If n is an integer between 1 and 96 (both included), what is the probability that n*(n+1)*(n+2) is divisible by 8.
a. 50%
b. 37.5%
c. 62.5%
d. 12.5%
e. none of these
Confused in solving this one from GRE.
If n is odd, then can/will n*(n+1)*(n+2) be divisible by 8? Why?If n is an integer between 1 and 96 (both included), what is the probability that n*(n+1)*(n+2) is divisible by 8.
a. 50%
b. 37.5%
c. 62.5%
d. 12.5%
e. none of these
Confused in solving this one from GRE.
If n is an integer between 1 and 96 (both included), what is the probability that n*(n+1)*(n+2) is divisible by 8.
a. 50%
b. 37.5%
c. 62.5%
d. 12.5%
e. none of these
Confused in solving this one from GRE.
If n is an integer between 1 and 96 (both included), what is the probability that n*(n+1)*(n+2) is divisible by 8.
a. 50% b. 37.5% c. 62.5%d. 12.5% e. none of these
Confused in solving this one from GRE.
As an old testing guy this question really puzzles me. Even having worked with the GRE this just does not seem usual unless it was a time waster.
There are 96/2 = 48 evens and 48 odds in the interval.
every even no. works (try)
Among the odds, only those that make n + 1 divisible by 8 work. Thus, n = 7, 15, 23, 31, 39, 47, 55, 63, 71, 79, 87, 95 (try)
adding the no. of evens and odds that work, 48 + 12 = 60
The probability that n*(n+1)*(n+2) is 60/96 = 5/8 = 62.5%
I wouldn't bother saying 5/8 = 62.5%.
The probability that n*(n+1)*(n+2) is 60/96 = 5/8 = 62.5%
Please don't give full solutions so quickly. It is better to let the OP try to figure it out themselves with hints. Thanks!
There are 96/2 = 48 evens and 48 odds in the interval.
every even no. works (try)
Among the odds, only those that make n + 1 divisible by 8 work. Thus, n = 7, 15, 23, 31, 39, 47, 55, 63, 71, 79, 87, 95 (try)
adding the no. of evens and odds that work, 48 + 12 = 60
The probability that n*(n+1)*(n+2) is 60/96 = 5/8 = 62.5%
Please don't give full solutions so quickly. It is better to let the OP try to figure it out themselves with hints. Thanks!
I suspect that pka simply forgot that "none of the above" was an option. All the other numeric options were obviously wrong once you are looking at 60/96 > 0.5.oh yes sorry. sure. but I don't get why not bother saying 5/8 or 60/96. Can you explain it a bit more? thank you.
That's what I thought.I suspect that pka simply forgot that "none of the above" was an option. All the other numeric options were obviously wrong once you are looking at 60/96 > 0.5.
You do need 60/96. If you take a test and get 60 points out of 96, then what would you exact to get if the test was out of a 100 points. The answer is that you would expect to get ((60/96)*4+ 60)/100~(2.5 + 60)% = 62.5%oh yes sorry. sure. but I don't get why not bother saying 5/8 or 60/96. Can you explain it a bit more? thank you.
You do need 60/96. If you take a test and get 60 points out of 96, then what would you exact to get if the test was out of a 100 points. The answer is that you would expect to get ((60/96)*4+ 60)/100~(2.5 + 60)% = 62.5%
The probability is 62.5% EXACTLY if we guess that the probability of choosing any specific number from 1 through 96 inclusive is equal to the probability of picking any other number.I am so sorry but that even made it complicated for me. sorry but I don't get it. I feel kinda dumb right now. haha. I just answered it the way I see it as probability. didn't really get why the closest answer is 62.5%, not actually 62.5%, or looking at the other choices , they're obviously wrong, so the answer would be none of the above. so sorry
The probability is 62.5% EXACTLY if we guess that the probability of choosing any specific number from 1 through 96 inclusive is equal to the probability of picking any other number.
