I hope you got a chance to try your hand at this week’s question. Let’s start by refreshing your memory:

## Question

How many three-digit numbers contain three primes that sum to an even number?

- 27
- 28
- 54
- 55
- 64

## Answer and Explanation

In order for three prime numbers to sum to an even number, either one of those three primes or all three of those primes has to be ‘2’. First, let’s count the instances that have exactly one ‘2’. Each of these instances can contain any of the single-digit prime numbers: 3, 5, and 7.

Ist case: 2 _ _

2^{nd} case: _ 2 _

3^{rd} case: _ _ 2

In each case, there can be three different prime numbers. Therefore, we get 3 x 3, 9 possibilities, for each case. Since there are three cases, we end up getting 27.

The final step is to account for that one instance in which all three numbers are 2, as in the number 222.

Therefore, the answer is (B) 28.

*How’d you do? If you’re up for another challenge (and I know you are) check back in two weeks for our next installment of Clemmonsdogpark GRE Brain Twister!*

Good question.really enjoyed working on it.:)

Glad to hear 🙂

@Anjana I believe yes – We know that one digit MUST be 2 and it can be in any of units,tens and hundreds place. So we have to choose 1 of three avalaible places for the 2 i.e., 3C1. Now the remaining two places can be filled by the remaining three numbers in 3×3 or 3^2 ways since repetition is allowed, so that leaves us with 3C1x3^2= 27 BUT that is not all, we get an even sum when 2 is filled in all three places as well, i.e., the number 222, so we need to add that occurence to 27 which will give us a sum of 28. Hope that helps!

@Chris Can you tell me how can I reach you folks if I need help with my strategies?

Hi Kavya,

Thanks for the elegant solution!

For help, just click the “help” tab when you are logged into the product.

Let me know if you have trouble finding it 🙂

I have a question- Is there a way we could do this using the concept of permutation and combination?

Anjana,

Check out Kavya’s excellent response above 🙂