Question

• find the smallest number by which 2028 must be multiplied so that the product become the perfect square . also find the square root of the perfect square so obtained

• find the smallest number by which 5445 must be divided so that it becomes a perfect square . also ,find the square root of the perfect square so obtained .

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Answer 1

QUESTION -

find the smallest number by which 2028 must be multiplied so that the product become the perfect square . also find the square root of the perfect square so obtained

ANSWER -

$\begin{array}{r | l}</p><p></p><p>2 & 2028 \\</p><p></p><p>2 & 1014 \\</p><p></p><p>3 & 507 \\</p><p></p><p>1 3 & 169 \\</p><p></p><p>13 & 13 \\</p><p></p><p> & 1</p><p></p><p>\end{array}$

By prime Factorisation , we get

2028 = 2 × 2 × 3 × 13 × 13

Clearly, on grouping the prime factors of 2028 into pairs of equal factors , we are left with the factor 3 , which can not be paired .

thus , to make 2028 a perfect square , it must be multiplied by 3

New number

→ 2028 × 3

→ 6084

→ 2 × 2 × 3 × 3 × 13 × 13

→ ( 2 × 3 × 13 ) ²

∴ Square root of the new number (perfect square) → 2 × 3 × 13 = 78

QUESTION -

find the smallest number by which 5445 must be divided so that it becomes a perfect square . also ,find the square root of the perfect square so obtained

ANSWER -

$\begin{array}{r | l}</p><p></p><p>3 & 5445 \\</p><p></p><p>3 & 1815 \\</p><p></p><p>5 & 605 \\</p><p></p><p>11 & 121 \\</p><p></p><p>11 & 11 \\</p><p></p><p> & 1</p><p></p><p>\end{array}$

By prime Factorisation , we get

5445 = 3 × 3 × 5 × 11 × 11

Clearly, on grouping the prime factors of 5445 into pairs of equal factors , we are left with the factor 5 , which can not be paired . So , 5445 must be divided by 5 , so that it becomes a perfect square .

New number

→ 5445 / 5

→ 1089

→ 3 × 3 × 11 × 11

→ ( 3 × 11 ) ²

∴ Square root of the new number (perfect square) → 3 × 11 = 33

Answer 2

QUESTION -

find the smallest number by which 2028 must be multiplied so that the product become the perfect square . also find the square root of the perfect square so obtained

ANSWER -

\begin{gathered}\begin{array}{r | l} < /p > < p > < /p > < p > 2 & 2028 \\ < /p > < p > < /p > < p > 2 & 1014 \\ < /p > < p > < /p > < p > 3 & 507 \\ < /p > < p > < /p > < p > 1 3 & 169 \\ < /p > < p > < /p > < p > 13 & 13 \\ < /p > < p > < /p > < p > & 1 < /p > < p > < /p > < p > \end{array}\end{gathered}

</p><p></p><p>2

</p><p></p><p>2

</p><p></p><p>3

</p><p></p><p>13

</p><p></p><p>13

</p><p></p><p>

2028

1014

507

169

13

1</p><p></p><p>

By prime Factorisation , we get

2028 = 2 × 2 × 3 × 13 × 13

Clearly, on grouping the prime factors of 2028 into pairs of equal factors , we are left with the factor 3 , which can not be paired .

thus , to make 2028 a perfect square , it must be multiplied by 3

New number

→ 2028 × 3

→ 6084

→ 2 × 2 × 3 × 3 × 13 × 13

→ ( 2 × 3 × 13 ) ²

∴ Square root of the new number (perfect square) → 2 × 3 × 13 = 78

QUESTION -

find the smallest number by which 5445 must be divided so that it becomes a perfect square . also ,find the square root of the perfect square so obtained

ANSWER -

\begin{gathered}\begin{array}{r | l} < /p > < p > < /p > < p > 3 & 5445 \\ < /p > < p > < /p > < p > 3 & 1815 \\ < /p > < p > < /p > < p > 5 & 605 \\ < /p > < p > < /p > < p > 11 & 121 \\ < /p > < p > < /p > < p > 11 & 11 \\ < /p > < p > < /p > < p > & 1 < /p > < p > < /p > < p > \end{array}\end{gathered}

</p><p></p><p>3

</p><p></p><p>3

</p><p></p><p>5

</p><p></p><p>11

</p><p></p><p>11

</p><p></p><p>

5445

1815

605

121

11

1</p><p></p><p>

By prime Factorisation , we get

5445 = 3 × 3 × 5 × 11 × 11

Clearly, on grouping the prime factors of 5445 into pairs of equal factors , we are left with the factor 5 , which can not be paired . So , 5445 must be divided by 5 , so that it becomes a perfect square .

New number

→ 5445 / 5

→ 1089

→ 3 × 3 × 11 × 11

→ ( 3 × 11 ) ²

∴ Square root of the new number (perfect square) → 3 × 11 = 33