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Re: If c is randomly chosen from the integers 20 to 99, inclusi [#permalink]
Can anyone explain this please?!
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Re: If c is randomly chosen from the integers 20 to 99, inclusi [#permalink]
Carcass wrote:
This is a really tough question.

Starting from what we do have

\(\frac{favorable outcomes}{total # of possibilities}\)

99-20+1=80 possibilities

Now a number of being divisible by 12 must contain in it 2*2*3.

Here is probably the most difficult thing to realize \(c^3 - c\) is the same of having \(c(c^2 - 1)\) = \((c-1)c(c+1)\)

This latter expression means you do have a set of 3 consecutive integers divisible by 12. A rule is: every 3 consecutive integers must be divisible by 3

Actually, the question really boils down: how many numbers you pick from are divisible by \(2*2=4\).

In this case, c can be 21,23,25.............95,97,99 so \(\frac{99-21}{2}\) + 1 = 40


OR 20,24,26...........92,96 so \(\frac{96-20}{4}\) + 1 = 20

Which means \(\frac{40+20}{80}\) = \(\frac{60}{80}\) = \(\frac{3}{4}\)

Ask if something is unclear.

Regards


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)
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Re: If c is randomly chosen from the integers 20 to 99, inclusi [#permalink]
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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 3

Now 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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Re: If c is randomly chosen from the integers 20 to 99, inclusi [#permalink]
Hi :)
may i kindly ask you to specify how did you arrive to conclusion that C^3 is c*(c^2-1)?
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Re: If c is randomly chosen from the integers 20 to 99, inclusi [#permalink]
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jullskl wrote:
Hi :)
may i kindly ask you to specify how did you arrive to conclusion that C^3 is c*(c^2-1)?




Can you plz check it again:

it is \(c^3 - c = c(c^2 -1)\)
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Re: If c is randomly chosen from the integers 20 to 99, inclusi [#permalink]
Why is it divisible by only 3. Please explain the divisibility part. The part where it has to be divisible by 2*2.
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Re: If c is randomly chosen from the integers 20 to 99, inclusi [#permalink]
Good question - thank you for posting this up !
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Re: If c is randomly chosen from the integers 20 to 99, inclusi [#permalink]
Dear Carcass, I have a concern here- do we need to determine whether c is divisible by 12 or we need to determine (C*3-C )is divisible by 12.

If its (C*3-C) then what ever number we chose between 20 to 99 its always divided by 12.
Like if we chose C=37 then three consecutive number 36*37*38 is always divided by 12 .
Correct me if wrong.
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Re: If c is randomly chosen from the integers 20 to 99, inclusi [#permalink]
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Expert Reply
Fixed to looks better

Two things, now:
1. There are total of 80 integers from 20 to 99, inclusive: 20, 21, ..., 99.
2. C^3-C=(C-1)*C*(C+1): we have the product of 3 consecutive integers, which is always divisible by 3. So the question basically is whether (C-1)*C*(C+1) is divisible by 4.

Next, the only way the product NOT to be divisible by 4 is C to be even but not a multiple of 4, in this case we would have (C-1)*C*(C+1)=odd*(even not multiple of 4)*odd.

Now, out of first the 4 integers {20, 21, 22, 23} there is only 1 even not multiple of 4: 22. All following groups of 4 will also have only 1 even not multiple of 4 (for example in {24, 25, 26, 27} it's 26, and in {96, 97, 98, 99} it's 98, always 3rd in the group) and as our 80 integers are entirely built with such groups of 4 then the overall probability that C is not divisible by 4 is 1/4. Hence the probability that it is divisible by 4 is \(1-\frac{1}{4}=\frac{3}{4}\).
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Re: If c is randomly chosen from the integers 20 to 99, inclusi [#permalink]
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Alternative

(C-1)C(C+1) should be divisible by 12.


Question is: How many of the integers from 20 to 99 are either ODD or Divisible by 4.

ODD=(99-21)/2+1=40
Divisible by 4= (96-20)/4+1=20
Total=99-20+1=80

P=Favorable/Total=(40+20)/80=60/80=3/4
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Re: If c is randomly chosen from the integers 20 to 99, inclusi [#permalink]
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Another method:
divide the number into equal segments:
20-27
28-35
between 20-27 the probability is 6/8
Similarly for 28-35 the probability is 6/8.
Hence the total probability is 6/8 = 3/4
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Re: If c is randomly chosen from the integers 20 to 99, inclusi [#permalink]
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Re: If c is randomly chosen from the integers 20 to 99, inclusi [#permalink]
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