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 (f
or 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.
The initial 5 entries (before the multiple of 12) have 4 entries which have a multiple of 12
Then their are 8 groups of 12, and each having a probability of being divisible by 12 = 9/12=3/4
.
.
35.36.37 47.48.49..............8 such groups..