What is the smallest number by which 7688 must be multiplied so the product is a perfect square number?

To do :

We have to find the smallest numbers by which the given numbers must be multiplied so that the products are perfect square.

Solution:

Perfect Square: A perfect square has each distinct prime factor occurring an even number of times.

(i) Prime factorisation of 23805 $=3\times3\times5\times23\times23$

$=(3)^2\times5\times(23)^2$

$=(3\times23)^2\times5$

$=(69)^2\times5$

In order to make the pairs an even number of pairs, we have to multiply 23805 by 5, then the product will be the perfect square.

Therefore, 5 is the smallest number by which 23805 must be multiplied so that the product is a perfect square.

(ii) Prime factorisation of 12150 $=2\times3\times3\times3\times3\times3\times5\times5$

$=2\times3\times(3)^2\times(3)^2\times(5)^2$

In order to make the pairs an even number of pairs, we have to multiply 12150 by $2\times3=6$, then the product will be the perfect square.

Therefore, 6 is the smallest number by which 12150 must be multiplied so that the product is a perfect square.

(iii) Prime factorisation of 7688 $=2\times2\times2\times31\times31$

$=2\times(2)^2\times(31)^2$

In order to make the pairs an even number of pairs, we have to multiply 7688 by $2$, then the product will be the perfect square.

Therefore, 2 is the smallest number by which 7688 must be multiplied so that the product is a perfect square.

(i) 23805

First find the prime factors for 23805

23805 = 3×3×23×23×5

By grouping the prime factors in equal pairs we get,

= (3×3) × (23×23) × 5

By observation, prime factor 5 is left out.

So, multiply by 5 we get,

23805 × 5 = (3×3) × (23×23) × (5×5)

= (3×5×23) × (3×5×23)

= 345 × 345

= (345)2

∴ Product is the square of 345.

(ii) 12150

First find the prime factors for 12150

12150 = 2×2×2×2×3×3×5×5×2

By grouping the prime factors in equal pairs we get,

= (2×2) × (2×2) × (3×3) × (5×5) × 2

By observation, prime factor 2 is left out.

So, multiply by 2 we get,

12150 × 2 = (2×2) × (2×2) × (3×3) × (5×5) × (2×2)

= (2×2×3×5×2) × (2×2×3×5×2)

= 120 × 120

= (120)2

∴ Product is the square of 120.

(iii) 7688

First find the prime factors for 7688

7688 = 2×2×31×31×2

By grouping the prime factors in equal pairs we get,

= (2×2) × (31×31) × 2

By observation, prime factor 2 is left out.

So, multiply by 2 we get,

7688 × 2 = (2×2) × (31×31)× (2×2)

= (2×31×2) × (2×31×2)

= 124 × 124

= (124)2

∴ Product is the square of 124.

Find the smallest number by which the given number must be multiplied so hat the product is a perfect square. i 23805 ii 12150 iii 7688

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(ii) $12150=2\times 3\times \underset{–}{3\times 3}\times \underset{–}{3\times 3}\times \underset{–}{5\times 5}$

Grouping the factors in pairs of equal factors, we see that factors 2 and 3 are left unpaired

$\therefore$ In order to complete the pairs, we have to multiply 12150 by $2\times 3=6$ i.e., then the product wil be the complete sqaure.

(iii) 7688 $=2\times \underset{–}{2\times 2}\times \underset{–}{31\times 31}$

$\therefore$ In order to complete the pairs we have to multiply 7688 by 2, then the product will be the complete square.