Therefore, if that guess is correct, the probability of picking a specific one of the numbers in the set is 1/96.
After that your reasoning was fine although perhaps a little lacking in detail.
How many numbers in that set satisfy the specified criterion?
Let p = n(n + 2) and m = n(n + 1)(n + 2) = p(n + 1).
Therefore if m is evenly divisible by 8, then (a) p is divisible by 8, (b) or (n + 1) is divisible by 8, or (c) p is divisible by 4 and (n + 1) is divisible by 2, or (d) p is divisible by 2 and (n + 1) is divisible by 4.
In cases c and d, n + 1 is even so n and n + 2 are odd so p is odd and thus not divisible by any power of 2. Thus cases c and d are impossible.
In any sequence of eight successive integers, exactly one is odd with its immediate successor being evenly divisible by 8. From 1 through 96 inclusive, there are 12 such sequences. So there are twelve instances of case b.
If n is evenly divisible by 4, then n = 4k and n + 2 = 4k + 2 so p = 16k^2 + 8k = 8(2k^2 + k) so p is evenly divisible by 8.
If n is even but not evenly divisible by four, we have
\(\displaystyle n = 2j = 2(2k - 1) = 4k - 2 \implies n + 2 = 4k \implies p = (4k - 2)(4k) = 16k^2 - 8k = 8(2k^2 - 1) \implies\)
p is evenly divisible by 8. Therefore whenever n is even, p is evenly divisible by 8. There are 48 even numbers in the set.
There is no overlap between cases a and b.
\(\displaystyle \therefore \text {the probability that } 8 \text { evenly divides } m \text { is } = \dfrac{12 + 48}{96} = 62.5\%.\)
EDIT: Most people will never use any math beyond basic algebra once they leave school. But everyone will have many problems to solve. One thing math teaches you is to formulate problems carefully and to consider all alternatives. It is unfortunate that the thread did not start with getting the original poster to phrase the question in a workable way. After that, all of us were working in the dark. So it was easy for any of us to get confused.
Of the integers 1 to 96 one-half are even. If \(\displaystyle n\) is even then \(\displaystyle n=2j \).I am so sorry but that even made it complicated for me. sorry but I don't get it. I feel kinda dumb right now. haha. I just answered it the way I see it as probability. didn't really get why the closest answer is 62.5%, not actually 62.5%, or looking at the other choices , they're obviously wrong, so the answer would be none of the above. so sorry
Define "easy to understand".
Of the integers 1 to 96 one-half are even. If \(\displaystyle n\) is even then \(\displaystyle n=2j \).
\(\displaystyle \begin{align*}n(n+1)(n+2)&=(2j)(2j+1)(2j+2)\\&=(2j)(4j^2+6j+2)\\&=2(2j)(2j^2+3j+1)\end{align*}\)
If \(\displaystyle j \) is even then \(\displaystyle 2(2j) \) is a multiple of eight.
If \(\displaystyle j \) is odd then \(\displaystyle (2j^2+3j+1)\) is even so that \(\displaystyle 2(2j)(2j^2+3j+1) \) is a multiple of eight.
Now consider the case in which \(\displaystyle n \) is odd. Thus \(\displaystyle n=2j-1,~j=1,2,\cdots,48 \).
\(\displaystyle n(n+1)(n+2)=(2j-1)(2j)(2j+1)\)
Noting the first factor & the third are odd, their product is odd therefore the expression is a multiple of eight if \(\displaystyle j \) is a multiple of four.
Among the integers \(\displaystyle 1\text{ to }48\) there are twelve multiples of four.
\(\displaystyle 48+12=60 \). But most importantly \(\displaystyle \dfrac{60}{96}=62.5\% \)
I did say that I saw on reason for this to have been on any GRE exam.
Not Denis, Inverse of Denis, -Denis and my favorite, 1/DenisDefine "easy to understand".