kruttikaaggarwal wrote:

I do not understand this method. All we need is c to be divisble by 12 because \(c^3 - c\) = \(c(c^2 - 1)\) . That means there are 7 out of the 80 integers divisble by 12 (24, 36, 48, 60, 72, 84 and 96)

Hi,

You are only considering C in this case and what about \(c^2-1\)? which is (c - 1)(c + 1)

If we simplify \(c(c^2 - 1)\) = (c - 1)c(c + 1)

i.e. we have 3 consecutive numbers and

it is always divisible by 3Now since it can be any number from 20 to 99 then we have to take both possibilities i.e.

1. If c is odd

(c = 21, 23, 25, ..., 95, 97,99)(even)(odd)(even)

(c - 1)c(c + 1)

SO The total number of ODD Integers

99 = 21 + (c-1) 2 or c = 40

2. if c is even

(c = 20, 24, 28, ..., 92, 96)(odd)(even)(odd)

(c - 1)c(c + 1)

Since we have 2 odd numbers and they are consecutive nos. so anyone will be divisible by 3, we are concern about the even number, which must be only divisible by 4, because 26 is also even but odd*26*odd is not divisible by 4 and the question becomes null, so the question to satisfy only even multiple of 4 is to be considered

so The total Numbers divisible by 4

96 = 20 + (c-1)4 or c = 20

Now the probability = \(\frac{(40 +20)}{80}= \frac{3}{4}\)

***Plz remember if the total numbers in a set is

EVEN then always half are odd nos. and half are even nos. in this case we had total of 80 nos. so half i.e. 40 are odd and 40 are even, but here we need even nos. that are multiple of 4 so we did the calculation.

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