Question

If n=23×34×54×7, then the number of consecutive zeros in n, where n is a natural number, is...?

Answer 1

Question :- If n=23×34×54×7, 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

→ 100000 = 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 = 23 * 34 * 54 * 7

Prime Factors of N will be

→ N = 23 * (2 * 17) * (2 * 3 * 3 * 3) * 7

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

Therefore, we can conclude that,

Total number of consecutive zeros in the last of N will be Null. ( No zero..)

Answer 2

correct answer is 3

Step-by-step explanation:

n=2³×3⁴×5⁴×7

n=2³×5⁴×3⁴×7

n=2³×5³×5×3⁴×7

n=(2×5)³×5×3⁴×7

n=5×3⁴×7×(2×5)³

n=5×3⁴×7×(10)³

so the number of consecutive zeros are 3 because 10 has power 3 and (10)³ = 1000 - having 3 consecutive zeros .