Prob. que

[n_adroja]

If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8? 25% 50% 62.5% 72.5% 75%

can someone pls help with an approach to this one

[forrestgump]

E = n*(n+1)*(n+2)

E is divisible by 8, if n is even.
No of even numbers ( between 1 and 96) = 48

E is divisible by 8, also when n = 8k - 1 ( k = 1,2,3,.....)
Such numbers total = 12( 7,15,....)

Favourable cases = 48+12 = 60.
Total cases = 96
P = 60/96 = 62.5

[GmatGnat]

Same question posted here: http://www.urch.com/forums/showthread.php?t=18036 (Probability que)

[n_adroja]

To calculate the
case 1 -numbers that are divisible by x between m to n inclusive and
case 2 between m-n
adroja

E = n*(n+1)*(n+2)

E is divisible by 8, if n is even.
No of even numbers ( between 1 and 96) = 48

E is divisible by 8, also when n = 8k - 1 ( k = 1,2,3,.....)
Such numbers total = 12( 7,15,....)

Favourable cases = 48+12 = 60.
Total cases = 96
P = 60/96 = 62.5