Question

If the prime factorization of a number is
$2 {}^{3} \times 3 {}^{2} \times 5 {}^{2} \times 7$
then how many
zeroes at the end of the number?
​

Answer 1

$2 {}^{3} \times 3 {}^{2} \times 5 {}^{2} \times 7$. =

2×2×3×3×5×5×7 = 6300 Or 6.3× 10³

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Answer 2

Answer :-

We are given that,

Prime factorization of a number = 2³ × 3² × 5² × 7

Now, we are asked to find the number of zeros at the end of the number.

For a number to end with zero, it must be multiple of 10 and factors of 10 are 2 and 5.

So, a number must have pair of 2 and 5 as its factor to end with zero.

For the given number, 2³ × 3² × 5² × 7

Power of 3 and 7 doesn't matter.

Now, we have to see that how many 10 can be form from 2³ × 5².

Here, as the least power is 2, there can be 2 pair possible to make product as 10.

So, number of zeros at the end of number = 2