Question

If the prime factorization of a natural number(n) is 2^3×3^2×5^2×7 how many zeroes will it have at the end

Answer 1

it will have one zero at the end because one pair of 2 and 5 makes one ten.

Answer 2

$\mathfrak{\huge{\underline{\blue{Points\: To\: Remember}}}}$

¶ The No. of Zeroes in prime factorisation of a No. or End of product is contributed by :

$2^{m} × 5^{n}$

{As we have product of 2 & 5 results a Zero placed at the End of the Number}

The No. of zeroes is equal to the power of 5 or 2 whichever is less.

i. e.,

¶ No. of zeroes in End of product(E.O.P)
= m ; if m < n
= n ; if n < m

$\mathbb{\underline{\blue{SOLUTION:}}}$

Given,
Prime factorisation of No. is
2^{3}×3^{2}×5^{2}×7

We only consider
2^{3} × 5^{2}

Here 2<3

•°• $\underline{\underline{ No.\:of\: zeroes\: in\: E.O.P\: of\: the\: given\: No.\: is\: 2. }}$

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$\mathcal{\huge{\orange{Hope\: it\: Helps}}}$