← The reduced denominator is what the box must supply · Divisibility and Remainder Reasoning

The reduced denominator is what the box must supply · 12 practice problems

6.NS.A.16.NS.B.4

Generated variants — 12

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

Variant 1 easy answer: 5

Find the natural number for $\blacksquare$ that makes the result the smallest possible whole number.

$6\frac{2}{5} \div 4 \times \blacksquare$

Show solution

1 · Understandwhat's really being asked

The expression 6 and 2/5 divided by 4, times a box, has to come out a whole number. We want the smallest natural number the box can hold.

Givens

The mixed number is 6 and 2/5.
It is divided by 4 before the box multiplies it.

Unknowns

The smallest natural number that makes the result whole.

Constraints

The box holds a natural number, so 0 and fractions are out.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #9 Solve an Easier Related Problem

Two operations are waiting on the box, so do the one that does not involve it first. Once the division leaves a single fraction, the question is only about what its denominator still needs.

3 · Execute4 carry out the plan

1Write the mixed number as one fraction

#7 Identify Subproblems 6.NS.A.1

6 wholes are 30 5ths, and 2 more makes 32.

6\frac{2}{5} = \frac{32}{5}

One fraction is easier to divide than a mixed number.

2Divide by 4

#7 Identify Subproblems 6.NS.A.1

Dividing by 4 multiplies the denominator by 4, and then the fraction reduces as far as it will go.

\frac{32}{5} \div 4 = \frac{32}{20} = \frac{8}{5}

Everything before the box is now 8/5.

3Ask what the box has to supply

#9 Solve an Easier Related Problem 6.NS.B.4

8/5 times the box is whole exactly when the box cancels the 5. Since 8 and 5 share no factor left, nothing in the numerator helps.

\frac{8}{5} \times \blacksquare = \text{whole}

The box has to be a multiple of 5.

4Take the smallest such multiple

#9 Solve an Easier Related Problem 6.NS.B.4

Multiples of 5 all work, and the smallest natural one is 5 itself.

\blacksquare = 5,\quad \frac{8}{5} \times 5 = 8

5 gives 8, and nothing smaller works.

Answer: 5

4 · Reviewdoes it hold up?

Try 4: 8/5 times 4 is 32/5, not a whole number. 5 is the first one that lands.

Another way: Reaching for 20 -- the denominator before reducing -- also gives a whole number, but it is 20, larger than 5. Reducing first is what makes the answer smallest.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Collapsing the mixed number and the division into one fraction.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Reading the reduced denominator as the factor the box must supply.

💡Takeaway. Simplify first, then look at the denominator: it names exactly what the missing number has to cancel.

Variant 2 easy answer: 8

Find the natural number for $\blacksquare$ that makes the result the smallest possible whole number.

$1\frac{7}{8} \div 5 \times \blacksquare$

Show solution

1 · Understandwhat's really being asked

The expression 1 and 7/8 divided by 5, times a box, has to come out a whole number. We want the smallest natural number the box can hold.

Givens

The mixed number is 1 and 7/8.
It is divided by 5 before the box multiplies it.

Unknowns

The smallest natural number that makes the result whole.

Constraints

The box holds a natural number, so 0 and fractions are out.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #9 Solve an Easier Related Problem

Two operations are waiting on the box, so do the one that does not involve it first. Once the division leaves a single fraction, the question is only about what its denominator still needs.

3 · Execute4 carry out the plan

1Write the mixed number as one fraction

#7 Identify Subproblems 6.NS.A.1

1 wholes are 8 8ths, and 7 more makes 15.

1\frac{7}{8} = \frac{15}{8}

One fraction is easier to divide than a mixed number.

2Divide by 5

#7 Identify Subproblems 6.NS.A.1

Dividing by 5 multiplies the denominator by 5, and then the fraction reduces as far as it will go.

\frac{15}{8} \div 5 = \frac{15}{40} = \frac{3}{8}

Everything before the box is now 3/8.

3Ask what the box has to supply

#9 Solve an Easier Related Problem 6.NS.B.4

3/8 times the box is whole exactly when the box cancels the 8. Since 3 and 8 share no factor left, nothing in the numerator helps.

