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rocky620
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11 Oct 2020, 07:13
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Bunuel, VeritasKarishma , chetan2u
Hi Sir/Mam, I Approached this ques as below:
Total numbers = 99-20+1=80
Since (c-1)*c*(c+1) is always divisible by 3, we need to find its divisibility by 4.
There are 20 multiples of 4 in this range, and these numbers can take any three positions of (c-1), c, (c+1). So, there will be 60 such cases. and hence the probability 60/80=3/4.
Is it a valid approach in all these kind of ques?
Hi Sir/Mam, I Approached this ques as below:
Total numbers = 99-20+1=80
Since (c-1)*c*(c+1) is always divisible by 3, we need to find its divisibility by 4.
There are 20 multiples of 4 in this range, and these numbers can take any three positions of (c-1), c, (c+1). So, there will be 60 such cases. and hence the probability 60/80=3/4.
Is it a valid approach in all these kind of ques?
11 Oct 2020, 09:32
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shashankp27
To be divisible by 12, c³-c must be a multiple of 3 and 4.
c³-c = c(c²-1) = c(c-1)(c+1).
c-1, c, and c+1 are three consecutive integers.
Of every three consecutive integers, exactly one is a multiple of 3.
Thus, (c-1)(c)(c+1) is a multiple of 3.
The question is whether (c-1)(c)(c+1) will be a multiple of 4.
If c is odd, then (c-1)(c)(c+1) = (even)(odd)(even).
In this case, the product will be a multiple of 4, since it includes two even factors.
If c is a multiple of 4, then (c-1)(c)(c+1) = (odd)(multiple of 4)(odd).
In this case, the product will be a multiple of 4, since one of its factors is a multiple of 4.
If c is an even integer that is NOT a multiple of 4, then (c-1)(c)(c+1) = (odd)(even non-multiple of 4)(odd).
In this case, (c-1)(c)(c+1) will NOT be a multiple of 4.
Between 20 and 99:
The total number of integers = 80.
Since half are even, the number of even integers = 40.
Since every other even integer is a multiple of 4, the number of even integers that are NOT multiples of 4 = 20.
Thus, P(even non-multiple of 4) = 20/80 = 1/4.
Since any other value for c will yield a multiple of 4 for (c-1)(c)(c+1), we get:
P(c³-c will be a multiple of 12) = 1 - 1/4 = 3/4.
11 Oct 2020, 10:53
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Total numbers from 20 to 99: 99 - 20 + 1 = 80
\(c^3\) - c is divisible by 12:
=> \(c^3 - c = c (c^2 - 1)\) = (c - 1) * c * ( c + 1)
As it is the product of three consecutive integers then it is divisible by '3'. So, we will consider all those numbers those are multiples of only '4' and only '12'
For example: 12: Multiple is 24: We will have '3' possible appearance of 24: (22,23,24), (23,24,25) and (24,25,26)
So, every multiple will have '3' appearances.Total multiple of '12' till 99 from 20 are: 24, 36, 48, 60, 72, 84, 96 = 7
Total number divisible: 7 * 3 = 21
Total multiples of '4' till 99 from 20 excluding common multiples of '12' and '4': 20, 28, 32, 40, 44, 52, 56, 64, 68, 76, 80, 88, 92 = 13
Total number divisible: 13 * 3 = 39
Overall total numbers: 21 + 39 = 60.
Probability: \(\frac{60 }{ 80} = \frac{3}{4}\)
Answer C
\(c^3\) - c is divisible by 12:
=> \(c^3 - c = c (c^2 - 1)\) = (c - 1) * c * ( c + 1)
As it is the product of three consecutive integers then it is divisible by '3'. So, we will consider all those numbers those are multiples of only '4' and only '12'
For example: 12: Multiple is 24: We will have '3' possible appearance of 24: (22,23,24), (23,24,25) and (24,25,26)
So, every multiple will have '3' appearances.Total multiple of '12' till 99 from 20 are: 24, 36, 48, 60, 72, 84, 96 = 7
Total number divisible: 7 * 3 = 21
Total multiples of '4' till 99 from 20 excluding common multiples of '12' and '4': 20, 28, 32, 40, 44, 52, 56, 64, 68, 76, 80, 88, 92 = 13
Total number divisible: 13 * 3 = 39
Overall total numbers: 21 + 39 = 60.
Probability: \(\frac{60 }{ 80} = \frac{3}{4}\)
Answer C
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