Practice · Smallest square that is a common multiple · Sensim Math

Variant 1 easy

Natural numbers $a$ , $b$ , and $c$ satisfy $45a = 60b = c^2$ . Find the smallest possible value of $c$ .

Your answer

Show solution

1 · Understandwhat's really being asked

$c^2$ equals $45$ times a natural number and also $60$ times a natural number. Find the smallest natural number $c$ for which this can happen.

Givens

$45a = 60b = c^2$
$a$ , $b$ , $c$ are natural numbers.

Unknowns

The smallest possible $c$ .

Constraints

$c^2$ must be a multiple of $45$ and of $60$ .
$c^2$ is a perfect square, so every prime in it has an even exponent.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems

Rewrite $45$ and $60$ as products of prime powers. Then $c^2$ must contain the least common multiple, and on top of that every exponent must be even. Raise each odd exponent by one and take the square root.

3 · Execute4 carry out the plan

1Factor both numbers into primes

#15 Organize Information in More Ways 6.EE.A.1

Break each coefficient into prime powers.

45 = 3^2 \times 5, \quad 60 = 2^2 \times 3 \times 5

Now the exponents can be compared prime by prime.

2 $c^2$ is a common multiple

#7 Identify Subproblems 6.NS.B.4

Because

c^2 = 45a = 60b

, it is a multiple of both numbers, so it is a multiple of their least common multiple. Take the larger exponent of each prime.

2^2 \times 3^2 \times 5 = 180

c^2

is a multiple of

180

.

3Make every exponent even

#15 Organize Information in More Ways 6.EE.A.1

A perfect square has even exponents. The exponent

1

on

5

is odd, so raise it to

2

; the exponent

2

on

2

and the exponent

2

on

3

are already even. This is the smallest such change.

c^2 = 2^2 \times 3^2 \times 5^2 = 900

This is the smallest square that is a multiple of

180

.

4Take the square root

#7 Identify Subproblems 6.EE.A.1

Halve each exponent.

c = 2 \times 3 \times 5 = 30

The smallest

c

is

30

.

Answer: 30

4 · Reviewdoes it hold up?

$30^2 = 900$ , and $900 \div 45 = 20$ and $900 \div 60 = 15$ are both natural numbers, so $a = 20$ , $b = 15$ work.

Another way: $c$ itself must make $c^2$ divisible by $2^2$ , $3^2$ , $5^1$ : so $c$ needs at least $2^1$ , $3^1$ , $5^1$ , giving $c = 2 \times 3 \times 5 = 30$ .

Standardsmin grade 6

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the least common multiple of $45$ and $60$.
6.EE.A.1 Write and evaluate expressions with whole-number exponents — Writing the numbers with exponents, making the exponents even, and evaluating $2 \times 3 \times 5$.

💡Takeaway. For a square, look at the exponents: every one must be even.

Variant 2 easy

Natural numbers $a$ , $b$ , and $c$ satisfy $72a = 120b = c^2$ . Find the smallest possible value of $c$ .

Your answer

Show solution

1 · Understandwhat's really being asked

$c^2$ equals $72$ times a natural number and also $120$ times a natural number. Find the smallest natural number $c$ for which this can happen.

Givens

$72a = 120b = c^2$
$a$ , $b$ , $c$ are natural numbers.

Unknowns

The smallest possible $c$ .

Constraints

$c^2$ must be a multiple of $72$ and of $120$ .
$c^2$ is a perfect square, so every prime in it has an even exponent.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems

Rewrite $72$ and $120$ as products of prime powers. Then $c^2$ must contain the least common multiple, and on top of that every exponent must be even. Raise each odd exponent by one and take the square root.

3 · Execute4 carry out the plan

1Factor both numbers into primes

#15 Organize Information in More Ways 6.EE.A.1

Break each coefficient into prime powers.

72 = 2^3 \times 3^2, \quad 120 = 2^3 \times 3 \times 5

Now the exponents can be compared prime by prime.

2 $c^2$ is a common multiple

#7 Identify Subproblems 6.NS.B.4

Because

c^2 = 72a = 120b

, it is a multiple of both numbers, so it is a multiple of their least common multiple. Take the larger exponent of each prime.

2^3 \times 3^2 \times 5 = 360

c^2

is a multiple of

360

.

