Practice · Smallest fraction: LCM of denominators over GCF · Sensim Math

Variant 1 easy

When a certain number is multiplied by $2\frac{4}{7}$ , the result is a natural number, and when the same number is divided by $\frac{2}{3}$ , the result is also a natural number. What is the smallest number it can be?

Your answer

Show solution

1 · Understandwhat's really being asked

A number times $2\frac{4}{7}$ is a whole number, and the same number divided by $\frac{2}{3}$ is a whole number. Find the smallest such number.

Givens

The number times $2\frac{4}{7}$ is a natural number.
The number divided by $\frac{2}{3}$ is a natural number.

Unknowns

The smallest possible value of the number.

Constraints

The number is positive, and it need not be a whole number itself.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra

Turn both conditions into multiplications by a fraction. Then write the unknown as a fraction in lowest terms and ask what its numerator and denominator must do for each product to come out whole.

3 · Execute3 carry out the plan

1Make both conditions multiplications

#7 Identify Subproblems 7.NS.A.2

Change the mixed number to an improper fraction, and replace dividing by

\frac{2}{3}

with multiplying by its reciprocal.

2\frac{4}{7} = \frac{18}{7}, \quad \div \frac{2}{3} = \times \frac{3}{2}

Now the number times

\frac{18}{7}

and times

\frac{3}{2}

must both be whole.

2Find what the numerator and denominator need

#13 Convert to Algebra 6.NS.B.4

Write the number as

\frac{p}{q}

in lowest terms. In

\frac{p}{q} \times \frac{18}{7}

the

7

can only cancel against

p

, and

q

can only cancel against

18

. So

p

is a multiple of

7

and

q

divides

18

. In the same way,

p

is a multiple of

2

and

q

divides

3

.

p: \text{multiple of } 7 \text{ and } 2, \quad q: \text{divisor of } 18 \text{ and } 3

The numerator is a common multiple; the denominator is a common divisor.

3Make the fraction as small as possible

#7 Identify Subproblems 6.NS.B.4

A fraction is smallest when its numerator is as small as possible and its denominator as large as possible: the least common multiple of

7

and

2

over the greatest common factor of

18

and

3

.

\frac{14}{3}

The LCM of

7

and

2

is

14

; the GCF of

18

and

3

is

3

.

Answer:

\frac{14}{3}

4 · Reviewdoes it hold up?

Check both products: $\frac{14}{3} \times \frac{18}{7} = \frac{252}{21} = 12$ and $\frac{14}{3} \times \frac{3}{2} = \frac{42}{6} = 7$ , both natural numbers.

Another way: Any valid number is $\frac{14}{3}$ times something: a smaller numerator would miss a factor $7$ or $2$ , and a larger denominator could not divide both $18$ and $3$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Taking the LCM of $7$ and $2$ and the GCF of $18$ and $3$.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Rewriting division by a fraction as multiplication and multiplying fractions.

💡Takeaway. To make two fraction products whole, the numerator collects the LCM and the denominator keeps the GCF.

Variant 2 easy

When a certain number is multiplied by $1\frac{7}{8}$ , the result is a natural number, and when the same number is divided by $\frac{5}{12}$ , the result is also a natural number. What is the smallest number it can be?

Your answer

Show solution

1 · Understandwhat's really being asked

A number times $1\frac{7}{8}$ is a whole number, and the same number divided by $\frac{5}{12}$ is a whole number. Find the smallest such number.

Givens

The number times $1\frac{7}{8}$ is a natural number.
The number divided by $\frac{5}{12}$ is a natural number.

Unknowns

The smallest possible value of the number.

Constraints

The number is positive, and it need not be a whole number itself.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra

Turn both conditions into multiplications by a fraction. Then write the unknown as a fraction in lowest terms and ask what its numerator and denominator must do for each product to come out whole.

3 · Execute3 carry out the plan

1Make both conditions multiplications

#7 Identify Subproblems 7.NS.A.2

Change the mixed number to an improper fraction, and replace dividing by

\frac{5}{12}

with multiplying by its reciprocal.

1\frac{7}{8} = \frac{15}{8}, \quad \div \frac{5}{12} = \times \frac{12}{5}

Now the number times

\frac{15}{8}

and times

\frac{12}{5}

must both be whole.

