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If integer C is randomly selected from 20 to 99, inclusive.

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Re: If integer C is randomly selected from 20 to 99, inclusive.  [#permalink]

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New post 15 Oct 2018, 13:51
Bunuel wrote:
If integer C is randomly selected from 20 to 99, inclusive. What is the probability that c^3-c is divisible by 12?
A. 1/2
B. 2/3
C. 3/4
D. 4/5
E. 1/3

Two things:
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 1-1/4=3/4.

Answer: C.


nkhosh wrote:
Would you please explain (1) The formulas you used for finding the # of ODD and "Divisible by 4" outcomes?
AND: (2) Why are we looking for the number of ODD integers within 20-99? I'm confused b/c 23 for example is not divisible by 12....

Thank you for the great post :)


1. There are even # of consecutive integers in our range - 80 (from 20 to 99, inclusive). Out of even number of consecutive integers half is always odd and half is always even, thus numbers of odd integers in the given range is 40.

2. \(# \ of \ multiples \ of \ x \ in \ the \ range =\)
\(=\frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).

So, for our case as 96 is the last multiple of 4 in the range and 20 is the first multiple of 4 in the range then total # of multiples of 4 in the range is (96-20)/4+1=20.

Or look at it in another way out of 20 consecutive even integers we have half will be even multiples of 4 and half will be even not multiples of 4.

Hope it's clear.


Hi Bunuel
Do you think this is a 740+ level GMAT question?
There are so many conditions to be thought of. Although your explanation is brilliant, I don't think I could do this under even 3 minutes.
Thanks.
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Re: If integer C is randomly selected from 20 to 99, inclusive.  [#permalink]

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New post 15 Oct 2018, 17:45
ankitamundhra28 wrote:
Bunuel wrote:
If integer C is randomly selected from 20 to 99, inclusive. What is the probability that c^3-c is divisible by 12?
A. 1/2
B. 2/3
C. 3/4
D. 4/5
E. 1/3

Two things:
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 1-1/4=3/4.

Answer: C.


nkhosh wrote:
Would you please explain (1) The formulas you used for finding the # of ODD and "Divisible by 4" outcomes?
AND: (2) Why are we looking for the number of ODD integers within 20-99? I'm confused b/c 23 for example is not divisible by 12....

Thank you for the great post :)


1. There are even # of consecutive integers in our range - 80 (from 20 to 99, inclusive). Out of even number of consecutive integers half is always odd and half is always even, thus numbers of odd integers in the given range is 40.

2. \(# \ of \ multiples \ of \ x \ in \ the \ range =\)
\(=\frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).

So, for our case as 96 is the last multiple of 4 in the range and 20 is the first multiple of 4 in the range then total # of multiples of 4 in the range is (96-20)/4+1=20.

Or look at it in another way out of 20 consecutive even integers we have half will be even multiples of 4 and half will be even not multiples of 4.

Hope it's clear.


Hi Bunuel
Do you think this is a 740+ level GMAT question?
There are so many conditions to be thought of. Although your explanation is brilliant, I don't think I could do this under even 3 minutes.
Thanks.

ankitamundhra28
There is no difficulty level bucket like 740+ but if you are asking if this question is very difficult question then you are probably right... :)
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Re: If integer C is randomly selected from 20 to 99, inclusive.  [#permalink]

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New post 11 Feb 2019, 18:18
First work out the givens, we have 20-99 (inclusive) consec ints, so 99-20+1 = 80 ints
Also, C^3-C can be broken down into C(C^2-1) --> (C-1)C(C+1).
We know that any 3 consecutive ints are divisible by 3, so the question really is: How many sets of 3 consec ints from 20 to 99 have enough factors of 2 (12=3*2^2)?

Case 1--> any 3 consec ints with 2 E# starting from 2 are divisible by 4, e.g. {20, 21, 22}
Out of 80 consec ints, half are even. So there are 40 sets of EOE where this is the case.

Case 2--> OEO consec ints when the single even number has at least 2 factors of 2.
e.g. {23,24,25}
So the middle even number needs to be a multiple of 4, Last multiple (96) - First multiple (20) = 76/4 + 1 = 19 + 1 = 20
40+20/80 = 60/80 = 2/3.

The NO Case is how many do NOT have enough factors of 2. These are the even numbers that come up in OEO that are not divisible by 4. e.g. {21, 22, 23}, {25,26,27} ... {97, 98, 99}
We can see that when we divide, 22/4 = 5R2 or 98/4 = 14R2, there's always a remainder of 2, which means that the next or previous even number is divisible by 4.

This might actually be a quicker way to determine the answer.
If half of the 80 is evens and half is odds, that's 40 of each.
We are taking 3 consecutive numbers, so EOE or OEO. When can we not divide the 3 numbers by 4?
When the single even number is not divisible by 4. I.e. Every other even number isn't divisible, so out of 40 OEO half would not be divisible by 4. e.g. 20, 22, 24, 26, 28, 30, 32, 34 ...
So 1/4 of all possible sets of consec ints have only 1 E# number that is not divisible by 4.

We can write out the sequences of 12 numbers where n = 20 through 31 to see the pattern:
n-1, n, n+1
{19,20,21} yes, case 2
{20,21,22} yes, case 1
{21,22,23} no
{22,23,24} yes, case 1
{23,24,25} yes, case 2
{24,25,26} yes, case 1
{25,26,27} no
{26,27,28} yes, case 1
{27,28,29} yes, case 2
{28,29,30} yes, case 1
{29,30,31} no
{30,31,32} yes, case 1

So 9/12=3/4 cases are YES, 3/12=1/4 are NO.

This pattern repeats 4 times completely (from n=20-31, 32-43, 44-55, 56-67, 68-79, 80-91) with 9*6 = 54 YES and 1 time partially (n=92-99) with 6 YES cases, 54+6 = 60 out of 80, which is the same as what we got above.
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Re: If integer C is randomly selected from 20 to 99, inclusive.   [#permalink] 11 Feb 2019, 18:18

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