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If c is randomly chosen from the integers 20 to 99, inclusi

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If c is randomly chosen from the integers 20 to 99, inclusi [#permalink] New post 26 Aug 2017, 02:19
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If c is randomly chosen from the integers 20 to 99, inclusive, what is the probability that \(c^3\) – c is divisible by 12?

enter your value as a fraction

[Reveal] Spoiler: OA
\(\frac{3}{4}\)

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Re: If c is randomly chosen from the integers 20 to 99, inclusi [#permalink] New post 30 Aug 2017, 07:09
Can anyone explain this please?!
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Re: If c is randomly chosen from the integers 20 to 99, inclusi [#permalink] New post 31 Aug 2017, 08:36
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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.

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Re: If c is randomly chosen from the integers 20 to 99, inclusi [#permalink] New post 16 Sep 2018, 18:18
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] New post 16 Sep 2018, 20:52
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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] New post 22 Dec 2018, 09:09
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] New post 27 Dec 2018, 09:48
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] New post 23 Mar 2019, 12:19
Why is it divisible by only 3. Please explain the divisibility part. The part where it has to be divisible by 2*2.
Re: If c is randomly chosen from the integers 20 to 99, inclusi   [#permalink] 23 Mar 2019, 12:19
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