2Find what the numerator and denominator need

#13 Convert to Algebra 6.NS.B.4

Write the number as

\frac{p}{q}

in lowest terms. In

\frac{p}{q} \times \frac{15}{8}

the

8

can only cancel against

p

, and

q

can only cancel against

15

. So

p

is a multiple of

8

and

q

divides

15

. In the same way,

p

is a multiple of

5

and

q

divides

12

.

p: \text{multiple of } 8 \text{ and } 5, \quad q: \text{divisor of } 15 \text{ and } 12

The numerator is a common multiple; the denominator is a common divisor.

3Make the fraction as small as possible

#7 Identify Subproblems 6.NS.B.4

A fraction is smallest when its numerator is as small as possible and its denominator as large as possible: the least common multiple of

8

and

5

over the greatest common factor of

15

and

12

.

\frac{40}{3}

The LCM of

8

and

5

is

40

; the GCF of

15

and

12

is

3

.

Answer:

\frac{40}{3}

4 · Reviewdoes it hold up?

Check both products: $\frac{40}{3} \times \frac{15}{8} = \frac{600}{24} = 25$ and $\frac{40}{3} \times \frac{12}{5} = \frac{480}{15} = 32$ , both natural numbers.

Another way: Any valid number is $\frac{40}{3}$ times something: a smaller numerator would miss a factor $8$ or $5$ , and a larger denominator could not divide both $15$ and $12$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Taking the LCM of $8$ and $5$ and the GCF of $15$ and $12$.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Rewriting division by a fraction as multiplication and multiplying fractions.

💡Takeaway. To make two fraction products whole, the numerator collects the LCM and the denominator keeps the GCF.

Variant 3 easy

When a certain number is multiplied by $1\frac{1}{9}$ , the result is a natural number, and when the same number is divided by $\frac{5}{14}$ , the result is also a natural number. What is the smallest number it can be?

Your answer

Show solution

1 · Understandwhat's really being asked

A number times $1\frac{1}{9}$ is a whole number, and the same number divided by $\frac{5}{14}$ is a whole number. Find the smallest such number.

Givens

The number times $1\frac{1}{9}$ is a natural number.
The number divided by $\frac{5}{14}$ is a natural number.

Unknowns

The smallest possible value of the number.

Constraints

The number is positive, and it need not be a whole number itself.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra

Turn both conditions into multiplications by a fraction. Then write the unknown as a fraction in lowest terms and ask what its numerator and denominator must do for each product to come out whole.

3 · Execute3 carry out the plan

1Make both conditions multiplications

#7 Identify Subproblems 7.NS.A.2

Change the mixed number to an improper fraction, and replace dividing by

\frac{5}{14}

with multiplying by its reciprocal.

1\frac{1}{9} = \frac{10}{9}, \quad \div \frac{5}{14} = \times \frac{14}{5}

Now the number times

\frac{10}{9}

and times

\frac{14}{5}

must both be whole.

2Find what the numerator and denominator need

#13 Convert to Algebra 6.NS.B.4

Write the number as

\frac{p}{q}

in lowest terms. In

\frac{p}{q} \times \frac{10}{9}

the

9

can only cancel against

p

, and

q

can only cancel against

10

. So

p

is a multiple of

9

and

q

divides

10

. In the same way,

p

is a multiple of

5

and

q

divides

14

.

p: \text{multiple of } 9 \text{ and } 5, \quad q: \text{divisor of } 10 \text{ and } 14

The numerator is a common multiple; the denominator is a common divisor.

3Make the fraction as small as possible

#7 Identify Subproblems 6.NS.B.4

A fraction is smallest when its numerator is as small as possible and its denominator as large as possible: the least common multiple of

9

and

5

over the greatest common factor of

10

and

14

.

\frac{45}{2}

The LCM of

9

and

5

is

45

; the GCF of

10

and

14

is

2

.

Answer:

\frac{45}{2}

4 · Reviewdoes it hold up?

Check both products: $\frac{45}{2} \times \frac{10}{9} = \frac{450}{18} = 25$ and $\frac{45}{2} \times \frac{14}{5} = \frac{630}{10} = 63$ , both natural numbers.

Another way: Any valid number is $\frac{45}{2}$ times something: a smaller numerator would miss a factor $9$ or $5$ , and a larger denominator could not divide both $10$ and $14$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Taking the LCM of $9$ and $5$ and the GCF of $10$ and $14$.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Rewriting division by a fraction as multiplication and multiplying fractions.

💡Takeaway. To make two fraction products whole, the numerator collects the LCM and the denominator keeps the GCF.

