Question

smallest number by which each of the following numbers must be multiplied to get a perfect squre number also find the squre root the number is 4500​

Answer 1

1) We first find the prime factors of 252 By prime
factorization.

Hence,

252 = 2 x 2 x 3 x 3 x 7

= (2×2)×(3×3)×7


Here, we see that 2 and 3 occur  in pairs but 7 needs a pair. Hence, the given
number is not a perfect square.



If We now multiply 252 by 7 then we get



252×7= 1764 = (2×2)×(3×3)×(7×7)



 Therefore,the
number 252 has 3 pairs of equal prime factors .



Hence, 1764 is a perfect square &

√1764= 2×3×7=42



Hence, the smallest number by which 1764 must
be multiplied so that the product is a perfect square is 7.

And the square root of the new number is
√1764=42.




2)

By prime factorization, we get, 180 = (3 x 3) x (2
x 2) x 5



Here, 3 and 2 are in pair but 5 needs a pair to
make 180 a perfect square.



So, 180 needs to be multiplied by 5 to become a
perfect square.

180×5 =(3 x 3) x (2 x 2) x (5×5)

Therefore, the number 180 has 3 pairs of equal
prime factors .

Hence, 900 is a perfect square & √ 900=
3×2×5=30

Hence, the smallest number by which 180 must be
multiplied so that the product is a perfect square is 5.

And the square root of the new number is √900=30.

 

 



3) By prime factorization of 1008, we get

 



1008 = (2 x 2) x (2 x 2) x (3 x 3 )x 7



Here, 2 and 3 are in pair, but 7 needs a pair to
make 1008 a perfect square.



Thus, 1008 needs to be multiplied by 7 to become
a perfect square.



1008 × 7 = (2 x 2 )x (2 x 2) x (3 x 3) x (7×7)

Therefore, the number 1008 has 4 pairs of equal
prime factors .

Hence 7056 is a perfect square & √ 7056= 2×2×3×7=84

 

Hence, the smallest number by which1008 must
be multiplied so that the product is a perfect square is 7.

And the square root of the new number is √7056=84.

 



4) By prime factorization of 2028, we get



2028 = 2 x 2 x 3 x 13 x 13



Here, 2 and 13 are in pair, but 3 needs a pair
to make 2028 a perfect square.



Thus, 2028 needs to be multiplied by 3 to become
a perfect square.

2028 ×3 = (2 x 2) x (3 x 3)×(13 x 13)

 

Therefore, the number 6084 has 3 pairs of equal
prime factors .

Hence, 6084 is a perfect square & √ 6084= 2×3×13=78

 

Hence, the smallest number by which 2028 must
be multiplied so that the product is a perfect square is 7.

And the square root of the new number is √6084=78.

 



5) By prime factorization of 1458, we get



1458 = 2 x 3 x 3 x 3 x 3 x 3 x 3



Here, 3 are in pair, but 2 needs a pair to make
1458 a perfect square.



So, 1458 needs to be multiplied by 2 to become a
perfect square.

1458 ×2 =(2× 2) x (3 x 3) x (3 x 3) x (3 x 3)

 

Therefore, the number  has 4 pairs of equal prime factors .

Hence, 2916 is a perfect square & √2916= 2×3×3×3=54

 

Hence, the smallest number by which 1458 must be multiplied
so that the product is a perfect square is 2.

And the square root of the new number is √2916=54.

 



6) By prime factorization of 768, we get

768= 2 x 2 x 2 x 2 x 2 x 2 x 2 x
2 x 3



Here, 2 are in pair, but 3 needs a pair to make
768 a perfect square.



So, 768 needs to be multiplied by 3 to become a
perfect square.

768 × 3=( 2 x 2 )x (2 x 2) x (2 x
2) x (2 x 2) x (3×3)

 

Therefore, the number 768 has 5 pairs of equal
prime factors .

Hence, 2304 is a perfect square & √2304= 2×2×2×2×3=48

 

Hence, the smallest number by which 768 must be
multiplied so that the product is a perfect square is 3

And the square root of the new number is √2304=48.