3Make every exponent even

#15 Organize Information in More Ways 6.EE.A.1

A perfect square has even exponents. The exponent

1

on

5

is odd, so raise it to

2

; the exponent

3

on

2

is odd, so raise it to

4

; the exponent

2

on

3

is already even. This is the smallest such change.

c^2 = 2^4 \times 3^2 \times 5^2 = 3600

This is the smallest square that is a multiple of

360

.

4Take the square root

#7 Identify Subproblems 6.EE.A.1

Halve each exponent.

c = 2^2 \times 3 \times 5 = 60

The smallest

c

is

60

.

Answer: 60

4 · Reviewdoes it hold up?

$60^2 = 3600$ , and $3600 \div 72 = 50$ and $3600 \div 120 = 30$ are both natural numbers, so $a = 50$ , $b = 30$ work.

Another way: $c$ itself must make $c^2$ divisible by $2^3$ , $3^2$ , $5^1$ : so $c$ needs at least $2^2$ , $3^1$ , $5^1$ , giving $c = 4 \times 3 \times 5 = 60$ .

Standardsmin grade 6

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the least common multiple of $72$ and $120$.
6.EE.A.1 Write and evaluate expressions with whole-number exponents — Writing the numbers with exponents, making the exponents even, and evaluating $2^2 \times 3 \times 5$.

💡Takeaway. For a square, look at the exponents: every one must be even.

Variant 3 easy

Natural numbers $a$ , $b$ , and $c$ satisfy $84a = 126b = c^2$ . Find the smallest possible value of $c$ .

Your answer

Show solution

1 · Understandwhat's really being asked

$c^2$ equals $84$ times a natural number and also $126$ times a natural number. Find the smallest natural number $c$ for which this can happen.

Givens

$84a = 126b = c^2$
$a$ , $b$ , $c$ are natural numbers.

Unknowns

The smallest possible $c$ .

Constraints

$c^2$ must be a multiple of $84$ and of $126$ .
$c^2$ is a perfect square, so every prime in it has an even exponent.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems

Rewrite $84$ and $126$ as products of prime powers. Then $c^2$ must contain the least common multiple, and on top of that every exponent must be even. Raise each odd exponent by one and take the square root.

3 · Execute4 carry out the plan

1Factor both numbers into primes

#15 Organize Information in More Ways 6.EE.A.1

Break each coefficient into prime powers.

84 = 2^2 \times 3 \times 7, \quad 126 = 2 \times 3^2 \times 7

Now the exponents can be compared prime by prime.

2 $c^2$ is a common multiple

#7 Identify Subproblems 6.NS.B.4

Because

c^2 = 84a = 126b

, it is a multiple of both numbers, so it is a multiple of their least common multiple. Take the larger exponent of each prime.

2^2 \times 3^2 \times 7 = 252

c^2

is a multiple of

252

.

3Make every exponent even

#15 Organize Information in More Ways 6.EE.A.1

A perfect square has even exponents. The exponent

1

on

7

is odd, so raise it to

2

; the exponent

2

on

2

and the exponent

2

on

3

are already even. This is the smallest such change.

c^2 = 2^2 \times 3^2 \times 7^2 = 1764

This is the smallest square that is a multiple of

252

.

4Take the square root

#7 Identify Subproblems 6.EE.A.1

Halve each exponent.

c = 2 \times 3 \times 7 = 42

The smallest

c

is

42

.

Answer: 42

4 · Reviewdoes it hold up?

$42^2 = 1764$ , and $1764 \div 84 = 21$ and $1764 \div 126 = 14$ are both natural numbers, so $a = 21$ , $b = 14$ work.

Another way: $c$ itself must make $c^2$ divisible by $2^2$ , $3^2$ , $7^1$ : so $c$ needs at least $2^1$ , $3^1$ , $7^1$ , giving $c = 2 \times 3 \times 7 = 42$ .

Standardsmin grade 6

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the least common multiple of $84$ and $126$.
6.EE.A.1 Write and evaluate expressions with whole-number exponents — Writing the numbers with exponents, making the exponents even, and evaluating $2 \times 3 \times 7$.

💡Takeaway. For a square, look at the exponents: every one must be even.