\frac{3}{8} \times \blacksquare = \text{whole}

The box has to be a multiple of 8.

4Take the smallest such multiple

#9 Solve an Easier Related Problem 6.NS.B.4

Multiples of 8 all work, and the smallest natural one is 8 itself.

\blacksquare = 8,\quad \frac{3}{8} \times 8 = 3

8 gives 3, and nothing smaller works.

Answer: 8

4 · Reviewdoes it hold up?

Try 7: 3/8 times 7 is 21/8, not a whole number. 8 is the first one that lands.

Another way: Reaching for 40 -- the denominator before reducing -- also gives a whole number, but it is 40, larger than 8. Reducing first is what makes the answer smallest.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Collapsing the mixed number and the division into one fraction.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Reading the reduced denominator as the factor the box must supply.

💡Takeaway. Simplify first, then look at the denominator: it names exactly what the missing number has to cancel.

Variant 3 easy answer: 8

Find the natural number for $\blacksquare$ that makes the result the smallest possible whole number.

$3\frac{1}{8} \div 5 \times \blacksquare$

Show solution

1 · Understandwhat's really being asked

The expression 3 and 1/8 divided by 5, times a box, has to come out a whole number. We want the smallest natural number the box can hold.

Givens

The mixed number is 3 and 1/8.
It is divided by 5 before the box multiplies it.

Unknowns

The smallest natural number that makes the result whole.

Constraints

The box holds a natural number, so 0 and fractions are out.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #9 Solve an Easier Related Problem

Two operations are waiting on the box, so do the one that does not involve it first. Once the division leaves a single fraction, the question is only about what its denominator still needs.

3 · Execute4 carry out the plan

1Write the mixed number as one fraction

#7 Identify Subproblems 6.NS.A.1

3 wholes are 24 8ths, and 1 more makes 25.

3\frac{1}{8} = \frac{25}{8}

One fraction is easier to divide than a mixed number.

2Divide by 5

#7 Identify Subproblems 6.NS.A.1

Dividing by 5 multiplies the denominator by 5, and then the fraction reduces as far as it will go.

\frac{25}{8} \div 5 = \frac{25}{40} = \frac{5}{8}

Everything before the box is now 5/8.

3Ask what the box has to supply

#9 Solve an Easier Related Problem 6.NS.B.4

5/8 times the box is whole exactly when the box cancels the 8. Since 5 and 8 share no factor left, nothing in the numerator helps.

\frac{5}{8} \times \blacksquare = \text{whole}

The box has to be a multiple of 8.

4Take the smallest such multiple

#9 Solve an Easier Related Problem 6.NS.B.4

Multiples of 8 all work, and the smallest natural one is 8 itself.

\blacksquare = 8,\quad \frac{5}{8} \times 8 = 5

8 gives 5, and nothing smaller works.

Answer: 8

4 · Reviewdoes it hold up?

Try 7: 5/8 times 7 is 35/8, not a whole number. 8 is the first one that lands.

Another way: Reaching for 40 -- the denominator before reducing -- also gives a whole number, but it is 40, larger than 8. Reducing first is what makes the answer smallest.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Collapsing the mixed number and the division into one fraction.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Reading the reduced denominator as the factor the box must supply.

💡Takeaway. Simplify first, then look at the denominator: it names exactly what the missing number has to cancel.

Variant 4 easy answer: 9

Find the natural number for $\blacksquare$ that makes the result the smallest possible whole number.

$2\frac{2}{9} \div 5 \times \blacksquare$

Show solution

1 · Understandwhat's really being asked

The expression 2 and 2/9 divided by 5, times a box, has to come out a whole number. We want the smallest natural number the box can hold.

Givens

The mixed number is 2 and 2/9.
It is divided by 5 before the box multiplies it.

Unknowns

The smallest natural number that makes the result whole.

Constraints

The box holds a natural number, so 0 and fractions are out.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #9 Solve an Easier Related Problem

Two operations are waiting on the box, so do the one that does not involve it first. Once the division leaves a single fraction, the question is only about what its denominator still needs.

3 · Execute4 carry out the plan

1Write the mixed number as one fraction

#7 Identify Subproblems 6.NS.A.1

2 wholes are 18 9ths, and 2 more makes 20.

