Work Backwards

Decimals make this look harder than it is. Count everything in hundredths and the condition becomes a plain remainder question about whole numbers -- and the smallest amount to remove is that remainder.

3 · Execute4 carry out the plan

1Count in hundredths

#8 Analyze the Units 6.NS.B.3

25 is 2500 hundredths, and the amount removed is a whole number of hundredths too.

25 = 2500 \times 0.01

Everything is now a whole number.

2Say what 'ends at the second place' means

#13 Convert to Algebra 6.NS.B.2

The quotient must be a whole number of hundredths, so what is left must be 13 times a whole number of hundredths -- that is, a multiple of 13 in hundredths.

(2500 - \square) \div 13 = \text{whole}

We need a multiple of 13.

3Divide 2500 by 13 and keep the remainder

#11 Work Backwards 6.NS.B.2

2500 leaves 4 over, so removing 4 hundredths lands exactly on the multiple below.

2500 \div 13 \rightarrow \text{remainder }4

4 hundredths is what is in the way.

4Write it back as a decimal

#8 Analyze the Units 6.NS.B.3

4 hundredths is 0.04, and nothing smaller works: anything less leaves a remainder behind.

4 \times 0.01 = 0.04

Take off 0.04.

Answer: 0.04

4 · Reviewdoes it hold up?

25 minus 0.04 is 24.96, and dividing that by 13 gives exactly 1.92 -- two decimal places and no remainder.

Another way: Trying 0.01, 0.02 and so on in order reaches 0.04 after 4 attempts; the remainder finds it in one.

Standardsmin grade 6

6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Moving between a decimal and a whole number of hundredths.
6.NS.B.2 Fluently divide multi-digit numbers — Using the remainder of a whole-number division.

💡Takeaway. Pick a unit small enough and the decimals turn into whole numbers. Then the leftover you need is just a remainder.

Variant 2 easy answer: 0.13

Dividing $36$ by $17$ does not come out exactly. When $0.\bigcirc\triangle$ is subtracted from $36$ first, the division by $17$ ends exactly at the second decimal place. Find the smallest such number $0.\bigcirc\triangle$ .

Show solution

1 · Understandwhat's really being asked

36 does not divide by 17 exactly. We want the smallest two-digit decimal that can be taken off 36 so that the division stops at the second decimal place.

Givens

The number is 36 and the divisor is 17.
The amount removed is written 0 point two digits.
After removing it, the division ends at the second decimal place.

Unknowns

The smallest such two-digit decimal.

Constraints

Ending at the second decimal place means the quotient is a whole number of hundredths.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #13 Convert to Algebra#11 Work Backwards

Decimals make this look harder than it is. Count everything in hundredths and the condition becomes a plain remainder question about whole numbers -- and the smallest amount to remove is that remainder.

3 · Execute4 carry out the plan

1Count in hundredths

#8 Analyze the Units 6.NS.B.3

36 is 3600 hundredths, and the amount removed is a whole number of hundredths too.

36 = 3600 \times 0.01

Everything is now a whole number.

2Say what 'ends at the second place' means

#13 Convert to Algebra 6.NS.B.2

The quotient must be a whole number of hundredths, so what is left must be 17 times a whole number of hundredths -- that is, a multiple of 17 in hundredths.

(3600 - \square) \div 17 = \text{whole}

We need a multiple of 17.

3Divide 3600 by 17 and keep the remainder

#11 Work Backwards 6.NS.B.2

3600 leaves 13 over, so removing 13 hundredths lands exactly on the multiple below.

3600 \div 17 \rightarrow \text{remainder }13

13 hundredths is what is in the way.

4Write it back as a decimal

#8 Analyze the Units 6.NS.B.3

13 hundredths is 0.13, and nothing smaller works: anything less leaves a remainder behind.

13 \times 0.01 = 0.13

Take off 0.13.

Answer: 0.13

4 · Reviewdoes it hold up?

36 minus 0.13 is 35.87, and dividing that by 17 gives exactly 2.11 -- two decimal places and no remainder.

Another way: Trying 0.01, 0.02 and so on in order reaches 0.13 after 13 attempts; the remainder finds it in one.

Standardsmin grade 6

6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Moving between a decimal and a whole number of hundredths.
6.NS.B.2 Fluently divide multi-digit numbers — Using the remainder of a whole-number division.