Variant 4 medium

Natural numbers $a$ , $b$ , and $c$ satisfy $63a = 150b = c^2$ . Find the smallest possible value of $c$ .

Your answer

Show solution

1 · Understandwhat's really being asked

$c^2$ equals $63$ times a natural number and also $150$ times a natural number. Find the smallest natural number $c$ for which this can happen.

Givens

$63a = 150b = c^2$
$a$ , $b$ , $c$ are natural numbers.

Unknowns

The smallest possible $c$ .

Constraints

$c^2$ must be a multiple of $63$ and of $150$ .
$c^2$ is a perfect square, so every prime in it has an even exponent.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems

Rewrite $63$ and $150$ as products of prime powers. Then $c^2$ must contain the least common multiple, and on top of that every exponent must be even. Raise each odd exponent by one and take the square root.

3 · Execute4 carry out the plan

1Factor both numbers into primes

#15 Organize Information in More Ways 6.EE.A.1

Break each coefficient into prime powers.

63 = 3^2 \times 7, \quad 150 = 2 \times 3 \times 5^2

Now the exponents can be compared prime by prime.

2 $c^2$ is a common multiple

#7 Identify Subproblems 6.NS.B.4

Because

c^2 = 63a = 150b

, it is a multiple of both numbers, so it is a multiple of their least common multiple. Take the larger exponent of each prime.

2 \times 3^2 \times 5^2 \times 7 = 3150

c^2

is a multiple of

3150

.

3Make every exponent even

#15 Organize Information in More Ways 6.EE.A.1

A perfect square has even exponents. The exponents

1

on

2

and on

7

are odd, so raise each to

2

; the exponent

2

on

3

and the exponent

2

on

5

are already even. This is the smallest such change.

c^2 = 2^2 \times 3^2 \times 5^2 \times 7^2 = 44100

This is the smallest square that is a multiple of

3150

.

4Take the square root

#7 Identify Subproblems 6.EE.A.1

Halve each exponent.

c = 2 \times 3 \times 5 \times 7 = 210

The smallest

c

is

210

.

Answer: 210

4 · Reviewdoes it hold up?

$210^2 = 44100$ , and $44100 \div 63 = 700$ and $44100 \div 150 = 294$ are both natural numbers, so $a = 700$ , $b = 294$ work.

Another way: $c$ itself must make $c^2$ divisible by $2^1$ , $3^2$ , $5^2$ , $7^1$ : so $c$ needs at least $2^1$ , $3^1$ , $5^1$ , $7^1$ , giving $c = 2 \times 3 \times 5 \times 7 = 210$ .

Standardsmin grade 6

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the least common multiple of $63$ and $150$.
6.EE.A.1 Write and evaluate expressions with whole-number exponents — Writing the numbers with exponents, making the exponents even, and evaluating $2 \times 3 \times 5 \times 7$.

💡Takeaway. For a square, look at the exponents: every one must be even.

Variant 5 easy

Natural numbers $a$ , $b$ , and $c$ satisfy $108a = 150b = c^2$ . Find the smallest possible value of $c$ .

Your answer

Show solution

1 · Understandwhat's really being asked

$c^2$ equals $108$ times a natural number and also $150$ times a natural number. Find the smallest natural number $c$ for which this can happen.

Givens

$108a = 150b = c^2$
$a$ , $b$ , $c$ are natural numbers.

Unknowns

The smallest possible $c$ .

Constraints

$c^2$ must be a multiple of $108$ and of $150$ .
$c^2$ is a perfect square, so every prime in it has an even exponent.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems

Rewrite $108$ and $150$ as products of prime powers. Then $c^2$ must contain the least common multiple, and on top of that every exponent must be even. Raise each odd exponent by one and take the square root.

3 · Execute4 carry out the plan

1Factor both numbers into primes

#15 Organize Information in More Ways 6.EE.A.1

Break each coefficient into prime powers.

108 = 2^2 \times 3^3, \quad 150 = 2 \times 3 \times 5^2

Now the exponents can be compared prime by prime.

2 $c^2$ is a common multiple

#7 Identify Subproblems 6.NS.B.4

Because

c^2 = 108a = 150b

, it is a multiple of both numbers, so it is a multiple of their least common multiple. Take the larger exponent of each prime.