2\frac{2}{9} = \frac{20}{9}

One fraction is easier to divide than a mixed number.

2Divide by 5

#7 Identify Subproblems 6.NS.A.1

Dividing by 5 multiplies the denominator by 5, and then the fraction reduces as far as it will go.

\frac{20}{9} \div 5 = \frac{20}{45} = \frac{4}{9}

Everything before the box is now 4/9.

3Ask what the box has to supply

#9 Solve an Easier Related Problem 6.NS.B.4

4/9 times the box is whole exactly when the box cancels the 9. Since 4 and 9 share no factor left, nothing in the numerator helps.

\frac{4}{9} \times \blacksquare = \text{whole}

The box has to be a multiple of 9.

4Take the smallest such multiple

#9 Solve an Easier Related Problem 6.NS.B.4

Multiples of 9 all work, and the smallest natural one is 9 itself.

\blacksquare = 9,\quad \frac{4}{9} \times 9 = 4

9 gives 4, and nothing smaller works.

Answer: 9

4 · Reviewdoes it hold up?

Try 8: 4/9 times 8 is 32/9, not a whole number. 9 is the first one that lands.

Another way: Reaching for 45 -- the denominator before reducing -- also gives a whole number, but it is 45, larger than 9. Reducing first is what makes the answer smallest.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Collapsing the mixed number and the division into one fraction.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Reading the reduced denominator as the factor the box must supply.

💡Takeaway. Simplify first, then look at the denominator: it names exactly what the missing number has to cancel.

Variant 5 medium answer: 10

Find the natural number for $\blacksquare$ that makes the result the smallest possible whole number.

$3\frac{3}{10} \div 11 \times \blacksquare$

Show solution

1 · Understandwhat's really being asked

The expression 3 and 3/10 divided by 11, times a box, has to come out a whole number. We want the smallest natural number the box can hold.

Givens

The mixed number is 3 and 3/10.
It is divided by 11 before the box multiplies it.

Unknowns

The smallest natural number that makes the result whole.

Constraints

The box holds a natural number, so 0 and fractions are out.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #9 Solve an Easier Related Problem

Two operations are waiting on the box, so do the one that does not involve it first. Once the division leaves a single fraction, the question is only about what its denominator still needs.

3 · Execute4 carry out the plan

1Write the mixed number as one fraction

#7 Identify Subproblems 6.NS.A.1

3 wholes are 30 10ths, and 3 more makes 33.

3\frac{3}{10} = \frac{33}{10}

One fraction is easier to divide than a mixed number.

2Divide by 11

#7 Identify Subproblems 6.NS.A.1

Dividing by 11 multiplies the denominator by 11, and then the fraction reduces as far as it will go.

\frac{33}{10} \div 11 = \frac{33}{110} = \frac{3}{10}

Everything before the box is now 3/10.

3Ask what the box has to supply

#9 Solve an Easier Related Problem 6.NS.B.4

3/10 times the box is whole exactly when the box cancels the 10. Since 3 and 10 share no factor left, nothing in the numerator helps.

\frac{3}{10} \times \blacksquare = \text{whole}

The box has to be a multiple of 10.

4Take the smallest such multiple

#9 Solve an Easier Related Problem 6.NS.B.4

Multiples of 10 all work, and the smallest natural one is 10 itself.

\blacksquare = 10,\quad \frac{3}{10} \times 10 = 3

10 gives 3, and nothing smaller works.

Answer: 10

4 · Reviewdoes it hold up?

Try 9: 3/10 times 9 is 27/10, not a whole number. 10 is the first one that lands.

Another way: Reaching for 110 -- the denominator before reducing -- also gives a whole number, but it is 110, larger than 10. Reducing first is what makes the answer smallest.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Collapsing the mixed number and the division into one fraction.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Reading the reduced denominator as the factor the box must supply.

💡Takeaway. Simplify first, then look at the denominator: it names exactly what the missing number has to cancel.

Variant 6 medium answer: 6

Find the natural number for $\blacksquare$ that makes the result the smallest possible whole number.

$1\frac{5}{6} \div 11 \times \blacksquare$

Show solution

1 · Understandwhat's really being asked

The expression 1 and 5/6 divided by 11, times a box, has to come out a whole number. We want the smallest natural number the box can hold.