Variant 4 easy

When a certain number is multiplied by $1\frac{4}{5}$ , the result is a natural number, and when the same number is divided by $\frac{4}{15}$ , the result is also a natural number. What is the smallest number it can be?

Your answer

Show solution

1 · Understandwhat's really being asked

A number times $1\frac{4}{5}$ is a whole number, and the same number divided by $\frac{4}{15}$ is a whole number. Find the smallest such number.

Givens

The number times $1\frac{4}{5}$ is a natural number.
The number divided by $\frac{4}{15}$ is a natural number.

Unknowns

The smallest possible value of the number.

Constraints

The number is positive, and it need not be a whole number itself.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra

Turn both conditions into multiplications by a fraction. Then write the unknown as a fraction in lowest terms and ask what its numerator and denominator must do for each product to come out whole.

3 · Execute3 carry out the plan

1Make both conditions multiplications

#7 Identify Subproblems 7.NS.A.2

Change the mixed number to an improper fraction, and replace dividing by

\frac{4}{15}

with multiplying by its reciprocal.

1\frac{4}{5} = \frac{9}{5}, \quad \div \frac{4}{15} = \times \frac{15}{4}

Now the number times

\frac{9}{5}

and times

\frac{15}{4}

must both be whole.

2Find what the numerator and denominator need

#13 Convert to Algebra 6.NS.B.4

Write the number as

\frac{p}{q}

in lowest terms. In

\frac{p}{q} \times \frac{9}{5}

the

5

can only cancel against

p

, and

q

can only cancel against

9

. So

p

is a multiple of

5

and

q

divides

9

. In the same way,

p

is a multiple of

4

and

q

divides

15

.

p: \text{multiple of } 5 \text{ and } 4, \quad q: \text{divisor of } 9 \text{ and } 15

The numerator is a common multiple; the denominator is a common divisor.

3Make the fraction as small as possible

#7 Identify Subproblems 6.NS.B.4

A fraction is smallest when its numerator is as small as possible and its denominator as large as possible: the least common multiple of

5

and

4

over the greatest common factor of

9

and

15

.

\frac{20}{3}

The LCM of

5

and

4

is

20

; the GCF of

9

and

15

is

3

.

Answer:

\frac{20}{3}

4 · Reviewdoes it hold up?

Check both products: $\frac{20}{3} \times \frac{9}{5} = \frac{180}{15} = 12$ and $\frac{20}{3} \times \frac{15}{4} = \frac{300}{12} = 25$ , both natural numbers.

Another way: Any valid number is $\frac{20}{3}$ times something: a smaller numerator would miss a factor $5$ or $4$ , and a larger denominator could not divide both $9$ and $15$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Taking the LCM of $5$ and $4$ and the GCF of $9$ and $15$.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Rewriting division by a fraction as multiplication and multiplying fractions.

💡Takeaway. To make two fraction products whole, the numerator collects the LCM and the denominator keeps the GCF.

Variant 5 medium

When a certain number is multiplied by $3\frac{7}{11}$ , the result is a natural number, and when the same number is divided by $\frac{7}{16}$ , the result is also a natural number. What is the smallest number it can be?

Your answer

Show solution

1 · Understandwhat's really being asked

A number times $3\frac{7}{11}$ is a whole number, and the same number divided by $\frac{7}{16}$ is a whole number. Find the smallest such number.

Givens

The number times $3\frac{7}{11}$ is a natural number.
The number divided by $\frac{7}{16}$ is a natural number.

Unknowns

The smallest possible value of the number.

Constraints

The number is positive, and it need not be a whole number itself.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra

Turn both conditions into multiplications by a fraction. Then write the unknown as a fraction in lowest terms and ask what its numerator and denominator must do for each product to come out whole.

3 · Execute3 carry out the plan

1Make both conditions multiplications

#7 Identify Subproblems 7.NS.A.2

Change the mixed number to an improper fraction, and replace dividing by

\frac{7}{16}

with multiplying by its reciprocal.

3\frac{7}{11} = \frac{40}{11}, \quad \div \frac{7}{16} = \times \frac{16}{7}

Now the number times

\frac{40}{11}

and times

\frac{16}{7}

must both be whole.