2^2 \times 3^3 \times 5^2 = 2700

c^2

is a multiple of

2700

.

3Make every exponent even

#15 Organize Information in More Ways 6.EE.A.1

A perfect square has even exponents. The exponent

3

on

3

is odd, so raise it to

4

; the exponent

2

on

2

and the exponent

2

on

5

are already even. This is the smallest such change.

c^2 = 2^2 \times 3^4 \times 5^2 = 8100

This is the smallest square that is a multiple of

2700

.

4Take the square root

#7 Identify Subproblems 6.EE.A.1

Halve each exponent.

c = 2 \times 3^2 \times 5 = 90

The smallest

c

is

90

.

Answer: 90

4 · Reviewdoes it hold up?

$90^2 = 8100$ , and $8100 \div 108 = 75$ and $8100 \div 150 = 54$ are both natural numbers, so $a = 75$ , $b = 54$ work.

Another way: $c$ itself must make $c^2$ divisible by $2^2$ , $3^3$ , $5^2$ : so $c$ needs at least $2^1$ , $3^2$ , $5^1$ , giving $c = 2 \times 9 \times 5 = 90$ .

Standardsmin grade 6

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the least common multiple of $108$ and $150$.
6.EE.A.1 Write and evaluate expressions with whole-number exponents — Writing the numbers with exponents, making the exponents even, and evaluating $2 \times 3^2 \times 5$.

💡Takeaway. For a square, look at the exponents: every one must be even.

Variant 6 medium

Natural numbers $a$ , $b$ , and $c$ satisfy $250a = 140b = c^2$ . Find the smallest possible value of $c$ .

Your answer

Show solution

1 · Understandwhat's really being asked

$c^2$ equals $250$ times a natural number and also $140$ times a natural number. Find the smallest natural number $c$ for which this can happen.

Givens

$250a = 140b = c^2$
$a$ , $b$ , $c$ are natural numbers.

Unknowns

The smallest possible $c$ .

Constraints

$c^2$ must be a multiple of $250$ and of $140$ .
$c^2$ is a perfect square, so every prime in it has an even exponent.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems

Rewrite $250$ and $140$ as products of prime powers. Then $c^2$ must contain the least common multiple, and on top of that every exponent must be even. Raise each odd exponent by one and take the square root.

3 · Execute4 carry out the plan

1Factor both numbers into primes

#15 Organize Information in More Ways 6.EE.A.1

Break each coefficient into prime powers.

250 = 2 \times 5^3, \quad 140 = 2^2 \times 5 \times 7

Now the exponents can be compared prime by prime.

2 $c^2$ is a common multiple

#7 Identify Subproblems 6.NS.B.4

Because

c^2 = 250a = 140b

, it is a multiple of both numbers, so it is a multiple of their least common multiple. Take the larger exponent of each prime.

2^2 \times 5^3 \times 7 = 3500

c^2

is a multiple of

3500

.

3Make every exponent even

#15 Organize Information in More Ways 6.EE.A.1

A perfect square has even exponents. The exponent

1

on

7

is odd, so raise it to

2

; the exponent

3

on

5

is odd, so raise it to

4

; the exponent

2

on

2

is already even. This is the smallest such change.

c^2 = 2^2 \times 5^4 \times 7^2 = 122500

This is the smallest square that is a multiple of

3500

.

4Take the square root

#7 Identify Subproblems 6.EE.A.1

Halve each exponent.

c = 2 \times 5^2 \times 7 = 350

The smallest

c

is

350

.

Answer: 350

4 · Reviewdoes it hold up?

$350^2 = 122500$ , and $122500 \div 250 = 490$ and $122500 \div 140 = 875$ are both natural numbers, so $a = 490$ , $b = 875$ work.

Another way: $c$ itself must make $c^2$ divisible by $2^2$ , $5^3$ , $7^1$ : so $c$ needs at least $2^1$ , $5^2$ , $7^1$ , giving $c = 2 \times 25 \times 7 = 350$ .

Standardsmin grade 6

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the least common multiple of $250$ and $140$.
6.EE.A.1 Write and evaluate expressions with whole-number exponents — Writing the numbers with exponents, making the exponents even, and evaluating $2 \times 5^2 \times 7$.

💡Takeaway. For a square, look at the exponents: every one must be even.