Givens

The mixed number is 1 and 5/6.
It is divided by 11 before the box multiplies it.

Unknowns

The smallest natural number that makes the result whole.

Constraints

The box holds a natural number, so 0 and fractions are out.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #9 Solve an Easier Related Problem

Two operations are waiting on the box, so do the one that does not involve it first. Once the division leaves a single fraction, the question is only about what its denominator still needs.

3 · Execute4 carry out the plan

1Write the mixed number as one fraction

#7 Identify Subproblems 6.NS.A.1

1 wholes are 6 6ths, and 5 more makes 11.

1\frac{5}{6} = \frac{11}{6}

One fraction is easier to divide than a mixed number.

2Divide by 11

#7 Identify Subproblems 6.NS.A.1

Dividing by 11 multiplies the denominator by 11, and then the fraction reduces as far as it will go.

\frac{11}{6} \div 11 = \frac{11}{66} = \frac{1}{6}

Everything before the box is now 1/6.

3Ask what the box has to supply

#9 Solve an Easier Related Problem 6.NS.B.4

1/6 times the box is whole exactly when the box cancels the 6. Since 1 and 6 share no factor left, nothing in the numerator helps.

\frac{1}{6} \times \blacksquare = \text{whole}

The box has to be a multiple of 6.

4Take the smallest such multiple

#9 Solve an Easier Related Problem 6.NS.B.4

Multiples of 6 all work, and the smallest natural one is 6 itself.

\blacksquare = 6,\quad \frac{1}{6} \times 6 = 1

6 gives 1, and nothing smaller works.

Answer: 6

4 · Reviewdoes it hold up?

Try 5: 1/6 times 5 is 5/6, not a whole number. 6 is the first one that lands.

Another way: Reaching for 66 -- the denominator before reducing -- also gives a whole number, but it is 66, larger than 6. Reducing first is what makes the answer smallest.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Collapsing the mixed number and the division into one fraction.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Reading the reduced denominator as the factor the box must supply.

💡Takeaway. Simplify first, then look at the denominator: it names exactly what the missing number has to cancel.

Variant 7 medium answer: 10

Find the natural number for $\blacksquare$ that makes the result the smallest possible whole number.

$1\frac{1}{10} \div 11 \times \blacksquare$

Show solution

1 · Understandwhat's really being asked

The expression 1 and 1/10 divided by 11, times a box, has to come out a whole number. We want the smallest natural number the box can hold.

Givens

The mixed number is 1 and 1/10.
It is divided by 11 before the box multiplies it.

Unknowns

The smallest natural number that makes the result whole.

Constraints

The box holds a natural number, so 0 and fractions are out.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #9 Solve an Easier Related Problem

Two operations are waiting on the box, so do the one that does not involve it first. Once the division leaves a single fraction, the question is only about what its denominator still needs.

3 · Execute4 carry out the plan

1Write the mixed number as one fraction

#7 Identify Subproblems 6.NS.A.1

1 wholes are 10 10ths, and 1 more makes 11.

1\frac{1}{10} = \frac{11}{10}

One fraction is easier to divide than a mixed number.

2Divide by 11

#7 Identify Subproblems 6.NS.A.1

Dividing by 11 multiplies the denominator by 11, and then the fraction reduces as far as it will go.

\frac{11}{10} \div 11 = \frac{11}{110} = \frac{1}{10}

Everything before the box is now 1/10.

3Ask what the box has to supply

#9 Solve an Easier Related Problem 6.NS.B.4

1/10 times the box is whole exactly when the box cancels the 10. Since 1 and 10 share no factor left, nothing in the numerator helps.

\frac{1}{10} \times \blacksquare = \text{whole}

The box has to be a multiple of 10.

4Take the smallest such multiple

#9 Solve an Easier Related Problem 6.NS.B.4

Multiples of 10 all work, and the smallest natural one is 10 itself.

\blacksquare = 10,\quad \frac{1}{10} \times 10 = 1

10 gives 1, and nothing smaller works.

Answer: 10

4 · Reviewdoes it hold up?

Try 9: 1/10 times 9 is 9/10, not a whole number. 10 is the first one that lands.

