Probability

Question

If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8? A. 1/4 B. 3/8 C. 1/2 D. 5/8 E. 3/4

parallel_chase · Answer

reachac wrote:If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

I think its 1/2. Whats the OA?

n(n+1)(n+2) for this to be divisible by 8 the condition is n should be even

[96-2]/ 2 + 1 = 48

48/96 = 1/2

Mani_mba · Answer

IMO, the answer is B 3/8.

If n(n+1)(n+2) to be divisible by 8, either
a) n should be divisible by 8
b) n+1 should be divisible by 8
c) n+2 should be divisible by 8.

Taking the first case. n can take values
8,16,24,32,40,48,56,64,72,80,88,96( totally 12 )
Taking the second case, n+1 can take values
7,15,.......................................,95 (totally 12)
Taking the third case, n+2 can take values
6,14,.......................................,94 (totally 12)

so totally 36 values out of 96 , n can have to satisfy the condition.
36/96 => 3/8.

parallel_chase · Answer

Mani_mba wrote:IMO, the answer is B 3/8.

If n(n+1)(n+2) to be divisible by 8, either
a) n should be divisible by 8
b) n+1 should be divisible by 8
c) n+2 should be divisible by 8.

Taking the first case. n can take values
8,16,24,32,40,48,56,64,72,80,88,96( totally 12 )
Taking the second case, n+1 can take values
7,15,.......................................,95 (totally 12)
Taking the third case, n+2 can take values
6,14,.......................................,94 (totally 12)

so totally 36 values out of 96 , n can have to satisfy the condition.
36/96 => 3/8.

I think you have exaggerated the problem. You have consider the product of n(n+1)(n+2) not individually. Let me know what you think.

classic · Answer

When n is even it will always be divided by 8

e.g when n=2 then we have 2*3*4

So all together we have 48 even umbers

Now when n= (K*8 -1) where k is natural number it will be divisible by 8

e.g n=7 we will have 7*8*9

so Prob= 60/96=5/8

Mani_mba · Answer

I agree to you Parallel_chase. I missed it.

reachac · Answer

I made the same mistake, got 1/2 as the ans. Correct ans is 5/8. Thanks @classic... Thanks ppl

parallel_chase · Answer

reachac wrote:I made the same mistake, got 1/2 as the ans.
Correct ans is 5/8.

Thanks @classic...
Thanks ppl

The answer is indeed 5/8. That is the most silliest mistake I have ever made.

Thanks for the question.

evansbd · Answer

How are we getting 60 total numbers that satisfy this problem?

parallel_chase · Answer

evansbd wrote:How are we getting 60 total numbers that satisfy this problem?

Integer which satisfies condition n(n+1)(n+2) is all the even numbers between 1-96

There are total of 48 even integers between 1-96

also you have to consider the possibility what if n is 7 then 7*8*9 will also be divisible by 8

Therefore 8*12 = 96 meaning there are 12 odd integers which satisfies the above condition.

Hence 12+48 = 60

60/96 = 5/8

Hope this helps.

evansbd · Answer

I see the solution that you need to find odd numbers that satisfy the criteria. However, on a real GMAT, I easily saw the 48 even numbers but I don't see (intuitively) how ppl are seeing that starting with 7 you can see other odds that will meet the criteria then you end up with 12.

What are the 12 numbers that meet the odd criteria and is there an intuitive way to find them quickly?

parallel_chase · Answer

There is no intuitive way of doing anything on GMAT. Always rely on the details given in the question. Everything has a reason and logic.
It is very necessary to understand the logic and reasoning behind the question.

Now coming to your question.

Odd numbers are
7 (7*8*9)
15(15*16*17)
23(23*24*25)
31(31*32*33)
+
+
+
95(95*96*97)

do you see the method in the above sequence, every multiple of 8 and the number before that.

Hope this helps.

Brent@GMATPrepNow · Answer

reachac wrote:If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

First recognize that n, n+1 and n+2 are 3 CONSECUTIVE INTEGERS.

Now let's make some observations:

When n = 1, we get: (1)(2)(3), which is NOT divisible by 8
n = 2, we get: (2)(3)(4), which is DIVISIBLE BY 8
n = 3, we get: (3)(4)(5), which is NOT divisible by 8
(4)(5)(6), which is DIVISIBLE BY 8
(5)(6)(7), which is NOT divisible by 8
(6)(7)(8), which is DIVISIBLE BY 8
(7)(8)(9), which is DIVISIBLE BY 8
(8)(9)(10), which is DIVISIBLE BY 8
-----------------------------
(9)(10)(11), which is NOT divisible by 8
(10)(11)(12), which is DIVISIBLE BY 8
(11)(12)(13), which is NOT divisible by 8
(12)(13)(14), which is DIVISIBLE BY 8
(13)(14)(15), which is NOT divisible by 8
(14)(15)(16), which is DIVISIBLE BY 8
(15)(16)(17), which is DIVISIBLE BY 8
(16)(17)(18)which is DIVISIBLE BY 8
-----------------------------
.
.
.
The pattern tells us that 5 out of every 8 products is divisible by 8.
So, [spoiler]5/8[/spoiler] of the 96 products will be divisible by 8.
This means that the probability is 5/8 that a given product will be divisible by 8.

Answer: D

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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