Variant 7 medium

Natural numbers $a$ , $b$ , and $c$ satisfy $72a = 312b = c^2$ . Find the smallest possible value of $c$ .

Your answer

Show solution

1 · Understandwhat's really being asked

$c^2$ equals $72$ times a natural number and also $312$ times a natural number. Find the smallest natural number $c$ for which this can happen.

Givens

$72a = 312b = c^2$
$a$ , $b$ , $c$ are natural numbers.

Unknowns

The smallest possible $c$ .

Constraints

$c^2$ must be a multiple of $72$ and of $312$ .
$c^2$ is a perfect square, so every prime in it has an even exponent.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems

Rewrite $72$ and $312$ as products of prime powers. Then $c^2$ must contain the least common multiple, and on top of that every exponent must be even. Raise each odd exponent by one and take the square root.

3 · Execute4 carry out the plan

1Factor both numbers into primes

#15 Organize Information in More Ways 6.EE.A.1

Break each coefficient into prime powers.

72 = 2^3 \times 3^2, \quad 312 = 2^3 \times 3 \times 13

Now the exponents can be compared prime by prime.

2 $c^2$ is a common multiple

#7 Identify Subproblems 6.NS.B.4

Because

c^2 = 72a = 312b

, it is a multiple of both numbers, so it is a multiple of their least common multiple. Take the larger exponent of each prime.

2^3 \times 3^2 \times 13 = 936

c^2

is a multiple of

936

.

3Make every exponent even

#15 Organize Information in More Ways 6.EE.A.1

A perfect square has even exponents. The exponent

1

on

13

is odd, so raise it to

2

; the exponent

3

on

2

is odd, so raise it to

4

; the exponent

2

on

3

is already even. This is the smallest such change.

c^2 = 2^4 \times 3^2 \times 13^2 = 24336

This is the smallest square that is a multiple of

936

.

4Take the square root

#7 Identify Subproblems 6.EE.A.1

Halve each exponent.

c = 2^2 \times 3 \times 13 = 156

The smallest

c

is

156

.

Answer: 156

4 · Reviewdoes it hold up?

$156^2 = 24336$ , and $24336 \div 72 = 338$ and $24336 \div 312 = 78$ are both natural numbers, so $a = 338$ , $b = 78$ work.

Another way: $c$ itself must make $c^2$ divisible by $2^3$ , $3^2$ , $13^1$ : so $c$ needs at least $2^2$ , $3^1$ , $13^1$ , giving $c = 4 \times 3 \times 13 = 156$ .

Standardsmin grade 6

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the least common multiple of $72$ and $312$.
6.EE.A.1 Write and evaluate expressions with whole-number exponents — Writing the numbers with exponents, making the exponents even, and evaluating $2^2 \times 3 \times 13$.

💡Takeaway. For a square, look at the exponents: every one must be even.

Variant 8 medium

Natural numbers $a$ , $b$ , and $c$ satisfy $28a = 378b = c^2$ . Find the smallest possible value of $c$ .

Your answer

Show solution

1 · Understandwhat's really being asked

$c^2$ equals $28$ times a natural number and also $378$ times a natural number. Find the smallest natural number $c$ for which this can happen.

Givens

$28a = 378b = c^2$
$a$ , $b$ , $c$ are natural numbers.

Unknowns

The smallest possible $c$ .

Constraints

$c^2$ must be a multiple of $28$ and of $378$ .
$c^2$ is a perfect square, so every prime in it has an even exponent.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems

Rewrite $28$ and $378$ as products of prime powers. Then $c^2$ must contain the least common multiple, and on top of that every exponent must be even. Raise each odd exponent by one and take the square root.

3 · Execute4 carry out the plan

1Factor both numbers into primes

#15 Organize Information in More Ways 6.EE.A.1

Break each coefficient into prime powers.

28 = 2^2 \times 7, \quad 378 = 2 \times 3^3 \times 7

Now the exponents can be compared prime by prime.

2 $c^2$ is a common multiple

#7 Identify Subproblems 6.NS.B.4

Because

c^2 = 28a = 378b

, it is a multiple of both numbers, so it is a multiple of their least common multiple. Take the larger exponent of each prime.

2^2 \times 3^3 \times 7 = 756

c^2

is a multiple of

756

.

