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
Open in App
(ii) 12150=2×3×3×3––––––×3×3––––––×5×5––––––
Grouping the factors in pairs of equal factors, we see that factors 2 and 3 are left unpaired
∴ In order to complete the pairs, we have to multiply 12150 by 2×3=6 i.e., then the product wil be the complete sqaure.
∴ Required smallest number = 6
(iii) 7688 =2×2×2––––––×31×31––––––––
∴ In order to complete the pairs we have to multiply 7688 by 2, then the product will be the complete square.
∴ Required smallest number = 2
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