2Find what the numerator and denominator need

#13 Convert to Algebra 6.NS.B.4

Write the number as

\frac{p}{q}

in lowest terms. In

\frac{p}{q} \times \frac{40}{11}

the

11

can only cancel against

p

, and

q

can only cancel against

40

. So

p

is a multiple of

11

and

q

divides

40

. In the same way,

p

is a multiple of

7

and

q

divides

16

.

p: \text{multiple of } 11 \text{ and } 7, \quad q: \text{divisor of } 40 \text{ and } 16

The numerator is a common multiple; the denominator is a common divisor.

3Make the fraction as small as possible

#7 Identify Subproblems 6.NS.B.4

A fraction is smallest when its numerator is as small as possible and its denominator as large as possible: the least common multiple of

11

and

7

over the greatest common factor of

40

and

16

.

\frac{77}{8}

The LCM of

11

and

7

is

77

; the GCF of

40

and

16

is

8

.

Answer:

\frac{77}{8}

4 · Reviewdoes it hold up?

Check both products: $\frac{77}{8} \times \frac{40}{11} = \frac{3080}{88} = 35$ and $\frac{77}{8} \times \frac{16}{7} = \frac{1232}{56} = 22$ , both natural numbers.

Another way: Any valid number is $\frac{77}{8}$ times something: a smaller numerator would miss a factor $11$ or $7$ , and a larger denominator could not divide both $40$ and $16$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Taking the LCM of $11$ and $7$ and the GCF of $40$ and $16$.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Rewriting division by a fraction as multiplication and multiplying fractions.

💡Takeaway. To make two fraction products whole, the numerator collects the LCM and the denominator keeps the GCF.

Variant 6 medium

When a certain number is multiplied by $2\frac{5}{6}$ , the result is a natural number, and when the same number is divided by $\frac{9}{17}$ , the result is also a natural number. What is the smallest number it can be?

Your answer

Show solution

1 · Understandwhat's really being asked

A number times $2\frac{5}{6}$ is a whole number, and the same number divided by $\frac{9}{17}$ is a whole number. Find the smallest such number.

Givens

The number times $2\frac{5}{6}$ is a natural number.
The number divided by $\frac{9}{17}$ is a natural number.

Unknowns

The smallest possible value of the number.

Constraints

The number is positive, and it need not be a whole number itself.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra

Turn both conditions into multiplications by a fraction. Then write the unknown as a fraction in lowest terms and ask what its numerator and denominator must do for each product to come out whole.

3 · Execute3 carry out the plan

1Make both conditions multiplications

#7 Identify Subproblems 7.NS.A.2

Change the mixed number to an improper fraction, and replace dividing by

\frac{9}{17}

with multiplying by its reciprocal.

2\frac{5}{6} = \frac{17}{6}, \quad \div \frac{9}{17} = \times \frac{17}{9}

Now the number times

\frac{17}{6}

and times

\frac{17}{9}

must both be whole.

2Find what the numerator and denominator need

#13 Convert to Algebra 6.NS.B.4

Write the number as

\frac{p}{q}

in lowest terms. In

\frac{p}{q} \times \frac{17}{6}

the

6

can only cancel against

p

, and

q

can only cancel against

17

. So

p

is a multiple of

6

and

q

divides

17

. In the same way,

p

is a multiple of

9

and

q

divides

17

.

p: \text{multiple of } 6 \text{ and } 9, \quad q: \text{divisor of } 17 \text{ and } 17

The numerator is a common multiple; the denominator is a common divisor.

3Make the fraction as small as possible

#7 Identify Subproblems 6.NS.B.4

A fraction is smallest when its numerator is as small as possible and its denominator as large as possible: the least common multiple of

6

and

9

over the greatest common factor of

17

and

17

.

\frac{18}{17}

The LCM of

6

and

9

is

18

; the GCF of

17

and

17

is

17

.

Answer:

\frac{18}{17}

4 · Reviewdoes it hold up?

Check both products: $\frac{18}{17} \times \frac{17}{6} = \frac{306}{102} = 3$ and $\frac{18}{17} \times \frac{17}{9} = \frac{306}{153} = 2$ , both natural numbers.

Another way: Any valid number is $\frac{18}{17}$ times something: a smaller numerator would miss a factor $6$ or $9$ , and a larger denominator could not divide both $17$ and $17$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Taking the LCM of $6$ and $9$ and the GCF of $17$ and $17$.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Rewriting division by a fraction as multiplication and multiplying fractions.

💡Takeaway. To make two fraction products whole, the numerator collects the LCM and the denominator keeps the GCF.

