n(n + 1)(n + 2) is a product of 3 consecutive integers.gmat009 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
Oa is D
Thanks for explanationMorgoth wrote:n(n + 1)(n + 2) is a product of 3 consecutive integers.gmat009 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
Oa is D
if n is even, n(n + 1)(n + 2), will be divisible by 8.
even integers from 1-96 inclusive = (96-2)/ 2 = 94/2 = 47, 47+1 = 48
also if n is odd and 1 less than multiple 8, n(n + 1)(n + 2) will be divisible by 8, because this will have at least 1 multiple of 8.
multiples of 8 from 1-96 = (96-8)/8 = 11, 11+1 = 12
total probability = 48+12 = 60/96 = 10/16 = 5/8.
Hope this helps.
No. what I did was find the odd integers which lead to one 8 in n(n+1)(n+2)gmat009 wrote: Thanks for explanation
I am not clear with just 1 thing.
You calculated " even integers from 1-96 divisible by 2=48"
and "multiples of 8 from 1-96 =12"
Total 48+12 but they will not be having some common numbers which are even and divisible by 8 and we are counting them twice.
Don't we need to subtract those numbers.
can you explain why you do (96-2)/2? then you add one to the result?Morgoth wrote:n(n + 1)(n + 2) is a product of 3 consecutive integers.gmat009 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
Oa is D
if n is even, n(n + 1)(n + 2), will be divisible by 8.
even integers from 1-96 inclusive = (96-2)/ 2 = 94/2 = 47, 47+1 = 48
also if n is odd and 1 less than multiple 8, n(n + 1)(n + 2) will be divisible by 8, because this will have at least 1 multiple of 8.
multiples of 8 from 1-96 = (96-8)/8 = 11, 11+1 = 12
total probability = 48+12 = 60/96 = 10/16 = 5/8.
Hope this helps.
I did that only to be safe. Here I could have simply done the division and it still would have worked because the starting number was 1. Had it not been 1 or some other whole number starting from 30s or 40s, the simply division method would have missed 2 important values and would have lead to the wrong answer.bullshark wrote: can you explain why you do (96-2)/2? then you add one to the result?
why did you not simply do 96/2 = 48 and 96/8 = 12 ?
Thanks
I guess it just comes with experience with number properties. However, if you notice, the applicability of the 48 even numbers is derived completely from the N and (N+2) term. The (N+1) term isn't even a factor for this part.uw490 wrote:
How do we know to look for the "number that's a multiple of 8, minus 1. Like, 7, 15, 23, etc..." Still not clear why we know to look for this portion of the solution.
J
