Practice · Make every prime exponent even · Sensim Math

Variant 1 easy

Multiply the repeating decimal $0.\overline{2}$ by a natural number $A$ so that the product is the square of some natural number. Find the smallest possible value of $A$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Find the smallest natural number $A$ that makes $0.\overline{2} \times A$ equal to a perfect square of a natural number.

Givens

The repeating decimal is $0.\overline{2}$ .
$A$ is a natural number.

Unknowns

The smallest possible $A$ .

Constraints

The product must be the square of a natural number, so it must be a whole number first.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #13 Convert to Algebra#6 Guess and Check

A repeating decimal is hard to multiply, so change it to a fraction. Then write the product with letters: the denominator $9$ has to cancel and the leftover factor $2$ needs a partner to make an even exponent.

3 · Execute3 carry out the plan

1Change the decimal to a fraction

#15 Organize Information in More Ways 8.NS.A.1

One repeating digit after the point means divide by

9

.

0.\overline{2} = \frac{2}{9}

Now the product is

\frac{2}{9} \times A

.

2Write the square condition with letters

#13 Convert to Algebra 4.OA.B.4

We need

\frac{2A}{9} = m^2

for a natural number

m

. Since

2

and

9

share no factor,

9

must divide

A

; write

A = 9B

. Then

2B = m^2

, so

m^2

is even,

m

is even, and

B

must be

2

times a square.

A = 9B, \quad 2B = m^2 \Rightarrow B = 2k^2 \Rightarrow A = 18k^2

Every prime in a perfect square appears an even number of times, so the lone

2

needs another

2

.

3Take the smallest case and check

#6 Guess and Check 8.NS.A.1

The smallest choice is

k = 1

, giving

A = 18

. Check the product.

\frac{2}{9} \times 18 = 4 = 2^2

The product

4

is a perfect square, so

A = 18

works.

Answer: 18

4 · Reviewdoes it hold up?

Smaller candidates fail: $A$ must be a multiple of $9$ , and $A = 9$ gives $\frac{2}{9} \times 9 = 2$ , which is not a perfect square. So $18$ is the first that works.

Another way: Guess and check over the multiples of $9$ : $9 \to 2$ , $18 \to 4 = 2^2$ . The second multiple already gives a square.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Writing $0.\overline{2}$ as the fraction $\frac{2}{9}$ and checking the product.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Using factors and multiples to see that $A$ must be $9$ times $2$ times a square.

💡Takeaway. Turn the repeating decimal into a fraction, then pair up every prime factor.

Variant 2 easy

Multiply the repeating decimal $0.\overline{5}$ by a natural number $A$ so that the product is the square of some natural number. Find the smallest possible value of $A$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Find the smallest natural number $A$ that makes $0.\overline{5} \times A$ equal to a perfect square of a natural number.

Givens

The repeating decimal is $0.\overline{5}$ .
$A$ is a natural number.

Unknowns

The smallest possible $A$ .

Constraints

The product must be the square of a natural number, so it must be a whole number first.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #13 Convert to Algebra#6 Guess and Check

A repeating decimal is hard to multiply, so change it to a fraction. Then write the product with letters: the denominator $9$ has to cancel and the leftover factor $5$ needs a partner to make an even exponent.

3 · Execute3 carry out the plan

1Change the decimal to a fraction

#15 Organize Information in More Ways 8.NS.A.1

One repeating digit after the point means divide by

9

.

0.\overline{5} = \frac{5}{9}

Now the product is

\frac{5}{9} \times A

.

2Write the square condition with letters

#13 Convert to Algebra 4.OA.B.4

We need

\frac{5A}{9} = m^2

for a natural number

m

. Since

5

and

9

share no factor,

9

must divide

A

; write

A = 9B

. Then

5B = m^2

, so

5

divides

m^2

and therefore

m

, which makes

B

5

times a square.

A = 9B, \quad 5B = m^2 \Rightarrow B = 5k^2 \Rightarrow A = 45k^2

Every prime in a perfect square appears an even number of times, so the lone

5

needs another

5

.

3Take the smallest case and check

#6 Guess and Check 8.NS.A.1

The smallest choice is

k = 1

, giving

A = 45

. Check the product.

\frac{5}{9} \times 45 = 25 = 5^2

The product

25

is a perfect square, so

A = 45

works.

Answer: 45

4 · Reviewdoes it hold up?