Variant 7 medium

When a certain number is multiplied by $1\frac{1}{5}$ , the result is a natural number, and when the same number is divided by $\frac{13}{18}$ , the result is also a natural number. What is the smallest number it can be?

Your answer

Show solution

1 · Understandwhat's really being asked

A number times $1\frac{1}{5}$ is a whole number, and the same number divided by $\frac{13}{18}$ is a whole number. Find the smallest such number.

Givens

The number times $1\frac{1}{5}$ is a natural number.
The number divided by $\frac{13}{18}$ is a natural number.

Unknowns

The smallest possible value of the number.

Constraints

The number is positive, and it need not be a whole number itself.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra

Turn both conditions into multiplications by a fraction. Then write the unknown as a fraction in lowest terms and ask what its numerator and denominator must do for each product to come out whole.

3 · Execute3 carry out the plan

1Make both conditions multiplications

#7 Identify Subproblems 7.NS.A.2

Change the mixed number to an improper fraction, and replace dividing by

\frac{13}{18}

with multiplying by its reciprocal.

1\frac{1}{5} = \frac{6}{5}, \quad \div \frac{13}{18} = \times \frac{18}{13}

Now the number times

\frac{6}{5}

and times

\frac{18}{13}

must both be whole.

2Find what the numerator and denominator need

#13 Convert to Algebra 6.NS.B.4

Write the number as

\frac{p}{q}

in lowest terms. In

\frac{p}{q} \times \frac{6}{5}

the

5

can only cancel against

p

, and

q

can only cancel against

6

. So

p

is a multiple of

5

and

q

divides

6

. In the same way,

p

is a multiple of

13

and

q

divides

18

.

p: \text{multiple of } 5 \text{ and } 13, \quad q: \text{divisor of } 6 \text{ and } 18

The numerator is a common multiple; the denominator is a common divisor.

3Make the fraction as small as possible

#7 Identify Subproblems 6.NS.B.4

A fraction is smallest when its numerator is as small as possible and its denominator as large as possible: the least common multiple of

5

and

13

over the greatest common factor of

6

and

18

.

\frac{65}{6}

The LCM of

5

and

13

is

65

; the GCF of

6

and

18

is

6

.

Answer:

\frac{65}{6}

4 · Reviewdoes it hold up?

Check both products: $\frac{65}{6} \times \frac{6}{5} = \frac{390}{30} = 13$ and $\frac{65}{6} \times \frac{18}{13} = \frac{1170}{78} = 15$ , both natural numbers.

Another way: Any valid number is $\frac{65}{6}$ times something: a smaller numerator would miss a factor $5$ or $13$ , and a larger denominator could not divide both $6$ and $18$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Taking the LCM of $5$ and $13$ and the GCF of $6$ and $18$.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Rewriting division by a fraction as multiplication and multiplying fractions.

💡Takeaway. To make two fraction products whole, the numerator collects the LCM and the denominator keeps the GCF.

Variant 8 hard

When a certain number is multiplied by $2\frac{8}{11}$ , the result is a natural number, and when the same number is divided by $\frac{13}{20}$ , the result is also a natural number. What is the smallest number it can be?

Your answer

Show solution

1 · Understandwhat's really being asked

A number times $2\frac{8}{11}$ is a whole number, and the same number divided by $\frac{13}{20}$ is a whole number. Find the smallest such number.

Givens

The number times $2\frac{8}{11}$ is a natural number.
The number divided by $\frac{13}{20}$ is a natural number.

Unknowns

The smallest possible value of the number.

Constraints

The number is positive, and it need not be a whole number itself.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra

Turn both conditions into multiplications by a fraction. Then write the unknown as a fraction in lowest terms and ask what its numerator and denominator must do for each product to come out whole.

3 · Execute3 carry out the plan

1Make both conditions multiplications

#7 Identify Subproblems 7.NS.A.2

Change the mixed number to an improper fraction, and replace dividing by

\frac{13}{20}

with multiplying by its reciprocal.

2\frac{8}{11} = \frac{30}{11}, \quad \div \frac{13}{20} = \times \frac{20}{13}

Now the number times

\frac{30}{11}

and times

\frac{20}{13}

must both be whole.

