Two integers will be randomly selected from sets A and B

Karishma,

I don't understand this part of your explanation:
"Notice that 5, 7, 11 and 13 are primes and none of their multiples are in either set. So ignore them. We just need to focus on 2 and 3 of set A and 2, 4 and 6 of set B."

Why do we have to do that? Thanks!

This is part of your step 2: Remove the products that appear more than once.
The logic here is that a product involving 5/7/11/13 will not appear more than once. So we ignore these numbers.

Say we select 5 from A. Now, when we select any number from set B, we get a distinct product i.e. we get 4 distinct products (5*2, 5*4, 5*6, 5*13)
Now think, can you select a number other than 5 from set A and some number from set B to make one of these 4 products? i.e. Without selecting 5 from set A, can you make a product of 10 or 20 or 30 or 65? No, because to make 10/20/30/65, you need a 5 but you have no other 5 or multiple of 5.
Same is the case with 7, 11 and 13 (primes that appear only once in one set). So the products made by these prime numbers will not appear more than once.

You don't really need to think all this during your test. Lots of practice and thorough analysis will make these things intuitive.

By looking at Set A, we can see that it's all primes. Thus, we should immediately break down the elements in Set B to their prime factors. That gives :

Set A = {2,3,5,7,11}

Set B = {2, 2x2, 3x2, 13}

Apart from 2x3x2 (taking 2 from set A) which is the same as 3x2x2(taking 3 from set A); there is nothing which can be repeated. Thus, the total unique product = 20-1 = 19.

Total values will be \(5C_1 * 4C_1\)

But there will one repetitive value as 12

20 - 1

19

Given:
A = {2, 3, 5, 7, 11}
B = {2, 4, 6, 13}

Two integers will be randomly selected from sets A and B, one integer from set A and one from set B, and then multiplied together.

Asked: How many different products can be obtained?

A*B = {4,6,8,10,12,14,18,20,22,26,28,30,39,42,44,65,66,91,143} : 19 different products

IMO C

Total possible pairs 5c1*4c1= 20
Common values which 2×6 & 3x4
Remove 1 as 12 comes twice
20-1;19
Option C



Posted from my mobile device

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.