If k is an integer from 48 to 84, inclusive, what is the probability that (k - 1)(k + 1)(k + 3) is divisible by 48?
A. 9/37
B. 12/37
C. 15/37
D. 18/37
E. 19/37
The denominator is the total number of possible outcomes, thus 84 - 48 + 1 = 37
Now numerator, desired outcomes -> given that (k-1)*(k+1)*(k+3) can be 3 consective odd or even numbers, depending wether k is odd or even.
For example, if k is 48 then -> 47*49*51 all are odd
if k is 49, which is odd then -> 48*50*52 all evens
For this product to be a multiple of 48 it has to be a multiple of 2^4*3^1 = which is the prime factorization of 48
now given that every 3rd consecutive number is a multiple of 3, odd or even; for example above, 51 is a multiple of 3 and 48 too, we know that every third number we have a multiple of 3 which coincides with our condition.
on the other hand, a multiple of 2^4 = 16 must be found. multiples of 16 are 16,32,48,64,80,96,... in the range only the bold ones.
(k - 1)(k + 1)(k + 3) is a sequence of 3 consecutive even or odd numbers, and for the purpose of finding multiples of 2^4 it k must be odd, so that (k - 1)(k + 1)(k + 3) contains at least 2 multiples of 4 and one multiple of 2 making it 2^5 and thus a multiple of 48.
Now for k to be odd in the range of 48 to 84, the common difference is 2 and the first and last odd numbers are 49 and 83, thus 83-49/ 2 +1 = 18 numbers
Answer (D) = 18/37