TS ICET 2015 Question Paper with Solutions for online practice

Question : 85 of 200

Marks: +1, -0

If m=

7^1 . 5^2 . 3^4 . 2^8

, then the greatest integer value of k such that

21^k

divides m is

Solution:

Given :

m=7^{1}.5^{2}.3^{4}.2^{8}

⇒{m}/{21^{k}}={7^{1}⋅5^{2}⋅3^{4}⋅2^{8}}/{21^{k}}={21.5^{2}⋅3^{3}⋅2^{8}}/{21^{k}}

∴ The required greatest value of k = 1

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