Smaller candidates fail: $A$ must be a multiple of $9$ , and $A = 9, 18, 27, 36$ give $5, 10, 15, 20$ , none of them a perfect square. So $45$ is the first that works.

Another way: Guess and check over the multiples of $9$ : $9 \to 5$ , $18 \to 10$ , $27 \to 15$ , $36 \to 20$ , $45 \to 25 = 5^2$ . The fifth multiple is the first to give a square.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Writing $0.\overline{5}$ as the fraction $\frac{5}{9}$ and checking the product.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Using factors and multiples to see that $A$ must be $9$ times $5$ times a square.

💡Takeaway. Turn the repeating decimal into a fraction, then pair up every prime factor.

Variant 3 easy

Multiply the repeating decimal $0.\overline{6}$ by a natural number $A$ so that the product is the square of some natural number. Find the smallest possible value of $A$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Find the smallest natural number $A$ that makes $0.\overline{6} \times A$ equal to a perfect square of a natural number.

Givens

The repeating decimal is $0.\overline{6}$ .
$A$ is a natural number.

Unknowns

The smallest possible $A$ .

Constraints

The product must be the square of a natural number, so it must be a whole number first.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #13 Convert to Algebra#6 Guess and Check

A repeating decimal is hard to multiply, so change it to a fraction. Then write the product with letters: the denominator $3$ has to cancel and the leftover factor $2$ needs a partner to make an even exponent.

3 · Execute3 carry out the plan

1Change the decimal to a fraction

#15 Organize Information in More Ways 8.NS.A.1

One repeating digit after the point means divide by

9

.

0.\overline{6} = \frac{6}{9} = \frac{2}{3}

Now the product is

\frac{2}{3} \times A

.

2Write the square condition with letters

#13 Convert to Algebra 4.OA.B.4

We need

\frac{2A}{3} = m^2

for a natural number

m

. Since

2

and

3

share no factor,

3

must divide

A

; write

A = 3B

. Then

2B = m^2

, so

m^2

is even,

m

is even, and

B

must be

2

times a square.

A = 3B, \quad 2B = m^2 \Rightarrow B = 2k^2 \Rightarrow A = 6k^2

Every prime in a perfect square appears an even number of times, so the lone

2

needs another

2

.

3Take the smallest case and check

#6 Guess and Check 8.NS.A.1

The smallest choice is

k = 1

, giving

A = 6

. Check the product.

\frac{2}{3} \times 6 = 4 = 2^2

The product

4

is a perfect square, so

A = 6

works.

Answer: 6

4 · Reviewdoes it hold up?

Smaller candidates fail: $A$ must be a multiple of $3$ , and $A = 3$ gives $\frac{2}{3} \times 3 = 2$ , which is not a perfect square. So $6$ is the first that works.

Another way: Guess and check over the multiples of $3$ : $3 \to 2$ , $6 \to 4 = 2^2$ . The second multiple already gives a square.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Writing $0.\overline{6}$ as the fraction $\frac{2}{3}$ and checking the product.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Using factors and multiples to see that $A$ must be $3$ times $2$ times a square.

💡Takeaway. Turn the repeating decimal into a fraction, then pair up every prime factor.

Variant 4 easy

Multiply the repeating decimal $0.\overline{7}$ by a natural number $A$ so that the product is the square of some natural number. Find the smallest possible value of $A$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Find the smallest natural number $A$ that makes $0.\overline{7} \times A$ equal to a perfect square of a natural number.

Givens

The repeating decimal is $0.\overline{7}$ .
$A$ is a natural number.

Unknowns

The smallest possible $A$ .

Constraints

The product must be the square of a natural number, so it must be a whole number first.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #13 Convert to Algebra#6 Guess and Check

A repeating decimal is hard to multiply, so change it to a fraction. Then write the product with letters: the denominator $9$ has to cancel and the leftover factor $7$ needs a partner to make an even exponent.

3 · Execute3 carry out the plan

1Change the decimal to a fraction

#15 Organize Information in More Ways 8.NS.A.1

One repeating digit after the point means divide by

9

.

0.\overline{7} = \frac{7}{9}

Now the product is

\frac{7}{9} \times A

.

2Write the square condition with letters

#13 Convert to Algebra 4.OA.B.4

We need

\frac{7A}{9} = m^2

for a natural number

m

. Since

7

and

9

share no factor,

9

must divide

A

; write

A = 9B

. Then

7B = m^2

, so

7

divides

m^2

and therefore

m

, which makes

B

7

times a square.

