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Question

Number of 5 digit numbers which are divisible by 5 and each number containing the digit 5, digits being all different is equal to K4!, then the value of K is


A

84

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B

168

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C

188

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D

208

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Solution

The correct option is B

168


Finding the value of k:

The number which is divisible by 5 when the digit at unit place of that number must be either 5 or 0

Given that,

Each 5 digit numbers which are divisible by 5 must contain the digit 5

Case 1: Consider 0 at unit place

Number of 5 digit numbers containing digit 5 in this case =4×8×7×6=1344

Case 2: Consider 5 at unit place

First digit cannot be the 0.

Number of 5 digit numbers in this case =8×8×7×6=2688

Therefore, total number of 5 digit numbers which are divisible by 5 must contain the digit 5=1344+2688=4032

Now,

K4!=4032K4!=168×24K4!=1684!

Equating both side

K=168

Hence, the value of K=168


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