Probability and Number Properties

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fangtray: Master | Next Rank: 500 Posts; Posts: 273; Joined: Thu Sep 08, 2011 6:50 amFollowed by: 3 members; Thanked: 5 times

by fangtray » Tue May 15, 2012 7:25 am

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

I got the right answer. Took 6 minutes. What's the under 2 minute way to do this?

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eagleeye: Legendary Member; Posts: 520; Joined: Sat Apr 28, 2012 9:12 pmFollowed by: 49 members; Thanked: 339 times; GMAT Score:770

by eagleeye » Tue May 15, 2012 10:32 am

How about 30 seconds?

This is how I did it. First realize that n(n+1)(n+2) are three consecutive numbers, and they can range from 1 to 96. Using pattern recognition.

Now I am going to write down the first 16 numbers, (I will tell you in a second why this makes it quick to do). I will be calling the product of three consecutive numbers a group.

1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16

Now we have to find the divisibility of groups of 3 consecutive numbers with 8.

1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16

Observe that groups starting with even numbers (2,4,6,8 etc) all are divisible by 8 (by groups I mean 2*3*4; 4*5*6 ; 6*7*8; 8*9*10);

1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16
Also observe that, for all odd numbers (1,3,5,7), the groups starting with them are not divisible by 8 except for the odd number just less than 8. That is 1*2*3, 3*4*5, 5*6*7 are not divisible by 8 but 7*8*9 is (because 8 itself is a factor).

Now observe that this pattern repeats itself again, (that is for groups of 9,11,13 not divisible, groups of 10,12,14,16 and 15 are divisible).

So we find a repeating pattern of 8 where 5 groups are divisible and 3 numbers are not. therefore the probability = 5/8.

Let me know if this helps

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Tue May 15, 2012 5:29 pm

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

Case 1: n(n+1)(n+2) = even*odd*even = multiple of 8:
Since every other even integer is a multiple of 4, the product here will always include an even integer and a multiple of 4, resulting in a multiple of 8.
Thus, n can be any even integer between 1 and 96.
96/2 = 48 favorable choices for n.

Case 2: n+1 is a multiple of 8:
The product will be a multiple of 8 if n+1 is a multiple of 8.
Number of multiples of 8 between 1 and 96 = 96/8 = 12.
Thus, there are 12 favorable choices for n+1, implying that there are 12 more favorable choices for n.

Total favorable choices for n = 48+12 = 60.
Favorable choices/Total choices = 60/96 = 5/8.

The correct answer is D.

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ihatemaths: Master | Next Rank: 500 Posts; Posts: 122; Joined: Tue May 01, 2012 6:05 am; Thanked: 3 times

by ihatemaths » Tue May 15, 2012 9:31 pm

still not clear people. what is the fun in dealing with cases of N then (N+1) how are you segregating ?

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Tue May 15, 2012 10:39 pm

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

I got the right answer. Took 6 minutes. What's the under 2 minute way to do this?

Case 1: n is even
Let us assume that n = 2k, for any integer k.
Then n(n + 1)(n + 2) = 2k(2k + 1)(2k + 2) = 4k(2k + 1)(k + 1)
Now either k or k + 1 will be even, so 8 will be a multiple of n(n + 1)(n + 2).

Number of even integers between 1 ans 96, inclusive = {(96 - 2)/2} + 1 = 48

Case 2: If n + 1 is divisible by 8
n + 1 = 8a, where a â‰¥ 1
n = 8a - 1
8a - 1 â‰¤ 96
8a â‰¤ 97
a â‰¤ 12.1 implies 12 integers.

Also, when n and n + 2 are even, n + 1 will be odd, and when n + 1 is divisible by 8, then n and n + 2 will be odd. This means, there is no repetition.

Total integers = 48 + 12 = 60

Therefore, probability that n(n + 1)(n + 2) will be divisible by 8 = 60/96 = [spoiler]5/8[/spoiler]

The correct answer is D.

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ankita1709: Master | Next Rank: 500 Posts; Posts: 109; Joined: Wed Feb 15, 2012 7:09 amFollowed by: 2 members; Thanked: 8 times

by ankita1709 » Wed May 16, 2012 2:32 am

ihatemaths wrote:still not clear people. what is the fun in dealing with cases of N then (N+1) how are you segregating ?

Seems like you really do hate maths

It's a oral question.
The number n(n+1)(n+2) to be divisible by 8. It should be either divisible by 2 and 4 or 8.

So lets look for all even numbers n = factor of 2 and n+2 = factor of 4
that means all even numbers included => 96/2 = 48

Secondly for all odd numbers n+1 = even but both n and n+2 will be odd. It can only be divisible by 8 if n+1 = factor of 8
that means all factors of 8 in 1-96 =>96/8 = 12

Total = 12+48 = 60
Probability = 60/96 = [spoiler]5/8[/spoiler]

Hence the answer

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amit28it: Senior | Next Rank: 100 Posts; Posts: 82; Joined: Fri Nov 04, 2011 12:35 amFollowed by: 2 members

by amit28it » Fri May 18, 2012 4:30 am

The answer is D,
In the end I have got the correct answer but I have taken around 10 minutes to solve this and I don't think that any method can do this question in 2 minutes.
pre calculus help

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