Scoretop 4 - Q9 - Number properties/Probability - The Beat The GMAT Forum

Scoretop 4 - Q9 - Number properties/Probability

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akay: Senior | Next Rank: 100 Posts; Posts: 53; Joined: Sun May 06, 2007 5:03 am

by akay » Tue Aug 28, 2007 7:11 pm

I have no clue on this one...

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

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Source: Beat The GMAT — Problem Solving | Tue Aug 28, 2007 7:11 pm

beny: Master | Next Rank: 500 Posts; Posts: 214; Joined: Fri Jul 20, 2007 1:35 am; Thanked: 3 times

by beny » Tue Aug 28, 2007 8:14 pm

These are three consecutive integers. For the product to be divisible by 8, at least two of these integers need to be even (the lowest combination is 2,3,4 and the highest combination is 96,97,98 ). Notice that as long as the first number (n) is even, you will have at least 2 even numbers out of the three, thus, the product is divisible by 8. Now you just need to know how many even integers there are between 1 and 96 inclusive (there are 48... you can either find this using counting methods or just know that the sequence starts odd, ends even, thus exactly half of the numbers must be even).

48/96 = 1/2

Answer is C.

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givemeanid: Master | Next Rank: 500 Posts; Posts: 277; Joined: Sun Jun 17, 2007 2:51 pmFollowed by: 1 member; Location: New York, NY; Thanked: 6 times

by givemeanid » Wed Aug 29, 2007 6:04 am

When (n+1) is straight divisible by 8, the product will be divisible by 8 (n will be odd in this case). There are 96/8 = 12 such numbers.

So, total = 48 + 12 = 60 numbers.

Prob = 60/96 = 5/8

So It Goes

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