Question

If the prime factorisation of a natural number n is 24 × 32 × 52 × 11, then write the number of consecutive

zeroes in n.​

Answer 1

Question :- If n = 24 * 32 * 52 * 11, then the number of consecutive zeros in last of n, where n is a natural number, is. ?

Solution :

To understand the problem, lets take few examples first.

→ 10 = 2*5

→ 100 = 2*2*5*5

→ 1000 = 2*2*2*5*5*5

→ 1000OO = 2*2*2*2*2*5*5*5*5*5

From this we can conclude that, Total number of zeros in the last of a number depend upon the 2 or 5 . Or we can say that, mainly on 5 . (As 4,6,8, also has factors 2).

if we have to find how many zeros in last of a natural number , we will see how many times 5 will come in factors of the number.

So,

→ N = 24 * 32 * 52 * 11

Prime Factors of N will be

→ N = (2*2*2*3) * (2*2*2*2*2) * (2*2*13) * (1*11)

As we can see here 2 comes ten times , but 5 comes zero times .

Therefore, we can conclude that,

Total number of consecutive zeros in the last of N will be zero...