💡Takeaway. Pick a unit small enough and the decimals turn into whole numbers. Then the leftover you need is just a remainder.

Variant 3 easy answer: 0.16

Dividing $43$ by $21$ does not come out exactly. When $0.\bigcirc\triangle$ is subtracted from $43$ first, the division by $21$ ends exactly at the second decimal place. Find the smallest such number $0.\bigcirc\triangle$ .

Show solution

1 · Understandwhat's really being asked

43 does not divide by 21 exactly. We want the smallest two-digit decimal that can be taken off 43 so that the division stops at the second decimal place.

Givens

The number is 43 and the divisor is 21.
The amount removed is written 0 point two digits.
After removing it, the division ends at the second decimal place.

Unknowns

The smallest such two-digit decimal.

Constraints

Ending at the second decimal place means the quotient is a whole number of hundredths.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #13 Convert to Algebra#11 Work Backwards

Decimals make this look harder than it is. Count everything in hundredths and the condition becomes a plain remainder question about whole numbers -- and the smallest amount to remove is that remainder.

3 · Execute4 carry out the plan

1Count in hundredths

#8 Analyze the Units 6.NS.B.3

43 is 4300 hundredths, and the amount removed is a whole number of hundredths too.

43 = 4300 \times 0.01

Everything is now a whole number.

2Say what 'ends at the second place' means

#13 Convert to Algebra 6.NS.B.2

The quotient must be a whole number of hundredths, so what is left must be 21 times a whole number of hundredths -- that is, a multiple of 21 in hundredths.

(4300 - \square) \div 21 = \text{whole}

We need a multiple of 21.

3Divide 4300 by 21 and keep the remainder

#11 Work Backwards 6.NS.B.2

4300 leaves 16 over, so removing 16 hundredths lands exactly on the multiple below.

4300 \div 21 \rightarrow \text{remainder }16

16 hundredths is what is in the way.

4Write it back as a decimal

#8 Analyze the Units 6.NS.B.3

16 hundredths is 0.16, and nothing smaller works: anything less leaves a remainder behind.

16 \times 0.01 = 0.16

Take off 0.16.

Answer: 0.16

4 · Reviewdoes it hold up?

43 minus 0.16 is 42.84, and dividing that by 21 gives exactly 2.04 -- two decimal places and no remainder.

Another way: Trying 0.01, 0.02 and so on in order reaches 0.16 after 16 attempts; the remainder finds it in one.

Standardsmin grade 6

6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Moving between a decimal and a whole number of hundredths.
6.NS.B.2 Fluently divide multi-digit numbers — Using the remainder of a whole-number division.

💡Takeaway. Pick a unit small enough and the decimals turn into whole numbers. Then the leftover you need is just a remainder.

Variant 4 easy answer: 0.12

Dividing $48$ by $19$ does not come out exactly. When $0.\bigcirc\triangle$ is subtracted from $48$ first, the division by $19$ ends exactly at the second decimal place. Find the smallest such number $0.\bigcirc\triangle$ .

Show solution

1 · Understandwhat's really being asked

48 does not divide by 19 exactly. We want the smallest two-digit decimal that can be taken off 48 so that the division stops at the second decimal place.

Givens

The number is 48 and the divisor is 19.
The amount removed is written 0 point two digits.
After removing it, the division ends at the second decimal place.

Unknowns

The smallest such two-digit decimal.

Constraints

Ending at the second decimal place means the quotient is a whole number of hundredths.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #13 Convert to Algebra#11 Work Backwards

Decimals make this look harder than it is. Count everything in hundredths and the condition becomes a plain remainder question about whole numbers -- and the smallest amount to remove is that remainder.

3 · Execute4 carry out the plan

1Count in hundredths

#8 Analyze the Units 6.NS.B.3

48 is 4800 hundredths, and the amount removed is a whole number of hundredths too.

48 = 4800 \times 0.01

Everything is now a whole number.

2Say what 'ends at the second place' means

#13 Convert to Algebra 6.NS.B.2

The quotient must be a whole number of hundredths, so what is left must be 19 times a whole number of hundredths -- that is, a multiple of 19 in hundredths.

(4800 - \square) \div 19 = \text{whole}

We need a multiple of 19.

3Divide 4800 by 19 and keep the remainder

#11 Work Backwards 6.NS.B.2

4800 leaves 12 over, so removing 12 hundredths lands exactly on the multiple below.

