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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. # If c is randomly chosen from the integers 20 to 99, inclusi  Question banks Downloads My Bookmarks Reviews Important topics
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TAGS: Founder  Joined: 18 Apr 2015
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If c is randomly chosen from the integers 20 to 99, inclusi [#permalink]
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Expert's post 00:00

Question Stats: 33% (02:24) correct 66% (05:00) wrong based on 27 sessions
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Re: If c is randomly chosen from the integers 20 to 99, inclusi [#permalink] Founder  Joined: 18 Apr 2015
Posts: 13940
GRE 1: Q160 V160 Followers: 316

Kudos [?]: 3690  , given: 12953

Re: If c is randomly chosen from the integers 20 to 99, inclusi [#permalink]
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Expert's post
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 $$\frac{40+20}{80}$$ = $$\frac{60}{80}$$ = $$\frac{3}{4}$$

Regards
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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 $$\frac{40+20}{80}$$ = $$\frac{60}{80}$$ = $$\frac{3}{4}$$

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) VP Joined: 20 Apr 2016
Posts: 1302
WE: Engineering (Energy and Utilities)
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Kudos [?]: 1342  , given: 251

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)?
VP Joined: 20 Apr 2016
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Re: If c is randomly chosen from the integers 20 to 99, inclusi [#permalink]
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. Re: If c is randomly chosen from the integers 20 to 99, inclusi   [#permalink] 23 Mar 2019, 12:19
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