knight247 wrote:If c is a randomly chosen from the integers 20 to 99, inclusive, what is the probability that
c³-c is divisible by 12?
Answer is [spoiler]3/4[/spoiler]
To be divisible by 12, c³-c must be a multiple of 3 and 4.
c³-c = c(c²-1) = c(c-1)(c+1).
c-1, c, and c+1 are three consecutive integers.
Of every three consecutive integers, exactly one is a multiple of 3.
Thus, (c-1)(c)(c+1) is a multiple of 3.
The question is whether (c-1)(c)(c+1) will be a multiple of 4.
If c is odd, then (c-1)(c)(c+1) = (even)(odd)(even).
In this case, the product will be a multiple of 4, since it includes 2 even factors.
If c is a multiple of 4, then (c-1)(c)(c+1) = (odd)(multiple of 4)(odd).
In this case, the product will be a multiple of 4, since one of its factors is a multiple of 4.
If c is an even integer that is NOT a multiple of 4, then (c-1)(c)(c+1) = (odd)(even non-multiple of 4)(odd).
In this case, (c-1)(c)(c+1) will NOT be a multiple of 4.
Between 20 and 99:
The total number of integers = 80.
Since half are even, the number of even integers = 40.
Since every other even integer is a multiple of 4, the number of even integers that are NOT multiples of 4 = 20.
Thus, P(even non-multiple of 4) = 20/80 = 1/4.
Since any other value for c will yield a multiple of 4 for (c-1)(c)(c+1):
P(c³-c will be a multiple of 12) = 1-1/4 = 3/4.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at
[email protected].
Student Review #1
Student Review #2
Student Review #3
Featured Course