4800 \div 19 \rightarrow \text{remainder }12

12 hundredths is what is in the way.

4Write it back as a decimal

#8 Analyze the Units 6.NS.B.3

12 hundredths is 0.12, and nothing smaller works: anything less leaves a remainder behind.

12 \times 0.01 = 0.12

Take off 0.12.

Answer: 0.12

4 · Reviewdoes it hold up?

48 minus 0.12 is 47.88, and dividing that by 19 gives exactly 2.52 -- two decimal places and no remainder.

Another way: Trying 0.01, 0.02 and so on in order reaches 0.12 after 12 attempts; the remainder finds it in one.

Standardsmin grade 6

6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Moving between a decimal and a whole number of hundredths.
6.NS.B.2 Fluently divide multi-digit numbers — Using the remainder of a whole-number division.

💡Takeaway. Pick a unit small enough and the decimals turn into whole numbers. Then the leftover you need is just a remainder.

Variant 5 medium answer: 0.02

Dividing $52$ by $23$ does not come out exactly. When $0.\bigcirc\triangle$ is subtracted from $52$ first, the division by $23$ ends exactly at the second decimal place. Find the smallest such number $0.\bigcirc\triangle$ .

Show solution

1 · Understandwhat's really being asked

52 does not divide by 23 exactly. We want the smallest two-digit decimal that can be taken off 52 so that the division stops at the second decimal place.

Givens

The number is 52 and the divisor is 23.
The amount removed is written 0 point two digits.
After removing it, the division ends at the second decimal place.

Unknowns

The smallest such two-digit decimal.

Constraints

Ending at the second decimal place means the quotient is a whole number of hundredths.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #13 Convert to Algebra#11 Work Backwards

Decimals make this look harder than it is. Count everything in hundredths and the condition becomes a plain remainder question about whole numbers -- and the smallest amount to remove is that remainder.

3 · Execute4 carry out the plan

1Count in hundredths

#8 Analyze the Units 6.NS.B.3

52 is 5200 hundredths, and the amount removed is a whole number of hundredths too.

52 = 5200 \times 0.01

Everything is now a whole number.

2Say what 'ends at the second place' means

#13 Convert to Algebra 6.NS.B.2

The quotient must be a whole number of hundredths, so what is left must be 23 times a whole number of hundredths -- that is, a multiple of 23 in hundredths.

(5200 - \square) \div 23 = \text{whole}

We need a multiple of 23.

3Divide 5200 by 23 and keep the remainder

#11 Work Backwards 6.NS.B.2

5200 leaves 2 over, so removing 2 hundredths lands exactly on the multiple below.

5200 \div 23 \rightarrow \text{remainder }2

2 hundredths is what is in the way.

4Write it back as a decimal

#8 Analyze the Units 6.NS.B.3

2 hundredths is 0.02, and nothing smaller works: anything less leaves a remainder behind.

2 \times 0.01 = 0.02

Take off 0.02.

Answer: 0.02

4 · Reviewdoes it hold up?

52 minus 0.02 is 51.98, and dividing that by 23 gives exactly 2.26 -- two decimal places and no remainder.

Another way: Trying 0.01, 0.02 and so on in order reaches 0.02 after 2 attempts; the remainder finds it in one.

Standardsmin grade 6

6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Moving between a decimal and a whole number of hundredths.
6.NS.B.2 Fluently divide multi-digit numbers — Using the remainder of a whole-number division.

💡Takeaway. Pick a unit small enough and the decimals turn into whole numbers. Then the leftover you need is just a remainder.

Variant 6 medium answer: 0.25

Dividing $58$ by $33$ does not come out exactly. When $0.\bigcirc\triangle$ is subtracted from $58$ first, the division by $33$ ends exactly at the second decimal place. Find the smallest such number $0.\bigcirc\triangle$ .

Show solution

1 · Understandwhat's really being asked

58 does not divide by 33 exactly. We want the smallest two-digit decimal that can be taken off 58 so that the division stops at the second decimal place.

Givens

The number is 58 and the divisor is 33.
The amount removed is written 0 point two digits.
After removing it, the division ends at the second decimal place.

Unknowns

The smallest such two-digit decimal.

Constraints

Ending at the second decimal place means the quotient is a whole number of hundredths.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #13 Convert to Algebra#11 Work Backwards

Decimals make this look harder than it is. Count everything in hundredths and the condition becomes a plain remainder question about whole numbers -- and the smallest amount to remove is that remainder.

