# If n is an integer between 1 and 96 (both included), what is the probability that n

#### hari

##### New member
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.

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#### Dr.Peterson

##### Elite Member
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.
Presumably, the intention is that n is chosen randomly from a uniform distribution. So they are really asking what percentage of those 96 numbers are such that n(n+1)(n+2) is a multiple of 8.

One way to start is to count how many are themselves multiples of 8, and for how many n+1 or n+2 is a multiple of 8. There will be more to think about after that.

What are your thoughts? If you haven't already read our guidelines for submission, please read here. We'd like to know where you need help, rather than just toss out ideas that you might already have.

#### Jomo

##### Elite Member
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 even, then can/will n*(n+1)*(n+2) be divisible by 8? Why?

#### jtayag0622

##### New member
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.
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%

#### pka

##### Elite Member
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.

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%
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.
Oh yes tayag0622 has the correct answer, but I must say that was the only answer to have marked if one were only guessing.

If one is looking for an analytical solution then note that for any number to be divisible by eight the number must contain a factor of $$\displaystyle 2^3$$ hence that explains why if $$\displaystyle n$$ is even then $$\displaystyle (n)(n+1)(n+2)$$ is a multiple of eight.
That said, the case where $$\displaystyle n$$ is odd is more problematic.

#### Jomo

##### Elite Member

The probability that n*(n+1)*(n+2) is 60/96 = 5/8 = 62.5%
I wouldn't bother saying 5/8 = 62.5%.
We have 60/96, ie 60 out of 94 which is a bit more than half, so you expect to get a bit more than half of the last 4 (to get out of 100). We know there is an answer (60/96) and the only reasonable answer is 62.5%

#### Jomo

##### Elite Member

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!

#### jtayag0622

##### New member
Please don't give full solutions so quickly. It is better to let the OP try to figure it out themselves with hints. Thanks!
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.

#### JeffM

##### Elite Member
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.
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.

#### Jomo

##### Elite Member
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.
That's what I thought.

#### Jomo

##### Elite Member
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%

#### jtayag0622

##### New member
I did a very quick calculation on excel (I attached it in jpg format and sorry if you need if you can't really see it or just try.). the first column are the numbers from 1 to 96 (n) inclusive. The second column are the numbers obtained when n is substituted. The third column is are the quotients when column 3 is divided by 8. I just used column 4 for the formula in division (just ignore it). I just used the last column to count how many of the numbers work. I did all the numbers from 1 to 96 (inclusive) and I got exactly 60 numbers for n.

so I think there's an answer, which is 62.5%.

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#### jtayag0622

##### New member
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%
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

#### JeffM

##### Elite Member
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.

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#### jtayag0622

##### New member
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.

I edited my post. my answer is actually 62.5%. for me, just don't make it complicated. if it's a simple probability, then solve it in simple probability. Your explanation is awesome. but i would rather make it simple to understand. my goal here is to make the explanation as simple as i can but understandable.

#### Denis

##### Senior Member
Define "easy to understand".

#### pka

##### Elite Member
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
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 no reason for this to have been on any GRE exam.

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#### jtayag0622

##### New member
Define "easy to understand".
not too long and not too complicated. A normal person like me would understand. putting explanations in layman's terms. not everyone can understand too complex explanations. not everyone on the web searching for some Math answers will understand explanations if its too complicated, especially when the problem is not that complex to solve, but make it too complicated.

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#### jtayag0622

##### New member
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.
Thank you so much pka

#### Jomo

##### Elite Member
Define "easy to understand".
Not Denis, Inverse of Denis, -Denis and my favorite, 1/Denis