2Find what the numerator and denominator need

#13 Convert to Algebra 6.NS.B.4

Write the number as

\frac{p}{q}

in lowest terms. In

\frac{p}{q} \times \frac{30}{11}

the

11

can only cancel against

p

, and

q

can only cancel against

30

. So

p

is a multiple of

11

and

q

divides

30

. In the same way,

p

is a multiple of

13

and

q

divides

20

.

p: \text{multiple of } 11 \text{ and } 13, \quad q: \text{divisor of } 30 \text{ and } 20

The numerator is a common multiple; the denominator is a common divisor.

3Make the fraction as small as possible

#7 Identify Subproblems 6.NS.B.4

A fraction is smallest when its numerator is as small as possible and its denominator as large as possible: the least common multiple of

11

and

13

over the greatest common factor of

30

and

20

.

\frac{143}{10}

The LCM of

11

and

13

is

143

; the GCF of

30

and

20

is

10

.

Answer:

\frac{143}{10}

4 · Reviewdoes it hold up?

Check both products: $\frac{143}{10} \times \frac{30}{11} = \frac{4290}{110} = 39$ and $\frac{143}{10} \times \frac{20}{13} = \frac{2860}{130} = 22$ , both natural numbers.

Another way: Any valid number is $\frac{143}{10}$ times something: a smaller numerator would miss a factor $11$ or $13$ , and a larger denominator could not divide both $30$ and $20$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Taking the LCM of $11$ and $13$ and the GCF of $30$ and $20$.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Rewriting division by a fraction as multiplication and multiplying fractions.

💡Takeaway. To make two fraction products whole, the numerator collects the LCM and the denominator keeps the GCF.

Variant 9 hard

When a certain number is multiplied by $2\frac{1}{7}$ , the result is a natural number, and when the same number is divided by $\frac{11}{20}$ , the result is also a natural number. What is the smallest number it can be?

Your answer

Show solution

1 · Understandwhat's really being asked

A number times $2\frac{1}{7}$ is a whole number, and the same number divided by $\frac{11}{20}$ is a whole number. Find the smallest such number.

Givens

The number times $2\frac{1}{7}$ is a natural number.
The number divided by $\frac{11}{20}$ is a natural number.

Unknowns

The smallest possible value of the number.

Constraints

The number is positive, and it need not be a whole number itself.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra

Turn both conditions into multiplications by a fraction. Then write the unknown as a fraction in lowest terms and ask what its numerator and denominator must do for each product to come out whole.

3 · Execute3 carry out the plan

1Make both conditions multiplications

#7 Identify Subproblems 7.NS.A.2

Change the mixed number to an improper fraction, and replace dividing by

\frac{11}{20}

with multiplying by its reciprocal.

2\frac{1}{7} = \frac{15}{7}, \quad \div \frac{11}{20} = \times \frac{20}{11}

Now the number times

\frac{15}{7}

and times

\frac{20}{11}

must both be whole.

2Find what the numerator and denominator need

#13 Convert to Algebra 6.NS.B.4

Write the number as

\frac{p}{q}

in lowest terms. In

\frac{p}{q} \times \frac{15}{7}

the

7

can only cancel against

p

, and

q

can only cancel against

15

. So

p

is a multiple of

7

and

q

divides

15

. In the same way,

p

is a multiple of

11

and

q

divides

20

.

p: \text{multiple of } 7 \text{ and } 11, \quad q: \text{divisor of } 15 \text{ and } 20

The numerator is a common multiple; the denominator is a common divisor.

3Make the fraction as small as possible

#7 Identify Subproblems 6.NS.B.4

A fraction is smallest when its numerator is as small as possible and its denominator as large as possible: the least common multiple of

7

and

11

over the greatest common factor of

15

and

20

.

\frac{77}{5}

The LCM of

7

and

11

is

77

; the GCF of

15

and

20

is

5

.

Answer:

\frac{77}{5}

4 · Reviewdoes it hold up?

Check both products: $\frac{77}{5} \times \frac{15}{7} = \frac{1155}{35} = 33$ and $\frac{77}{5} \times \frac{20}{11} = \frac{1540}{55} = 28$ , both natural numbers.

Another way: Any valid number is $\frac{77}{5}$ times something: a smaller numerator would miss a factor $7$ or $11$ , and a larger denominator could not divide both $15$ and $20$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Taking the LCM of $7$ and $11$ and the GCF of $15$ and $20$.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Rewriting division by a fraction as multiplication and multiplying fractions.

💡Takeaway. To make two fraction products whole, the numerator collects the LCM and the denominator keeps the GCF.