3 · Execute4 carry out the plan

1Count in hundredths

#8 Analyze the Units 6.NS.B.3

58 is 5800 hundredths, and the amount removed is a whole number of hundredths too.

58 = 5800 \times 0.01

Everything is now a whole number.

2Say what 'ends at the second place' means

#13 Convert to Algebra 6.NS.B.2

The quotient must be a whole number of hundredths, so what is left must be 33 times a whole number of hundredths -- that is, a multiple of 33 in hundredths.

(5800 - \square) \div 33 = \text{whole}

We need a multiple of 33.

3Divide 5800 by 33 and keep the remainder

#11 Work Backwards 6.NS.B.2

5800 leaves 25 over, so removing 25 hundredths lands exactly on the multiple below.

5800 \div 33 \rightarrow \text{remainder }25

25 hundredths is what is in the way.

4Write it back as a decimal

#8 Analyze the Units 6.NS.B.3

25 hundredths is 0.25, and nothing smaller works: anything less leaves a remainder behind.

25 \times 0.01 = 0.25

Take off 0.25.

Answer: 0.25

4 · Reviewdoes it hold up?

58 minus 0.25 is 57.75, and dividing that by 33 gives exactly 1.75 -- two decimal places and no remainder.

Another way: Trying 0.01, 0.02 and so on in order reaches 0.25 after 25 attempts; the remainder finds it in one.

Standardsmin grade 6

6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Moving between a decimal and a whole number of hundredths.
6.NS.B.2 Fluently divide multi-digit numbers — Using the remainder of a whole-number division.

💡Takeaway. Pick a unit small enough and the decimals turn into whole numbers. Then the leftover you need is just a remainder.

Variant 7 medium answer: 0.1

Dividing $61$ by $29$ does not come out exactly. When $0.\bigcirc\triangle$ is subtracted from $61$ first, the division by $29$ ends exactly at the second decimal place. Find the smallest such number $0.\bigcirc\triangle$ .

Show solution

1 · Understandwhat's really being asked

61 does not divide by 29 exactly. We want the smallest two-digit decimal that can be taken off 61 so that the division stops at the second decimal place.

Givens

The number is 61 and the divisor is 29.
The amount removed is written 0 point two digits.
After removing it, the division ends at the second decimal place.

Unknowns

The smallest such two-digit decimal.

Constraints

Ending at the second decimal place means the quotient is a whole number of hundredths.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #13 Convert to Algebra#11 Work Backwards

Decimals make this look harder than it is. Count everything in hundredths and the condition becomes a plain remainder question about whole numbers -- and the smallest amount to remove is that remainder.

3 · Execute4 carry out the plan

1Count in hundredths

#8 Analyze the Units 6.NS.B.3

61 is 6100 hundredths, and the amount removed is a whole number of hundredths too.

61 = 6100 \times 0.01

Everything is now a whole number.

2Say what 'ends at the second place' means

#13 Convert to Algebra 6.NS.B.2

The quotient must be a whole number of hundredths, so what is left must be 29 times a whole number of hundredths -- that is, a multiple of 29 in hundredths.

(6100 - \square) \div 29 = \text{whole}

We need a multiple of 29.

3Divide 6100 by 29 and keep the remainder

#11 Work Backwards 6.NS.B.2

6100 leaves 10 over, so removing 10 hundredths lands exactly on the multiple below.

6100 \div 29 \rightarrow \text{remainder }10

10 hundredths is what is in the way.

4Write it back as a decimal

#8 Analyze the Units 6.NS.B.3

10 hundredths is 0.1, and nothing smaller works: anything less leaves a remainder behind.

10 \times 0.01 = 0.1

Take off 0.1.

Answer: 0.1

4 · Reviewdoes it hold up?

61 minus 0.1 is 60.9, and dividing that by 29 gives exactly 2.1 -- two decimal places and no remainder.

Another way: Trying 0.01, 0.02 and so on in order reaches 0.1 after 10 attempts; the remainder finds it in one.

Standardsmin grade 6

6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Moving between a decimal and a whole number of hundredths.
6.NS.B.2 Fluently divide multi-digit numbers — Using the remainder of a whole-number division.

💡Takeaway. Pick a unit small enough and the decimals turn into whole numbers. Then the leftover you need is just a remainder.