A = 9B, \quad 7B = m^2 \Rightarrow B = 7k^2 \Rightarrow A = 63k^2

Every prime in a perfect square appears an even number of times, so the lone

7

needs another

7

.

3Take the smallest case and check

#6 Guess and Check 8.NS.A.1

The smallest choice is

k = 1

, giving

A = 63

. Check the product.

\frac{7}{9} \times 63 = 49 = 7^2

The product

49

is a perfect square, so

A = 63

works.

Answer: 63

4 · Reviewdoes it hold up?

Smaller candidates fail: $A$ must be a multiple of $9$ , and $A = 9, 18, 27, 36, 45, 54$ give $7, 14, 21, 28, 35, 42$ , none of them a perfect square. So $63$ is the first that works.

Another way: Guess and check over the multiples of $9$ : $9 \to 7$ , $18 \to 14$ , $27 \to 21$ , $36 \to 28$ , $45 \to 35$ , $54 \to 42$ , $63 \to 49 = 7^2$ . The seventh multiple is the first to give a square.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Writing $0.\overline{7}$ as the fraction $\frac{7}{9}$ and checking the product.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Using factors and multiples to see that $A$ must be $9$ times $7$ times a square.

💡Takeaway. Turn the repeating decimal into a fraction, then pair up every prime factor.

Variant 5 medium

Multiply the repeating decimal $0.\overline{8}$ by a natural number $A$ so that the product is the square of some natural number. Find the smallest possible value of $A$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Find the smallest natural number $A$ that makes $0.\overline{8} \times A$ equal to a perfect square of a natural number.

Givens

The repeating decimal is $0.\overline{8}$ .
$A$ is a natural number.

Unknowns

The smallest possible $A$ .

Constraints

The product must be the square of a natural number, so it must be a whole number first.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #13 Convert to Algebra#6 Guess and Check

A repeating decimal is hard to multiply, so change it to a fraction. Then write the product with letters: the denominator $9$ has to cancel and the leftover factor $2$ needs a partner to make an even exponent.

3 · Execute3 carry out the plan

1Change the decimal to a fraction

#15 Organize Information in More Ways 8.NS.A.1

One repeating digit after the point means divide by

9

.

0.\overline{8} = \frac{8}{9}

Now the product is

\frac{8}{9} \times A

.

2Write the square condition with letters

#13 Convert to Algebra 4.OA.B.4

We need

\frac{8A}{9} = m^2

for a natural number

m

. Since

8

and

9

share no factor,

9

must divide

A

; write

A = 9B

. Then

8B = m^2

, and in

8 = 2^3

the prime

2

appears an odd number of times, so

B

must be

2

times a square.

A = 9B, \quad 8B = m^2 \Rightarrow B = 2k^2 \Rightarrow A = 18k^2

Every prime in a perfect square appears an even number of times, so the unpaired

2

in

8

needs a partner.

3Take the smallest case and check

#6 Guess and Check 8.NS.A.1

The smallest choice is

k = 1

, giving

A = 18

. Check the product.

\frac{8}{9} \times 18 = 16 = 4^2

The product

16

is a perfect square, so

A = 18

works.

Answer: 18

4 · Reviewdoes it hold up?

Smaller candidates fail: $A$ must be a multiple of $9$ , and $A = 9$ gives $\frac{8}{9} \times 9 = 8$ , which is not a perfect square. So $18$ is the first that works.

Another way: Guess and check over the multiples of $9$ : $9 \to 8$ , $18 \to 16 = 4^2$ . The second multiple already gives a square.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Writing $0.\overline{8}$ as the fraction $\frac{8}{9}$ and checking the product.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Using factors and multiples to see that $A$ must be $9$ times $2$ times a square.

💡Takeaway. Turn the repeating decimal into a fraction, then pair up every prime factor.

Variant 6 medium

Multiply the repeating decimal $0.\overline{15}$ by a natural number $A$ so that the product is the square of some natural number. Find the smallest possible value of $A$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Find the smallest natural number $A$ that makes $0.\overline{15} \times A$ equal to a perfect square of a natural number.

Givens

The repeating decimal is $0.\overline{15}$ .
$A$ is a natural number.

Unknowns

The smallest possible $A$ .