Variant 10 hard

When a certain number is multiplied by $3\frac{1}{5}$ , the result is a natural number, and when the same number is divided by $\frac{11}{20}$ , the result is also a natural number. What is the smallest number it can be?

Your answer

Show solution

1 · Understandwhat's really being asked

A number times $3\frac{1}{5}$ is a whole number, and the same number divided by $\frac{11}{20}$ is a whole number. Find the smallest such number.

Givens

The number times $3\frac{1}{5}$ is a natural number.
The number divided by $\frac{11}{20}$ is a natural number.

Unknowns

The smallest possible value of the number.

Constraints

The number is positive, and it need not be a whole number itself.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra

Turn both conditions into multiplications by a fraction. Then write the unknown as a fraction in lowest terms and ask what its numerator and denominator must do for each product to come out whole.

3 · Execute3 carry out the plan

1Make both conditions multiplications

#7 Identify Subproblems 7.NS.A.2

Change the mixed number to an improper fraction, and replace dividing by

\frac{11}{20}

with multiplying by its reciprocal.

3\frac{1}{5} = \frac{16}{5}, \quad \div \frac{11}{20} = \times \frac{20}{11}

Now the number times

\frac{16}{5}

and times

\frac{20}{11}

must both be whole.

2Find what the numerator and denominator need

#13 Convert to Algebra 6.NS.B.4

Write the number as

\frac{p}{q}

in lowest terms. In

\frac{p}{q} \times \frac{16}{5}

the

5

can only cancel against

p

, and

q

can only cancel against

16

. So

p

is a multiple of

5

and

q

divides

16

. In the same way,

p

is a multiple of

11

and

q

divides

20

.

p: \text{multiple of } 5 \text{ and } 11, \quad q: \text{divisor of } 16 \text{ and } 20

The numerator is a common multiple; the denominator is a common divisor.

3Make the fraction as small as possible

#7 Identify Subproblems 6.NS.B.4

A fraction is smallest when its numerator is as small as possible and its denominator as large as possible: the least common multiple of

5

and

11

over the greatest common factor of

16

and

20

.

\frac{55}{4}

The LCM of

5

and

11

is

55

; the GCF of

16

and

20

is

4

.

Answer:

\frac{55}{4}

4 · Reviewdoes it hold up?

Check both products: $\frac{55}{4} \times \frac{16}{5} = \frac{880}{20} = 44$ and $\frac{55}{4} \times \frac{20}{11} = \frac{1100}{44} = 25$ , both natural numbers.

Another way: Any valid number is $\frac{55}{4}$ times something: a smaller numerator would miss a factor $5$ or $11$ , and a larger denominator could not divide both $16$ and $20$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Taking the LCM of $5$ and $11$ and the GCF of $16$ and $20$.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Rewriting division by a fraction as multiplication and multiplying fractions.

💡Takeaway. To make two fraction products whole, the numerator collects the LCM and the denominator keeps the GCF.

Variant 11 medium

When a certain number is multiplied by $1\frac{2}{3}$ , the result is a natural number, and when the same number is divided by $\frac{7}{20}$ , the result is also a natural number. What is the smallest number it can be?

Your answer

Show solution

1 · Understandwhat's really being asked

A number times $1\frac{2}{3}$ is a whole number, and the same number divided by $\frac{7}{20}$ is a whole number. Find the smallest such number.

Givens

The number times $1\frac{2}{3}$ is a natural number.
The number divided by $\frac{7}{20}$ is a natural number.

Unknowns

The smallest possible value of the number.

Constraints

The number is positive, and it need not be a whole number itself.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra

Turn both conditions into multiplications by a fraction. Then write the unknown as a fraction in lowest terms and ask what its numerator and denominator must do for each product to come out whole.

3 · Execute3 carry out the plan

1Make both conditions multiplications

#7 Identify Subproblems 7.NS.A.2

Change the mixed number to an improper fraction, and replace dividing by

\frac{7}{20}

with multiplying by its reciprocal.

1\frac{2}{3} = \frac{5}{3}, \quad \div \frac{7}{20} = \times \frac{20}{7}

Now the number times

\frac{5}{3}

and times

\frac{20}{7}

must both be whole.

2Find what the numerator and denominator need

#13 Convert to Algebra 6.NS.B.4

Write the number as

\frac{p}{q}

in lowest terms. In

\frac{p}{q} \times \frac{5}{3}

the

3

can only cancel against

p

, and

q

can only cancel against

5

. So

p

is a multiple of

3

and

q

divides

5

. In the same way,

p

is a multiple of

7

and

q

divides

20

.

p: \text{multiple of } 3 \text{ and } 7, \quad q: \text{divisor of } 5 \text{ and } 20

The numerator is a common multiple; the denominator is a common divisor.