Variant 8 medium answer: 0.04

Dividing $67$ by $27$ does not come out exactly. When $0.\bigcirc\triangle$ is subtracted from $67$ first, the division by $27$ ends exactly at the second decimal place. Find the smallest such number $0.\bigcirc\triangle$ .

Show solution

1 · Understandwhat's really being asked

67 does not divide by 27 exactly. We want the smallest two-digit decimal that can be taken off 67 so that the division stops at the second decimal place.

Givens

The number is 67 and the divisor is 27.
The amount removed is written 0 point two digits.
After removing it, the division ends at the second decimal place.

Unknowns

The smallest such two-digit decimal.

Constraints

Ending at the second decimal place means the quotient is a whole number of hundredths.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #13 Convert to Algebra#11 Work Backwards

Decimals make this look harder than it is. Count everything in hundredths and the condition becomes a plain remainder question about whole numbers -- and the smallest amount to remove is that remainder.

3 · Execute4 carry out the plan

1Count in hundredths

#8 Analyze the Units 6.NS.B.3

67 is 6700 hundredths, and the amount removed is a whole number of hundredths too.

67 = 6700 \times 0.01

Everything is now a whole number.

2Say what 'ends at the second place' means

#13 Convert to Algebra 6.NS.B.2

The quotient must be a whole number of hundredths, so what is left must be 27 times a whole number of hundredths -- that is, a multiple of 27 in hundredths.

(6700 - \square) \div 27 = \text{whole}

We need a multiple of 27.

3Divide 6700 by 27 and keep the remainder

#11 Work Backwards 6.NS.B.2

6700 leaves 4 over, so removing 4 hundredths lands exactly on the multiple below.

6700 \div 27 \rightarrow \text{remainder }4

4 hundredths is what is in the way.

4Write it back as a decimal

#8 Analyze the Units 6.NS.B.3

4 hundredths is 0.04, and nothing smaller works: anything less leaves a remainder behind.

4 \times 0.01 = 0.04

Take off 0.04.

Answer: 0.04

4 · Reviewdoes it hold up?

67 minus 0.04 is 66.96, and dividing that by 27 gives exactly 2.48 -- two decimal places and no remainder.

Another way: Trying 0.01, 0.02 and so on in order reaches 0.04 after 4 attempts; the remainder finds it in one.

Standardsmin grade 6

6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Moving between a decimal and a whole number of hundredths.
6.NS.B.2 Fluently divide multi-digit numbers — Using the remainder of a whole-number division.

💡Takeaway. Pick a unit small enough and the decimals turn into whole numbers. Then the leftover you need is just a remainder.

Variant 9 hard answer: 0.22

Dividing $74$ by $31$ does not come out exactly. When $0.\bigcirc\triangle$ is subtracted from $74$ first, the division by $31$ ends exactly at the second decimal place. Find the smallest such number $0.\bigcirc\triangle$ .

Show solution

1 · Understandwhat's really being asked

74 does not divide by 31 exactly. We want the smallest two-digit decimal that can be taken off 74 so that the division stops at the second decimal place.

Givens

The number is 74 and the divisor is 31.
The amount removed is written 0 point two digits.
After removing it, the division ends at the second decimal place.

Unknowns

The smallest such two-digit decimal.

Constraints

Ending at the second decimal place means the quotient is a whole number of hundredths.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #13 Convert to Algebra#11 Work Backwards

Decimals make this look harder than it is. Count everything in hundredths and the condition becomes a plain remainder question about whole numbers -- and the smallest amount to remove is that remainder.

3 · Execute4 carry out the plan

1Count in hundredths

#8 Analyze the Units 6.NS.B.3

74 is 7400 hundredths, and the amount removed is a whole number of hundredths too.

74 = 7400 \times 0.01

Everything is now a whole number.

2Say what 'ends at the second place' means

#13 Convert to Algebra 6.NS.B.2

The quotient must be a whole number of hundredths, so what is left must be 31 times a whole number of hundredths -- that is, a multiple of 31 in hundredths.

(7400 - \square) \div 31 = \text{whole}

We need a multiple of 31.

3Divide 7400 by 31 and keep the remainder

#11 Work Backwards 6.NS.B.2

7400 leaves 22 over, so removing 22 hundredths lands exactly on the multiple below.

7400 \div 31 \rightarrow \text{remainder }22

22 hundredths is what is in the way.

