TS ICET 23 May 2018 Shift 2 Paper

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Question : 110 of 200

Marks: +1, -0

If

\k_1,\k_2

are natural numbers such that
(i)

2^{\k_1}

|10! but

2^{\k_1 +1}

∤ 10 ! and

5^{\k_2}

|10! but

5^{\k_2 +1}

∤ 10 ! then

$\k_1 = \k_2$
$\k_1< \k_2$
$\k_1> \k_2$
$\k_1 = \k_2 + 3$

Solution:

110.10 !=2^{8} × 3^{4} × 5^{2} × 7

k_1

is the greatest index on 2 that divides

10! = 8

k_2

is the greatest index on 5 that divides

10! = 2

Clearly,

k_1 > k_2

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