Is it divisible by 3? -AMC 10
Probability
Level
3
Positive integers a, b, and c are randomly and independently selected with replacement from the set {1, 2, 3,...., 2010}. The the probability that abc + ab + a is divisible by 3 can be expressed as m/n, where m and n are co-prime. What is m+n?
The answer is 40.
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1 solution
Hi first time solution poster here so I'll start with this simple one.
So, first of all we need to realize that since the set starts at 1 mod 3 and ends at 0 mod 3, then 1/3 of the numbers are 0 mod 3, 1/3 are 1 mod 3, and 1/3 are 2 mod 3. That makes the problem much simpler to compute.
Now, factor abc + ab + a into a (bc + b +1) = a (b(c+1) + 1). The reason i did this is because you want fewer terms and multiplying is easier than adding here i think.
So, the first way you can have this thing divisible by 3 is if a is divisible by 3, or 0 mod 3. That probability is obviously 1/3. Note that once you find the other probabilities since they can occur at the same time as the first way you have to add them together and then subtract the probability that they both occur at the same time because you're counting it twice.
Next, the probability that b(c+1) + 1 is 0 mod 3 is equivalent to saying that b(c+1) is 2 mod 3. This means that one of the terms is 1 mod 3 and the other is 2 mod 3, so b is 2 mod 3 and c is 0 mod 3, making c+1 1 mod 3, or b is 1 mod 3 and c is 1 mod 3. These are obviously mutually exclusive and the probability of each one is 1/3 * 1/3 = 1/9. So, the probability of the other case we're looking for is 2/9.
So, 1/3 + 2/9 - 1/3 * 2/9 = 5/9 - 2/27 = 13/27 = m/n. 13 + 27 = 40
Sorry for making the solution twice as long as it takes you to solve the problem.