3Make every exponent even

#15 Organize Information in More Ways 6.EE.A.1

A perfect square has even exponents. The exponent

1

on

7

is odd, so raise it to

2

; the exponent

3

on

3

is odd, so raise it to

4

; the exponent

2

on

2

is already even. This is the smallest such change.

c^2 = 2^2 \times 3^4 \times 7^2 = 15876

This is the smallest square that is a multiple of

756

.

4Take the square root

#7 Identify Subproblems 6.EE.A.1

Halve each exponent.

c = 2 \times 3^2 \times 7 = 126

The smallest

c

is

126

.

Answer: 126

4 · Reviewdoes it hold up?

$126^2 = 15876$ , and $15876 \div 28 = 567$ and $15876 \div 378 = 42$ are both natural numbers, so $a = 567$ , $b = 42$ work.

Another way: $c$ itself must make $c^2$ divisible by $2^2$ , $3^3$ , $7^1$ : so $c$ needs at least $2^1$ , $3^2$ , $7^1$ , giving $c = 2 \times 9 \times 7 = 126$ .

Standardsmin grade 6

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the least common multiple of $28$ and $378$.
6.EE.A.1 Write and evaluate expressions with whole-number exponents — Writing the numbers with exponents, making the exponents even, and evaluating $2 \times 3^2 \times 7$.

💡Takeaway. For a square, look at the exponents: every one must be even.

Variant 9 hard

Natural numbers $a$ , $b$ , and $c$ satisfy $100a = 425b = c^2$ . Find the smallest possible value of $c$ .

Your answer

Show solution

1 · Understandwhat's really being asked

$c^2$ equals $100$ times a natural number and also $425$ times a natural number. Find the smallest natural number $c$ for which this can happen.

Givens

$100a = 425b = c^2$
$a$ , $b$ , $c$ are natural numbers.

Unknowns

The smallest possible $c$ .

Constraints

$c^2$ must be a multiple of $100$ and of $425$ .
$c^2$ is a perfect square, so every prime in it has an even exponent.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems

Rewrite $100$ and $425$ as products of prime powers. Then $c^2$ must contain the least common multiple, and on top of that every exponent must be even. Raise each odd exponent by one and take the square root.

3 · Execute4 carry out the plan

1Factor both numbers into primes

#15 Organize Information in More Ways 6.EE.A.1

Break each coefficient into prime powers.

100 = 2^2 \times 5^2, \quad 425 = 5^2 \times 17

Now the exponents can be compared prime by prime.

2 $c^2$ is a common multiple

#7 Identify Subproblems 6.NS.B.4

Because

c^2 = 100a = 425b

, it is a multiple of both numbers, so it is a multiple of their least common multiple. Take the larger exponent of each prime.

2^2 \times 5^2 \times 17 = 1700

c^2

is a multiple of

1700

.

3Make every exponent even

#15 Organize Information in More Ways 6.EE.A.1

A perfect square has even exponents. The exponent

1

on

17

is odd, so raise it to

2

; the exponent

2

on

2

and the exponent

2

on

5

are already even. This is the smallest such change.

c^2 = 2^2 \times 5^2 \times 17^2 = 28900

This is the smallest square that is a multiple of

1700

.

4Take the square root

#7 Identify Subproblems 6.EE.A.1

Halve each exponent.

c = 2 \times 5 \times 17 = 170

The smallest

c

is

170

.

Answer: 170

4 · Reviewdoes it hold up?

$170^2 = 28900$ , and $28900 \div 100 = 289$ and $28900 \div 425 = 68$ are both natural numbers, so $a = 289$ , $b = 68$ work.

Another way: $c$ itself must make $c^2$ divisible by $2^2$ , $5^2$ , $17^1$ : so $c$ needs at least $2^1$ , $5^1$ , $17^1$ , giving $c = 2 \times 5 \times 17 = 170$ .

Standardsmin grade 6

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the least common multiple of $100$ and $425$.
6.EE.A.1 Write and evaluate expressions with whole-number exponents — Writing the numbers with exponents, making the exponents even, and evaluating $2 \times 5 \times 17$.

💡Takeaway. For a square, look at the exponents: every one must be even.

Variant 10 hard

Natural numbers $a$ , $b$ , and $c$ satisfy $270a = 432b = c^2$ . Find the smallest possible value of $c$ .

