Q. 64.1( 18 Votes )

# If the prime factorization of a natural number n is 2^{3} × 3^{2} × 5^{2} × 7, write the number of consecutive zeros in n.

Answer :

If any number ends with the digit 0, it should be divisible by 10,

i.e. it will be divisible by 2 and 5.

Prime factorization of n is given as 2^{3} × 3^{2} × 5^{2} × 7.

It can be observed that there is (2 × 5) × (2 × 5)

⇒ 10 × 10 = 100

Thus, there are 2 zeros in n.

The number of consecutive zeros in n is 2.

Rate this question :

Find the LCM and HCF of the following pair of integers and verify that LCM X HCF = Product of two numbers :

902 and 1517

KC Sinha - MathematicsFind the LCM and HCF of the following pair of integers and verify that LCM X HCF = Product of two numbers :

36 and 64

KC Sinha - MathematicsFind LCM and HCF of the following integers by using prime factorization method:

48, 72 and 108

KC Sinha - MathematicsFind LCM and HCF of the following integers by using prime factorization method:

8, 9, and 25

KC Sinha - Mathematics