3Make the fraction as small as possible

#7 Identify Subproblems 6.NS.B.4

A fraction is smallest when its numerator is as small as possible and its denominator as large as possible: the least common multiple of

3

and

7

over the greatest common factor of

5

and

20

.

\frac{21}{5}

The LCM of

3

and

7

is

21

; the GCF of

5

and

20

is

5

.

Answer:

\frac{21}{5}

4 · Reviewdoes it hold up?

Check both products: $\frac{21}{5} \times \frac{5}{3} = \frac{105}{15} = 7$ and $\frac{21}{5} \times \frac{20}{7} = \frac{420}{35} = 12$ , both natural numbers.

Another way: Any valid number is $\frac{21}{5}$ times something: a smaller numerator would miss a factor $3$ or $7$ , and a larger denominator could not divide both $5$ and $20$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Taking the LCM of $3$ and $7$ and the GCF of $5$ and $20$.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Rewriting division by a fraction as multiplication and multiplying fractions.

💡Takeaway. To make two fraction products whole, the numerator collects the LCM and the denominator keeps the GCF.

Variant 12 hard

When a certain number is multiplied by $3\frac{1}{8}$ , the result is a natural number, and when the same number is divided by $\frac{11}{20}$ , the result is also a natural number. What is the smallest number it can be?

Your answer

Show solution

1 · Understandwhat's really being asked

A number times $3\frac{1}{8}$ is a whole number, and the same number divided by $\frac{11}{20}$ is a whole number. Find the smallest such number.

Givens

The number times $3\frac{1}{8}$ is a natural number.
The number divided by $\frac{11}{20}$ is a natural number.

Unknowns

The smallest possible value of the number.

Constraints

The number is positive, and it need not be a whole number itself.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra

Turn both conditions into multiplications by a fraction. Then write the unknown as a fraction in lowest terms and ask what its numerator and denominator must do for each product to come out whole.

3 · Execute3 carry out the plan

1Make both conditions multiplications

#7 Identify Subproblems 7.NS.A.2

Change the mixed number to an improper fraction, and replace dividing by

\frac{11}{20}

with multiplying by its reciprocal.

3\frac{1}{8} = \frac{25}{8}, \quad \div \frac{11}{20} = \times \frac{20}{11}

Now the number times

\frac{25}{8}

and times

\frac{20}{11}

must both be whole.

2Find what the numerator and denominator need

#13 Convert to Algebra 6.NS.B.4

Write the number as

\frac{p}{q}

in lowest terms. In

\frac{p}{q} \times \frac{25}{8}

the

8

can only cancel against

p

, and

q

can only cancel against

25

. So

p

is a multiple of

8

and

q

divides

25

. In the same way,

p

is a multiple of

11

and

q

divides

20

.

p: \text{multiple of } 8 \text{ and } 11, \quad q: \text{divisor of } 25 \text{ and } 20

The numerator is a common multiple; the denominator is a common divisor.

3Make the fraction as small as possible

#7 Identify Subproblems 6.NS.B.4

A fraction is smallest when its numerator is as small as possible and its denominator as large as possible: the least common multiple of

8

and

11

over the greatest common factor of

25

and

20

.

\frac{88}{5}

The LCM of

8

and

11

is

88

; the GCF of

25

and

20

is

5

.

Answer:

\frac{88}{5}

4 · Reviewdoes it hold up?

Check both products: $\frac{88}{5} \times \frac{25}{8} = \frac{2200}{40} = 55$ and $\frac{88}{5} \times \frac{20}{11} = \frac{1760}{55} = 32$ , both natural numbers.

Another way: Any valid number is $\frac{88}{5}$ times something: a smaller numerator would miss a factor $8$ or $11$ , and a larger denominator could not divide both $25$ and $20$ .

Standardsmin grade 7

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Taking the LCM of $8$ and $11$ and the GCF of $25$ and $20$.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Rewriting division by a fraction as multiplication and multiplying fractions.

💡Takeaway. To make two fraction products whole, the numerator collects the LCM and the denominator keeps the GCF.

Smallest fraction: LCM of denominators over GCF · 12 practice problems

Generated variants — 12

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