4Write it back as a decimal

#8 Analyze the Units 6.NS.B.3

22 hundredths is 0.22, and nothing smaller works: anything less leaves a remainder behind.

22 \times 0.01 = 0.22

Take off 0.22.

Answer: 0.22

4 · Reviewdoes it hold up?

74 minus 0.22 is 73.78, and dividing that by 31 gives exactly 2.38 -- two decimal places and no remainder.

Another way: Trying 0.01, 0.02 and so on in order reaches 0.22 after 22 attempts; the remainder finds it in one.

Standardsmin grade 6

6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Moving between a decimal and a whole number of hundredths.
6.NS.B.2 Fluently divide multi-digit numbers — Using the remainder of a whole-number division.

💡Takeaway. Pick a unit small enough and the decimals turn into whole numbers. Then the leftover you need is just a remainder.

Variant 10 hard answer: 0.22

Dividing $79$ by $39$ does not come out exactly. When $0.\bigcirc\triangle$ is subtracted from $79$ first, the division by $39$ ends exactly at the second decimal place. Find the smallest such number $0.\bigcirc\triangle$ .

Show solution

1 · Understandwhat's really being asked

79 does not divide by 39 exactly. We want the smallest two-digit decimal that can be taken off 79 so that the division stops at the second decimal place.

Givens

The number is 79 and the divisor is 39.
The amount removed is written 0 point two digits.
After removing it, the division ends at the second decimal place.

Unknowns

The smallest such two-digit decimal.

Constraints

Ending at the second decimal place means the quotient is a whole number of hundredths.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #13 Convert to Algebra#11 Work Backwards

Decimals make this look harder than it is. Count everything in hundredths and the condition becomes a plain remainder question about whole numbers -- and the smallest amount to remove is that remainder.

3 · Execute4 carry out the plan

1Count in hundredths

#8 Analyze the Units 6.NS.B.3

79 is 7900 hundredths, and the amount removed is a whole number of hundredths too.

79 = 7900 \times 0.01

Everything is now a whole number.

2Say what 'ends at the second place' means

#13 Convert to Algebra 6.NS.B.2

The quotient must be a whole number of hundredths, so what is left must be 39 times a whole number of hundredths -- that is, a multiple of 39 in hundredths.

(7900 - \square) \div 39 = \text{whole}

We need a multiple of 39.

3Divide 7900 by 39 and keep the remainder

#11 Work Backwards 6.NS.B.2

7900 leaves 22 over, so removing 22 hundredths lands exactly on the multiple below.

7900 \div 39 \rightarrow \text{remainder }22

22 hundredths is what is in the way.

4Write it back as a decimal

#8 Analyze the Units 6.NS.B.3

22 hundredths is 0.22, and nothing smaller works: anything less leaves a remainder behind.

22 \times 0.01 = 0.22

Take off 0.22.

Answer: 0.22

4 · Reviewdoes it hold up?

79 minus 0.22 is 78.78, and dividing that by 39 gives exactly 2.02 -- two decimal places and no remainder.

Another way: Trying 0.01, 0.02 and so on in order reaches 0.22 after 22 attempts; the remainder finds it in one.

Standardsmin grade 6

6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Moving between a decimal and a whole number of hundredths.
6.NS.B.2 Fluently divide multi-digit numbers — Using the remainder of a whole-number division.

💡Takeaway. Pick a unit small enough and the decimals turn into whole numbers. Then the leftover you need is just a remainder.

Variant 11 hard answer: 0.12

Dividing $83$ by $37$ does not come out exactly. When $0.\bigcirc\triangle$ is subtracted from $83$ first, the division by $37$ ends exactly at the second decimal place. Find the smallest such number $0.\bigcirc\triangle$ .

Show solution

1 · Understandwhat's really being asked

83 does not divide by 37 exactly. We want the smallest two-digit decimal that can be taken off 83 so that the division stops at the second decimal place.

Givens

The number is 83 and the divisor is 37.
The amount removed is written 0 point two digits.
After removing it, the division ends at the second decimal place.

Unknowns

The smallest such two-digit decimal.

Constraints

Ending at the second decimal place means the quotient is a whole number of hundredths.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #13 Convert to Algebra#11 Work Backwards

Decimals make this look harder than it is. Count everything in hundredths and the condition becomes a plain remainder question about whole numbers -- and the smallest amount to remove is that remainder.