Your answer

Show solution

1 · Understandwhat's really being asked

$c^2$ equals $270$ times a natural number and also $432$ times a natural number. Find the smallest natural number $c$ for which this can happen.

Givens

$270a = 432b = c^2$
$a$ , $b$ , $c$ are natural numbers.

Unknowns

The smallest possible $c$ .

Constraints

$c^2$ must be a multiple of $270$ and of $432$ .
$c^2$ is a perfect square, so every prime in it has an even exponent.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems

Rewrite $270$ and $432$ as products of prime powers. Then $c^2$ must contain the least common multiple, and on top of that every exponent must be even. Raise each odd exponent by one and take the square root.

3 · Execute4 carry out the plan

1Factor both numbers into primes

#15 Organize Information in More Ways 6.EE.A.1

Break each coefficient into prime powers.

270 = 2 \times 3^3 \times 5, \quad 432 = 2^4 \times 3^3

Now the exponents can be compared prime by prime.

2 $c^2$ is a common multiple

#7 Identify Subproblems 6.NS.B.4

Because

c^2 = 270a = 432b

, it is a multiple of both numbers, so it is a multiple of their least common multiple. Take the larger exponent of each prime.

2^4 \times 3^3 \times 5 = 2160

c^2

is a multiple of

2160

.

3Make every exponent even

#15 Organize Information in More Ways 6.EE.A.1

A perfect square has even exponents. The exponent

1

on

5

is odd, so raise it to

2

; the exponent

3

on

3

is odd, so raise it to

4

; the exponent

4

on

2

is already even. This is the smallest such change.

c^2 = 2^4 \times 3^4 \times 5^2 = 32400

This is the smallest square that is a multiple of

2160

.

4Take the square root

#7 Identify Subproblems 6.EE.A.1

Halve each exponent.

c = 2^2 \times 3^2 \times 5 = 180

The smallest

c

is

180

.

Answer: 180

4 · Reviewdoes it hold up?

$180^2 = 32400$ , and $32400 \div 270 = 120$ and $32400 \div 432 = 75$ are both natural numbers, so $a = 120$ , $b = 75$ work.

Another way: $c$ itself must make $c^2$ divisible by $2^4$ , $3^3$ , $5^1$ : so $c$ needs at least $2^2$ , $3^2$ , $5^1$ , giving $c = 4 \times 9 \times 5 = 180$ .

Standardsmin grade 6

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the least common multiple of $270$ and $432$.
6.EE.A.1 Write and evaluate expressions with whole-number exponents — Writing the numbers with exponents, making the exponents even, and evaluating $2^2 \times 3^2 \times 5$.

💡Takeaway. For a square, look at the exponents: every one must be even.

Variant 11 hard

Natural numbers $a$ , $b$ , and $c$ satisfy $306a = 578b = c^2$ . Find the smallest possible value of $c$ .

Your answer

Show solution

1 · Understandwhat's really being asked

$c^2$ equals $306$ times a natural number and also $578$ times a natural number. Find the smallest natural number $c$ for which this can happen.

Givens

$306a = 578b = c^2$
$a$ , $b$ , $c$ are natural numbers.

Unknowns

The smallest possible $c$ .

Constraints

$c^2$ must be a multiple of $306$ and of $578$ .
$c^2$ is a perfect square, so every prime in it has an even exponent.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems

Rewrite $306$ and $578$ as products of prime powers. Then $c^2$ must contain the least common multiple, and on top of that every exponent must be even. Raise each odd exponent by one and take the square root.

3 · Execute4 carry out the plan

1Factor both numbers into primes

#15 Organize Information in More Ways 6.EE.A.1

Break each coefficient into prime powers.

306 = 2 \times 3^2 \times 17, \quad 578 = 2 \times 17^2

Now the exponents can be compared prime by prime.

2 $c^2$ is a common multiple

#7 Identify Subproblems 6.NS.B.4

Because

c^2 = 306a = 578b

, it is a multiple of both numbers, so it is a multiple of their least common multiple. Take the larger exponent of each prime.

2 \times 3^2 \times 17^2 = 5202

c^2

is a multiple of

5202

.