Another way: Reaching for 110 -- the denominator before reducing -- also gives a whole number, but it is 110, larger than 10. Reducing first is what makes the answer smallest.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Collapsing the mixed number and the division into one fraction.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Reading the reduced denominator as the factor the box must supply.

💡Takeaway. Simplify first, then look at the denominator: it names exactly what the missing number has to cancel.

Variant 8 medium answer: 4

Find the natural number for $\blacksquare$ that makes the result the smallest possible whole number.

$2\frac{3}{4} \div 11 \times \blacksquare$

Show solution

1 · Understandwhat's really being asked

The expression 2 and 3/4 divided by 11, times a box, has to come out a whole number. We want the smallest natural number the box can hold.

Givens

The mixed number is 2 and 3/4.
It is divided by 11 before the box multiplies it.

Unknowns

The smallest natural number that makes the result whole.

Constraints

The box holds a natural number, so 0 and fractions are out.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #9 Solve an Easier Related Problem

Two operations are waiting on the box, so do the one that does not involve it first. Once the division leaves a single fraction, the question is only about what its denominator still needs.

3 · Execute4 carry out the plan

1Write the mixed number as one fraction

#7 Identify Subproblems 6.NS.A.1

2 wholes are 8 4ths, and 3 more makes 11.

2\frac{3}{4} = \frac{11}{4}

One fraction is easier to divide than a mixed number.

2Divide by 11

#7 Identify Subproblems 6.NS.A.1

Dividing by 11 multiplies the denominator by 11, and then the fraction reduces as far as it will go.

\frac{11}{4} \div 11 = \frac{11}{44} = \frac{1}{4}

Everything before the box is now 1/4.

3Ask what the box has to supply

#9 Solve an Easier Related Problem 6.NS.B.4

1/4 times the box is whole exactly when the box cancels the 4. Since 1 and 4 share no factor left, nothing in the numerator helps.

\frac{1}{4} \times \blacksquare = \text{whole}

The box has to be a multiple of 4.

4Take the smallest such multiple

#9 Solve an Easier Related Problem 6.NS.B.4

Multiples of 4 all work, and the smallest natural one is 4 itself.

\blacksquare = 4,\quad \frac{1}{4} \times 4 = 1

4 gives 1, and nothing smaller works.

Answer: 4

4 · Reviewdoes it hold up?

Try 3: 1/4 times 3 is 3/4, not a whole number. 4 is the first one that lands.

Another way: Reaching for 44 -- the denominator before reducing -- also gives a whole number, but it is 44, larger than 4. Reducing first is what makes the answer smallest.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Collapsing the mixed number and the division into one fraction.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Reading the reduced denominator as the factor the box must supply.

💡Takeaway. Simplify first, then look at the denominator: it names exactly what the missing number has to cancel.

Variant 9 hard answer: 4

Find the natural number for $\blacksquare$ that makes the result the smallest possible whole number.

$5\frac{3}{4} \div 23 \times \blacksquare$

Show solution

1 · Understandwhat's really being asked

The expression 5 and 3/4 divided by 23, times a box, has to come out a whole number. We want the smallest natural number the box can hold.

Givens

The mixed number is 5 and 3/4.
It is divided by 23 before the box multiplies it.

Unknowns

The smallest natural number that makes the result whole.

Constraints

The box holds a natural number, so 0 and fractions are out.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #9 Solve an Easier Related Problem

Two operations are waiting on the box, so do the one that does not involve it first. Once the division leaves a single fraction, the question is only about what its denominator still needs.

3 · Execute4 carry out the plan

1Write the mixed number as one fraction

#7 Identify Subproblems 6.NS.A.1

5 wholes are 20 4ths, and 3 more makes 23.

5\frac{3}{4} = \frac{23}{4}

One fraction is easier to divide than a mixed number.

2Divide by 23

#7 Identify Subproblems 6.NS.A.1

Dividing by 23 multiplies the denominator by 23, and then the fraction reduces as far as it will go.

\frac{23}{4} \div 23 = \frac{23}{92} = \frac{1}{4}

Everything before the box is now 1/4.