3 · Execute4 carry out the plan

1Count in hundredths

#8 Analyze the Units 6.NS.B.3

83 is 8300 hundredths, and the amount removed is a whole number of hundredths too.

83 = 8300 \times 0.01

Everything is now a whole number.

2Say what 'ends at the second place' means

#13 Convert to Algebra 6.NS.B.2

The quotient must be a whole number of hundredths, so what is left must be 37 times a whole number of hundredths -- that is, a multiple of 37 in hundredths.

(8300 - \square) \div 37 = \text{whole}

We need a multiple of 37.

3Divide 8300 by 37 and keep the remainder

#11 Work Backwards 6.NS.B.2

8300 leaves 12 over, so removing 12 hundredths lands exactly on the multiple below.

8300 \div 37 \rightarrow \text{remainder }12

12 hundredths is what is in the way.

4Write it back as a decimal

#8 Analyze the Units 6.NS.B.3

12 hundredths is 0.12, and nothing smaller works: anything less leaves a remainder behind.

12 \times 0.01 = 0.12

Take off 0.12.

Answer: 0.12

4 · Reviewdoes it hold up?

83 minus 0.12 is 82.88, and dividing that by 37 gives exactly 2.24 -- two decimal places and no remainder.

Another way: Trying 0.01, 0.02 and so on in order reaches 0.12 after 12 attempts; the remainder finds it in one.

Standardsmin grade 6

6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Moving between a decimal and a whole number of hundredths.
6.NS.B.2 Fluently divide multi-digit numbers — Using the remainder of a whole-number division.

💡Takeaway. Pick a unit small enough and the decimals turn into whole numbers. Then the leftover you need is just a remainder.

Variant 12 hard answer: 0.29

Dividing $95$ by $41$ does not come out exactly. When $0.\bigcirc\triangle$ is subtracted from $95$ first, the division by $41$ ends exactly at the second decimal place. Find the smallest such number $0.\bigcirc\triangle$ .

Show solution

1 · Understandwhat's really being asked

95 does not divide by 41 exactly. We want the smallest two-digit decimal that can be taken off 95 so that the division stops at the second decimal place.

Givens

The number is 95 and the divisor is 41.
The amount removed is written 0 point two digits.
After removing it, the division ends at the second decimal place.

Unknowns

The smallest such two-digit decimal.

Constraints

Ending at the second decimal place means the quotient is a whole number of hundredths.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #13 Convert to Algebra#11 Work Backwards

Decimals make this look harder than it is. Count everything in hundredths and the condition becomes a plain remainder question about whole numbers -- and the smallest amount to remove is that remainder.

3 · Execute4 carry out the plan

1Count in hundredths

#8 Analyze the Units 6.NS.B.3

95 is 9500 hundredths, and the amount removed is a whole number of hundredths too.

95 = 9500 \times 0.01

Everything is now a whole number.

2Say what 'ends at the second place' means

#13 Convert to Algebra 6.NS.B.2

The quotient must be a whole number of hundredths, so what is left must be 41 times a whole number of hundredths -- that is, a multiple of 41 in hundredths.

(9500 - \square) \div 41 = \text{whole}

We need a multiple of 41.

3Divide 9500 by 41 and keep the remainder

#11 Work Backwards 6.NS.B.2

9500 leaves 29 over, so removing 29 hundredths lands exactly on the multiple below.

9500 \div 41 \rightarrow \text{remainder }29

29 hundredths is what is in the way.

4Write it back as a decimal

#8 Analyze the Units 6.NS.B.3

29 hundredths is 0.29, and nothing smaller works: anything less leaves a remainder behind.

29 \times 0.01 = 0.29

Take off 0.29.

Answer: 0.29

4 · Reviewdoes it hold up?

95 minus 0.29 is 94.71, and dividing that by 41 gives exactly 2.31 -- two decimal places and no remainder.

Another way: Trying 0.01, 0.02 and so on in order reaches 0.29 after 29 attempts; the remainder finds it in one.

Standardsmin grade 6

6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Moving between a decimal and a whole number of hundredths.
6.NS.B.2 Fluently divide multi-digit numbers — Using the remainder of a whole-number division.

💡Takeaway. Pick a unit small enough and the decimals turn into whole numbers. Then the leftover you need is just a remainder.

Count in hundredths and an exact division is a remainder question · 12 practice problems

Generated variants — 12

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints