Question

If n= $2^{3}*3^{4}*5^{4}*7$, then the number of consecutive zeros in n,where n is a natural number, is
(a) 2
(b) 3
(c) 4
(d) 7

Answer 1

Option (b) is Correct : 3  

The number of consecutive zeros in n, is 3 (three) .

SOLUTION :  

GIVEN : Prime factorization of a natural number n is 2³ × 3⁴ × 5⁴ × 7

n = 2³ × 3⁴ × 5⁴ × 7

n = 2³ × 3⁴ × 5³ × 5 × 7

n = (2³ × 5³) × 3⁴ × 5 × 7

n = (2 × 5)³ × 3⁴ × 5 × 7

n = 10³ × 3⁴ × 5 × 7

Hence, the number of consecutive zeros in n, is 3 (three) .

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Answer 2

Hi there!
Here's the answer:

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¶ NO. OF ZEROES IN END OF PRODUCT

- Zeroes at the end of product are contributed only by 2 and 5 , as 2×5 places a zero at the end of Result.



When the No. is resolved into prime factors,
CASE - 1: If exponent of 2 < Exponent of 5

No. of zeroes at the end is
= Exponent of 2

CASE - 2: If exponent of 5 < Exponent of 2

No. of zeroes at the end is
= Exponent of 5

In other words,
No. of zeroes = Exponent of 2 or 5 whichever is less

•°•°•°•°<><><<><>><><>°•°•°•

SOLUTION:

Given,

n= $2^{3}*3^{4}*5^{4}*7$

Exponent of 2 = 3
Exponent of 5 = 4
3 < 4

•°• No. of zeroes at the End of product = 3


This answer exists in $\underline{\underline{Option\: (b)}}$

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Hope it helps