Constraints

The product must be the square of a natural number, so it must be a whole number first.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #13 Convert to Algebra#6 Guess and Check

A repeating decimal is hard to multiply, so change it to a fraction. Then write the product with letters: the denominator $33$ has to cancel and the leftover factor $5$ needs a partner to make an even exponent.

3 · Execute3 carry out the plan

1Change the decimal to a fraction

#15 Organize Information in More Ways 8.NS.A.1

Two repeating digits after the point means divide by

99

.

0.\overline{15} = \frac{15}{99} = \frac{5}{33}

Now the product is

\frac{5}{33} \times A

.

2Write the square condition with letters

#13 Convert to Algebra 4.OA.B.4

We need

\frac{5A}{33} = m^2

for a natural number

m

. Since

5

and

33

share no factor,

33

must divide

A

; write

A = 33B

. Then

5B = m^2

, so

5

divides

m^2

and therefore

m

, which makes

B

5

times a square.

A = 33B, \quad 5B = m^2 \Rightarrow B = 5k^2 \Rightarrow A = 165k^2

Every prime in a perfect square appears an even number of times, so the lone

5

needs another

5

.

3Take the smallest case and check

#6 Guess and Check 8.NS.A.1

The smallest choice is

k = 1

, giving

A = 165

. Check the product.

\frac{5}{33} \times 165 = 25 = 5^2

The product

25

is a perfect square, so

A = 165

works.

Answer: 165

4 · Reviewdoes it hold up?

Smaller candidates fail: $A$ must be a multiple of $33$ , and $A = 33, 66, 99, 132$ give $5, 10, 15, 20$ , none of them a perfect square. So $165$ is the first that works.

Another way: Guess and check over the multiples of $33$ : $33 \to 5$ , $66 \to 10$ , $99 \to 15$ , $132 \to 20$ , $165 \to 25 = 5^2$ . The fifth multiple is the first to give a square.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Writing $0.\overline{15}$ as the fraction $\frac{5}{33}$ and checking the product.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Using factors and multiples to see that $A$ must be $33$ times $5$ times a square.

💡Takeaway. Turn the repeating decimal into a fraction, then pair up every prime factor.

Variant 7 medium

Multiply the repeating decimal $0.\overline{18}$ by a natural number $A$ so that the product is the square of some natural number. Find the smallest possible value of $A$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Find the smallest natural number $A$ that makes $0.\overline{18} \times A$ equal to a perfect square of a natural number.

Givens

The repeating decimal is $0.\overline{18}$ .
$A$ is a natural number.

Unknowns

The smallest possible $A$ .

Constraints

The product must be the square of a natural number, so it must be a whole number first.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #13 Convert to Algebra#6 Guess and Check

A repeating decimal is hard to multiply, so change it to a fraction. Then write the product with letters: the denominator $11$ has to cancel and the leftover factor $2$ needs a partner to make an even exponent.

3 · Execute3 carry out the plan

1Change the decimal to a fraction

#15 Organize Information in More Ways 8.NS.A.1

Two repeating digits after the point means divide by

99

.

0.\overline{18} = \frac{18}{99} = \frac{2}{11}

Now the product is

\frac{2}{11} \times A

.

2Write the square condition with letters

#13 Convert to Algebra 4.OA.B.4

We need

\frac{2A}{11} = m^2

for a natural number

m

. Since

2

and

11

share no factor,

11

must divide

A

; write

A = 11B

. Then

2B = m^2

, so

m^2

is even,

m

is even, and

B

must be

2

times a square.

A = 11B, \quad 2B = m^2 \Rightarrow B = 2k^2 \Rightarrow A = 22k^2

Every prime in a perfect square appears an even number of times, so the lone

2

needs another

2

.

3Take the smallest case and check

#6 Guess and Check 8.NS.A.1

The smallest choice is

k = 1

, giving

A = 22

. Check the product.

\frac{2}{11} \times 22 = 4 = 2^2

The product

4

is a perfect square, so

A = 22

works.

Answer: 22

4 · Reviewdoes it hold up?

Smaller candidates fail: $A$ must be a multiple of $11$ , and $A = 11$ gives $\frac{2}{11} \times 11 = 2$ , which is not a perfect square. So $22$ is the first that works.

Another way: Guess and check over the multiples of $11$ : $11 \to 2$ , $22 \to 4 = 2^2$ . The second multiple already gives a square.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Writing $0.\overline{18}$ as the fraction $\frac{2}{11}$ and checking the product.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Using factors and multiples to see that $A$ must be $11$ times $2$ times a square.