3Ask what the box has to supply

#9 Solve an Easier Related Problem 6.NS.B.4

1/4 times the box is whole exactly when the box cancels the 4. Since 1 and 4 share no factor left, nothing in the numerator helps.

\frac{1}{4} \times \blacksquare = \text{whole}

The box has to be a multiple of 4.

4Take the smallest such multiple

#9 Solve an Easier Related Problem 6.NS.B.4

Multiples of 4 all work, and the smallest natural one is 4 itself.

\blacksquare = 4,\quad \frac{1}{4} \times 4 = 1

4 gives 1, and nothing smaller works.

Answer: 4

4 · Reviewdoes it hold up?

Try 3: 1/4 times 3 is 3/4, not a whole number. 4 is the first one that lands.

Another way: Reaching for 92 -- the denominator before reducing -- also gives a whole number, but it is 92, larger than 4. Reducing first is what makes the answer smallest.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Collapsing the mixed number and the division into one fraction.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Reading the reduced denominator as the factor the box must supply.

💡Takeaway. Simplify first, then look at the denominator: it names exactly what the missing number has to cancel.

Variant 10 hard answer: 6

Find the natural number for $\blacksquare$ that makes the result the smallest possible whole number.

$4\frac{1}{6} \div 25 \times \blacksquare$

Show solution

1 · Understandwhat's really being asked

The expression 4 and 1/6 divided by 25, times a box, has to come out a whole number. We want the smallest natural number the box can hold.

Givens

The mixed number is 4 and 1/6.
It is divided by 25 before the box multiplies it.

Unknowns

The smallest natural number that makes the result whole.

Constraints

The box holds a natural number, so 0 and fractions are out.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #9 Solve an Easier Related Problem

Two operations are waiting on the box, so do the one that does not involve it first. Once the division leaves a single fraction, the question is only about what its denominator still needs.

3 · Execute4 carry out the plan

1Write the mixed number as one fraction

#7 Identify Subproblems 6.NS.A.1

4 wholes are 24 6ths, and 1 more makes 25.

4\frac{1}{6} = \frac{25}{6}

One fraction is easier to divide than a mixed number.

2Divide by 25

#7 Identify Subproblems 6.NS.A.1

Dividing by 25 multiplies the denominator by 25, and then the fraction reduces as far as it will go.

\frac{25}{6} \div 25 = \frac{25}{150} = \frac{1}{6}

Everything before the box is now 1/6.

3Ask what the box has to supply

#9 Solve an Easier Related Problem 6.NS.B.4

1/6 times the box is whole exactly when the box cancels the 6. Since 1 and 6 share no factor left, nothing in the numerator helps.

\frac{1}{6} \times \blacksquare = \text{whole}

The box has to be a multiple of 6.

4Take the smallest such multiple

#9 Solve an Easier Related Problem 6.NS.B.4

Multiples of 6 all work, and the smallest natural one is 6 itself.

\blacksquare = 6,\quad \frac{1}{6} \times 6 = 1

6 gives 1, and nothing smaller works.

Answer: 6

4 · Reviewdoes it hold up?

Try 5: 1/6 times 5 is 5/6, not a whole number. 6 is the first one that lands.

Another way: Reaching for 150 -- the denominator before reducing -- also gives a whole number, but it is 150, larger than 6. Reducing first is what makes the answer smallest.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Collapsing the mixed number and the division into one fraction.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Reading the reduced denominator as the factor the box must supply.

💡Takeaway. Simplify first, then look at the denominator: it names exactly what the missing number has to cancel.

Variant 11 hard answer: 14

Find the natural number for $\blacksquare$ that makes the result the smallest possible whole number.

$2\frac{1}{14} \div 29 \times \blacksquare$

Show solution

1 · Understandwhat's really being asked

The expression 2 and 1/14 divided by 29, times a box, has to come out a whole number. We want the smallest natural number the box can hold.

Givens

The mixed number is 2 and 1/14.
It is divided by 29 before the box multiplies it.

Unknowns

The smallest natural number that makes the result whole.

Constraints

The box holds a natural number, so 0 and fractions are out.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #9 Solve an Easier Related Problem

Two operations are waiting on the box, so do the one that does not involve it first. Once the division leaves a single fraction, the question is only about what its denominator still needs.

