manik11 wrote:Experts..could you guys please show me a time efficient way to approach this question?

Two distinct numbers a and b are chosen randomly from the set of first 20 positive integers. What is the probability that |a-b| would not be divisible by 3?
49/380
21/95
49/190
42/95
133/190

Look for a PATTERN and BALLPARK.
Options for the value of |a-b}:
|a-b| = 1
|a-b| = 2
|a-b| = 3
|a-b| = 4
|a-b} = 5
|a-b| = 6
|a-b| = 7
|a-b} = 8
|a-b| = 9
And so on.

The options in red suggest that, of every 3 options for |a-b|, approximately 2 will NOT be divisible by 3.
Thus, the correct answer choice must be close to 2/3.
Only E is viable:
133/190 ≈ 120/180 = 2/3.

The correct answer is E.

Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.


Two distinct numbers a and b are chosen randomly from the set of first 20 positive integers. What is the probability that |a-b| would not be divisible by 3?
49/380
21/95
49/190
42/95
133/190




Let's divide the first 20 positive integers into 3 sets according to the remainder as dividing by 3. namely
O = {3 , 6 , 9 , ... , 18} ---> the number of elements of O = 6
I = {1 , 4 , 7 , ... , 19} ---> the number of elements of I = 7
II = {2 , 5 , 8 , ... , 20} ---> the number of elements of II = 7

If we choose a, b among the set O then the remainder of a-b would be 0.
Similarly if we choose a, b among the set I or among the set II the remainder of a-b would be 0. So we should choose a, b set O and set I(--> the number of cases 6*7) or set O and set II(--> the number of cases 6*7) or set I and set II(--> the number of cases 7*7).

The total number of choosing two different numbers among 20 numbers is 20C2= 190. So the probability is (42+42+49)/190 = 113/190.

The answer is, therefore, (E)

Another efficient testday approach here is approximation. Since 1/3 of the integers are divisible by 3, (2/3) of them are NOT divisible by 3. We're picking randomly from a set of consecutive integers, so our answer should be somewhere in the neighborhood of 2/3, and only 133/190 (which reduces to 7/10) is close. (Everything else is less than 1/2, so none of these answers is promising.)

A proper approach (longer, but more edifying) would look something like this:

If a and b have the same remainder when divided by 3, then |a - b| will be divisible by 3. We can see this algebraically. Suppose a = 3*something + 2 and b also = 3*(something else) + 2. Then

a - b =
(3*something + 2) - (3*something else + 2) =
3*something - 3*something else =
a multiple of 3

This will work if the remainders are both 1 as well (or both 0, obviously), so we'll always get a multiple of 3.

The question then reduces to "What is the probability that a and b have the SAME remainder when divided by 3?" From here, I'd just count the possibilities.

remainder 0: 3, 6, 9, ..., 18 (6 options)
remainder 1: 1, 4, ..., 19 (7 options)
remainder 2: 2, 5, ..., 20 (7 options)

So there are (6*5)/2 ways of choosing two numbers with remainder 0, (7*6)/2 ways of choosing two numbers with remainder 1, and (7*6)/2 ways of choosing two numbers with remainder 2.

That gives us 15 + 21 + 21 = 57 ways of choosing two numbers with the same remainder. There are (20*19)/2, or 190 ways of picking ANY pair of these numbers, so

Prob(divisible by 3) = 57/190
and
Prob(not divisible by 3) = 1 - 57/190 = 133/190