3Make every exponent even

#15 Organize Information in More Ways 6.EE.A.1

A perfect square has even exponents. The exponent

1

on

2

is odd, so raise it to

2

; the exponent

2

on

3

and the exponent

2

on

17

are already even. This is the smallest such change.

c^2 = 2^2 \times 3^2 \times 17^2 = 10404

This is the smallest square that is a multiple of

5202

.

4Take the square root

#7 Identify Subproblems 6.EE.A.1

Halve each exponent.

c = 2 \times 3 \times 17 = 102

The smallest

c

is

102

.

Answer: 102

4 · Reviewdoes it hold up?

$102^2 = 10404$ , and $10404 \div 306 = 34$ and $10404 \div 578 = 18$ are both natural numbers, so $a = 34$ , $b = 18$ work.

Another way: $c$ itself must make $c^2$ divisible by $2^1$ , $3^2$ , $17^2$ : so $c$ needs at least $2^1$ , $3^1$ , $17^1$ , giving $c = 2 \times 3 \times 17 = 102$ .

Standardsmin grade 6

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the least common multiple of $306$ and $578$.
6.EE.A.1 Write and evaluate expressions with whole-number exponents — Writing the numbers with exponents, making the exponents even, and evaluating $2 \times 3 \times 17$.

💡Takeaway. For a square, look at the exponents: every one must be even.

Variant 12 hard

Natural numbers $a$ , $b$ , and $c$ satisfy $360a = 1500b = c^2$ . Find the smallest possible value of $c$ .

Your answer

Show solution

1 · Understandwhat's really being asked

$c^2$ equals $360$ times a natural number and also $1500$ times a natural number. Find the smallest natural number $c$ for which this can happen.

Givens

$360a = 1500b = c^2$
$a$ , $b$ , $c$ are natural numbers.

Unknowns

The smallest possible $c$ .

Constraints

$c^2$ must be a multiple of $360$ and of $1500$ .
$c^2$ is a perfect square, so every prime in it has an even exponent.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems

Rewrite $360$ and $1500$ as products of prime powers. Then $c^2$ must contain the least common multiple, and on top of that every exponent must be even. Raise each odd exponent by one and take the square root.

3 · Execute4 carry out the plan

1Factor both numbers into primes

#15 Organize Information in More Ways 6.EE.A.1

Break each coefficient into prime powers.

360 = 2^3 \times 3^2 \times 5, \quad 1500 = 2^2 \times 3 \times 5^3

Now the exponents can be compared prime by prime.

2 $c^2$ is a common multiple

#7 Identify Subproblems 6.NS.B.4

Because

c^2 = 360a = 1500b

, it is a multiple of both numbers, so it is a multiple of their least common multiple. Take the larger exponent of each prime.

2^3 \times 3^2 \times 5^3 = 9000

c^2

is a multiple of

9000

.

3Make every exponent even

#15 Organize Information in More Ways 6.EE.A.1

A perfect square has even exponents. The exponents

3

on

2

and on

5

are odd, so raise each to

4

; the exponent

2

on

3

is already even. This is the smallest such change.

c^2 = 2^4 \times 3^2 \times 5^4 = 90000

This is the smallest square that is a multiple of

9000

.

4Take the square root

#7 Identify Subproblems 6.EE.A.1

Halve each exponent.

c = 2^2 \times 3 \times 5^2 = 300

The smallest

c

is

300

.

Answer: 300

4 · Reviewdoes it hold up?

$300^2 = 90000$ , and $90000 \div 360 = 250$ and $90000 \div 1500 = 60$ are both natural numbers, so $a = 250$ , $b = 60$ work.

Another way: $c$ itself must make $c^2$ divisible by $2^3$ , $3^2$ , $5^3$ : so $c$ needs at least $2^2$ , $3^1$ , $5^2$ , giving $c = 4 \times 3 \times 25 = 300$ .

Standardsmin grade 6

6.NS.B.4 Find GCF and LCM; distributive property with a common factor — Finding the least common multiple of $360$ and $1500$.
6.EE.A.1 Write and evaluate expressions with whole-number exponents — Writing the numbers with exponents, making the exponents even, and evaluating $2^2 \times 3 \times 5^2$.

💡Takeaway. For a square, look at the exponents: every one must be even.

Smallest square that is a common multiple · 12 practice problems

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