3 · Execute4 carry out the plan

1Write the mixed number as one fraction

#7 Identify Subproblems 6.NS.A.1

2 wholes are 28 14ths, and 1 more makes 29.

2\frac{1}{14} = \frac{29}{14}

One fraction is easier to divide than a mixed number.

2Divide by 29

#7 Identify Subproblems 6.NS.A.1

Dividing by 29 multiplies the denominator by 29, and then the fraction reduces as far as it will go.

\frac{29}{14} \div 29 = \frac{29}{406} = \frac{1}{14}

Everything before the box is now 1/14.

3Ask what the box has to supply

#9 Solve an Easier Related Problem 6.NS.B.4

1/14 times the box is whole exactly when the box cancels the 14. Since 1 and 14 share no factor left, nothing in the numerator helps.

\frac{1}{14} \times \blacksquare = \text{whole}

The box has to be a multiple of 14.

4Take the smallest such multiple

#9 Solve an Easier Related Problem 6.NS.B.4

Multiples of 14 all work, and the smallest natural one is 14 itself.

\blacksquare = 14,\quad \frac{1}{14} \times 14 = 1

14 gives 1, and nothing smaller works.

Answer: 14

4 · Reviewdoes it hold up?

Try 13: 1/14 times 13 is 13/14, not a whole number. 14 is the first one that lands.

Another way: Reaching for 406 -- the denominator before reducing -- also gives a whole number, but it is 406, larger than 14. Reducing first is what makes the answer smallest.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Collapsing the mixed number and the division into one fraction.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Reading the reduced denominator as the factor the box must supply.

💡Takeaway. Simplify first, then look at the denominator: it names exactly what the missing number has to cancel.

Variant 12 hard answer: 12

Find the natural number for $\blacksquare$ that makes the result the smallest possible whole number.

$2\frac{5}{12} \div 29 \times \blacksquare$

Show solution

1 · Understandwhat's really being asked

The expression 2 and 5/12 divided by 29, times a box, has to come out a whole number. We want the smallest natural number the box can hold.

Givens

The mixed number is 2 and 5/12.
It is divided by 29 before the box multiplies it.

Unknowns

The smallest natural number that makes the result whole.

Constraints

The box holds a natural number, so 0 and fractions are out.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #9 Solve an Easier Related Problem

Two operations are waiting on the box, so do the one that does not involve it first. Once the division leaves a single fraction, the question is only about what its denominator still needs.

3 · Execute4 carry out the plan

1Write the mixed number as one fraction

#7 Identify Subproblems 6.NS.A.1

2 wholes are 24 12ths, and 5 more makes 29.

2\frac{5}{12} = \frac{29}{12}

One fraction is easier to divide than a mixed number.

2Divide by 29

#7 Identify Subproblems 6.NS.A.1

Dividing by 29 multiplies the denominator by 29, and then the fraction reduces as far as it will go.

\frac{29}{12} \div 29 = \frac{29}{348} = \frac{1}{12}

Everything before the box is now 1/12.

3Ask what the box has to supply

#9 Solve an Easier Related Problem 6.NS.B.4

1/12 times the box is whole exactly when the box cancels the 12. Since 1 and 12 share no factor left, nothing in the numerator helps.

\frac{1}{12} \times \blacksquare = \text{whole}

The box has to be a multiple of 12.

4Take the smallest such multiple

#9 Solve an Easier Related Problem 6.NS.B.4

Multiples of 12 all work, and the smallest natural one is 12 itself.

\blacksquare = 12,\quad \frac{1}{12} \times 12 = 1

12 gives 1, and nothing smaller works.

Answer: 12

4 · Reviewdoes it hold up?

Try 11: 1/12 times 11 is 11/12, not a whole number. 12 is the first one that lands.

Another way: Reaching for 348 -- the denominator before reducing -- also gives a whole number, but it is 348, larger than 12. Reducing first is what makes the answer smallest.

Standardsmin grade 6

6.NS.A.1 Divide fractions by fractions — Collapsing the mixed number and the division into one fraction.
6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Reading the reduced denominator as the factor the box must supply.

💡Takeaway. Simplify first, then look at the denominator: it names exactly what the missing number has to cancel.