💡Takeaway. Turn the repeating decimal into a fraction, then pair up every prime factor.

Variant 8 medium

Multiply the repeating decimal $0.\overline{21}$ by a natural number $A$ so that the product is the square of some natural number. Find the smallest possible value of $A$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Find the smallest natural number $A$ that makes $0.\overline{21} \times A$ equal to a perfect square of a natural number.

Givens

The repeating decimal is $0.\overline{21}$ .
$A$ is a natural number.

Unknowns

The smallest possible $A$ .

Constraints

The product must be the square of a natural number, so it must be a whole number first.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #13 Convert to Algebra#6 Guess and Check

A repeating decimal is hard to multiply, so change it to a fraction. Then write the product with letters: the denominator $33$ has to cancel and the leftover factor $7$ needs a partner to make an even exponent.

3 · Execute3 carry out the plan

1Change the decimal to a fraction

#15 Organize Information in More Ways 8.NS.A.1

Two repeating digits after the point means divide by

99

.

0.\overline{21} = \frac{21}{99} = \frac{7}{33}

Now the product is

\frac{7}{33} \times A

.

2Write the square condition with letters

#13 Convert to Algebra 4.OA.B.4

We need

\frac{7A}{33} = m^2

for a natural number

m

. Since

7

and

33

share no factor,

33

must divide

A

; write

A = 33B

. Then

7B = m^2

, so

7

divides

m^2

and therefore

m

, which makes

B

7

times a square.

A = 33B, \quad 7B = m^2 \Rightarrow B = 7k^2 \Rightarrow A = 231k^2

Every prime in a perfect square appears an even number of times, so the lone

7

needs another

7

.

3Take the smallest case and check

#6 Guess and Check 8.NS.A.1

The smallest choice is

k = 1

, giving

A = 231

. Check the product.

\frac{7}{33} \times 231 = 49 = 7^2

The product

49

is a perfect square, so

A = 231

works.

Answer: 231

4 · Reviewdoes it hold up?

Smaller candidates fail: $A$ must be a multiple of $33$ , and $A = 33, 66, 99, 132, 165, 198$ give $7, 14, 21, 28, 35, 42$ , none of them a perfect square. So $231$ is the first that works.

Another way: Guess and check over the multiples of $33$ : $33 \to 7$ , $66 \to 14$ , $99 \to 21$ , $132 \to 28$ , $165 \to 35$ , $198 \to 42$ , $231 \to 49 = 7^2$ . The seventh multiple is the first to give a square.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Writing $0.\overline{21}$ as the fraction $\frac{7}{33}$ and checking the product.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Using factors and multiples to see that $A$ must be $33$ times $7$ times a square.

💡Takeaway. Turn the repeating decimal into a fraction, then pair up every prime factor.

Variant 9 hard

Multiply the repeating decimal $0.\overline{24}$ by a natural number $A$ so that the product is the square of some natural number. Find the smallest possible value of $A$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Find the smallest natural number $A$ that makes $0.\overline{24} \times A$ equal to a perfect square of a natural number.

Givens

The repeating decimal is $0.\overline{24}$ .
$A$ is a natural number.

Unknowns

The smallest possible $A$ .

Constraints

The product must be the square of a natural number, so it must be a whole number first.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #13 Convert to Algebra#6 Guess and Check

A repeating decimal is hard to multiply, so change it to a fraction. Then write the product with letters: the denominator $33$ has to cancel and the leftover factor $2$ needs a partner to make an even exponent.

3 · Execute3 carry out the plan

1Change the decimal to a fraction

#15 Organize Information in More Ways 8.NS.A.1

Two repeating digits after the point means divide by

99

.

0.\overline{24} = \frac{24}{99} = \frac{8}{33}

Now the product is

\frac{8}{33} \times A

.

2Write the square condition with letters

#13 Convert to Algebra 4.OA.B.4

We need

\frac{8A}{33} = m^2

for a natural number

m

. Since

8

and

33

share no factor,

33

must divide

A

; write

A = 33B

. Then

8B = m^2

, and in

8 = 2^3

the prime

2

appears an odd number of times, so

B

must be

2

times a square.

A = 33B, \quad 8B = m^2 \Rightarrow B = 2k^2 \Rightarrow A = 66k^2

Every prime in a perfect square appears an even number of times, so the unpaired

2

in

8

needs a partner.

