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by shashank.ism » Tue Feb 09, 2010 1:15 pm
If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8?

25%
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62.5%
72.5%
75%
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by fibbonnaci » Tue Feb 09, 2010 11:45 pm
for the term to be divisible by 8, the term should contain atleast 3 -2's coz 2*2*2 make 8

now we will get atleast 3- 2's only when n is a multiple of 2.
if you have a doubt then plug the values and check. you will find that whenever n is a multiple of 2 the term is divisible by 8.
now there are 48 multiples of 2 between 1 and 96 both inclusive.

Total number of ways of choosing n= 96.

Probability= desired outcomes/ total number of outcomes.

ie. 48/96 => 1/2 => 50% - B
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by Ian Stewart » Wed Feb 10, 2010 12:51 am
fibbonnaci wrote:for the term to be divisible by 8, the term should contain atleast 3 -2's coz 2*2*2 make 8

now we will get atleast 3- 2's only when n is a multiple of 2.
if you have a doubt then plug the values and check. you will find that whenever n is a multiple of 2 the term is divisible by 8.
now there are 48 multiples of 2 between 1 and 96 both inclusive.

Total number of ways of choosing n= 96.

Probability= desired outcomes/ total number of outcomes.

ie. 48/96 => 1/2 => 50% - B
When n is even, then n and n+2 are consecutive even numbers. Given any two consecutive even numbers, one will also be a multiple of 4, so (n)(n+2) will be a multiple of 8. So for the 48 even values of n between 1 and 96, we will find that n(n+1)(n+2) is divisible by 8.

However, there are also odd values of n which will work. If n+1 is a multiple of 8, then n(n+1)(n+2) is divisible by 8. For example, if n=7, then (n)(n+1)(n+2) = (7)(8)(9), which is certainly a multiple of 8. There will be 12 such values of n in the range given (n = 7, 15, 23, etc - one eighth of all of the values).

So of the 96 possible values of n, 48+12 = 60 will give us a multiple of 8, and the answer is 60/96 = 5/8 = 62.5%.
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