3Take the smallest case and check

#6 Guess and Check 8.NS.A.1

The smallest choice is

k = 1

, giving

A = 66

. Check the product.

\frac{8}{33} \times 66 = 16 = 4^2

The product

16

is a perfect square, so

A = 66

works.

Answer: 66

4 · Reviewdoes it hold up?

Smaller candidates fail: $A$ must be a multiple of $33$ , and $A = 33$ gives $\frac{8}{33} \times 33 = 8$ , which is not a perfect square. So $66$ is the first that works.

Another way: Guess and check over the multiples of $33$ : $33 \to 8$ , $66 \to 16 = 4^2$ . The second multiple already gives a square.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Writing $0.\overline{24}$ as the fraction $\frac{8}{33}$ and checking the product.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Using factors and multiples to see that $A$ must be $33$ times $2$ times a square.

💡Takeaway. Turn the repeating decimal into a fraction, then pair up every prime factor.

Variant 10 hard

Multiply the repeating decimal $0.\overline{27}$ by a natural number $A$ so that the product is the square of some natural number. Find the smallest possible value of $A$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Find the smallest natural number $A$ that makes $0.\overline{27} \times A$ equal to a perfect square of a natural number.

Givens

The repeating decimal is $0.\overline{27}$ .
$A$ is a natural number.

Unknowns

The smallest possible $A$ .

Constraints

The product must be the square of a natural number, so it must be a whole number first.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #13 Convert to Algebra#6 Guess and Check

A repeating decimal is hard to multiply, so change it to a fraction. Then write the product with letters: the denominator $11$ has to cancel and the leftover factor $3$ needs a partner to make an even exponent.

3 · Execute3 carry out the plan

1Change the decimal to a fraction

#15 Organize Information in More Ways 8.NS.A.1

Two repeating digits after the point means divide by

99

.

0.\overline{27} = \frac{27}{99} = \frac{3}{11}

Now the product is

\frac{3}{11} \times A

.

2Write the square condition with letters

#13 Convert to Algebra 4.OA.B.4

We need

\frac{3A}{11} = m^2

for a natural number

m

. Since

3

and

11

share no factor,

11

must divide

A

; write

A = 11B

. Then

3B = m^2

, so

3

divides

m^2

and therefore

m

, which makes

B

3

times a square.

A = 11B, \quad 3B = m^2 \Rightarrow B = 3k^2 \Rightarrow A = 33k^2

Every prime in a perfect square appears an even number of times, so the lone

3

needs another

3

.

3Take the smallest case and check

#6 Guess and Check 8.NS.A.1

The smallest choice is

k = 1

, giving

A = 33

. Check the product.

\frac{3}{11} \times 33 = 9 = 3^2

The product

9

is a perfect square, so

A = 33

works.

Answer: 33

4 · Reviewdoes it hold up?

Smaller candidates fail: $A$ must be a multiple of $11$ , and $A = 11, 22$ give $3, 6$ , none of them a perfect square. So $33$ is the first that works.

Another way: Guess and check over the multiples of $11$ : $11 \to 3$ , $22 \to 6$ , $33 \to 9 = 3^2$ . The third multiple is the first to give a square.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Writing $0.\overline{27}$ as the fraction $\frac{3}{11}$ and checking the product.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Using factors and multiples to see that $A$ must be $11$ times $3$ times a square.

💡Takeaway. Turn the repeating decimal into a fraction, then pair up every prime factor.

Variant 11 hard

Multiply the repeating decimal $0.\overline{45}$ by a natural number $A$ so that the product is the square of some natural number. Find the smallest possible value of $A$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Find the smallest natural number $A$ that makes $0.\overline{45} \times A$ equal to a perfect square of a natural number.

Givens

The repeating decimal is $0.\overline{45}$ .
$A$ is a natural number.

Unknowns

The smallest possible $A$ .

Constraints

The product must be the square of a natural number, so it must be a whole number first.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #13 Convert to Algebra#6 Guess and Check

A repeating decimal is hard to multiply, so change it to a fraction. Then write the product with letters: the denominator $11$ has to cancel and the leftover factor $5$ needs a partner to make an even exponent.

3 · Execute3 carry out the plan

1Change the decimal to a fraction

#15 Organize Information in More Ways 8.NS.A.1

Two repeating digits after the point means divide by

99

.

0.\overline{45} = \frac{45}{99} = \frac{5}{11}

Now the product is

\frac{5}{11} \times A

.

2Write the square condition with letters

#13 Convert to Algebra 4.OA.B.4

We need

\frac{5A}{11} = m^2

for a natural number

m

. Since

5

and

11

share no factor,

11

must divide

A

; write

A = 11B

. Then

5B = m^2

, so

5

divides

m^2

and therefore

m

, which makes

B

5

times a square.

A = 11B, \quad 5B = m^2 \Rightarrow B = 5k^2 \Rightarrow A = 55k^2

Every prime in a perfect square appears an even number of times, so the lone

5

needs another

5

.

3Take the smallest case and check

#6 Guess and Check 8.NS.A.1

The smallest choice is

k = 1

, giving

A = 55

. Check the product.

\frac{5}{11} \times 55 = 25 = 5^2

The product

25

is a perfect square, so

A = 55

works.

Answer: 55

4 · Reviewdoes it hold up?

Smaller candidates fail: $A$ must be a multiple of $11$ , and $A = 11, 22, 33, 44$ give $5, 10, 15, 20$ , none of them a perfect square. So $55$ is the first that works.

Another way: Guess and check over the multiples of $11$ : $11 \to 5$ , $22 \to 10$ , $33 \to 15$ , $44 \to 20$ , $55 \to 25 = 5^2$ . The fifth multiple is the first to give a square.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Writing $0.\overline{45}$ as the fraction $\frac{5}{11}$ and checking the product.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Using factors and multiples to see that $A$ must be $11$ times $5$ times a square.

💡Takeaway. Turn the repeating decimal into a fraction, then pair up every prime factor.

Variant 12 hard

Multiply the repeating decimal $0.\overline{63}$ by a natural number $A$ so that the product is the square of some natural number. Find the smallest possible value of $A$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Find the smallest natural number $A$ that makes $0.\overline{63} \times A$ equal to a perfect square of a natural number.

Givens

The repeating decimal is $0.\overline{63}$ .
$A$ is a natural number.

Unknowns

The smallest possible $A$ .

Constraints

The product must be the square of a natural number, so it must be a whole number first.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #13 Convert to Algebra#6 Guess and Check

A repeating decimal is hard to multiply, so change it to a fraction. Then write the product with letters: the denominator $11$ has to cancel and the leftover factor $7$ needs a partner to make an even exponent.

3 · Execute3 carry out the plan

1Change the decimal to a fraction

#15 Organize Information in More Ways 8.NS.A.1

Two repeating digits after the point means divide by

99

.

0.\overline{63} = \frac{63}{99} = \frac{7}{11}

Now the product is

\frac{7}{11} \times A

.

2Write the square condition with letters

#13 Convert to Algebra 4.OA.B.4

We need

\frac{7A}{11} = m^2

for a natural number

m

. Since

7

and

11

share no factor,

11

must divide

A

; write

A = 11B

. Then

7B = m^2

, so

7

divides

m^2

and therefore

m

, which makes

B

7

times a square.

A = 11B, \quad 7B = m^2 \Rightarrow B = 7k^2 \Rightarrow A = 77k^2

Every prime in a perfect square appears an even number of times, so the lone

7

needs another

7

.

3Take the smallest case and check

#6 Guess and Check 8.NS.A.1

The smallest choice is

k = 1

, giving

A = 77

. Check the product.

\frac{7}{11} \times 77 = 49 = 7^2

The product

49

is a perfect square, so

A = 77

works.

Answer: 77

4 · Reviewdoes it hold up?

Smaller candidates fail: $A$ must be a multiple of $11$ , and $A = 11, 22, 33, 44, 55, 66$ give $7, 14, 21, 28, 35, 42$ , none of them a perfect square. So $77$ is the first that works.

Another way: Guess and check over the multiples of $11$ : $11 \to 7$ , $22 \to 14$ , $33 \to 21$ , $44 \to 28$ , $55 \to 35$ , $66 \to 42$ , $77 \to 49 = 7^2$ . The seventh multiple is the first to give a square.

Standardsmin grade 8

8.NS.A.1 Know that numbers that are not rational are called irrational numbers — Writing $0.\overline{63}$ as the fraction $\frac{7}{11}$ and checking the product.
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite — Using factors and multiples to see that $A$ must be $11$ times $7$ times a square.

💡Takeaway. Turn the repeating decimal into a fraction, then pair up every prime factor.

Make every prime exponent even · 12 practice problems

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