Ready for this week’s brain twisting math challenge question?

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

- 27
- 28
- 54
- 55
- 64

*Be sure to check back on Thursday for the answer and explanation! Good luck. ðŸ™‚*

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Ready for this week’s brain twisting math challenge question?

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

- 27
- 28
- 54
- 55
- 64

*Be sure to check back on Thursday for the answer and explanation! Good luck. ðŸ™‚*

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It is easy man.

Prime number which are lesser than 10: 2, 3, 5, 7

Even number=2

Odd Numbers=3, 5, 7

and we know that:

Odd+Odd+Odd=Even —————> 3*3*3=27

Even+Even+Even=Even————->1 (when our number is 222)

Even+Odd+Odd=Even—————>1*3*3=9

Odd+Even+Odd=Even—————>3*1*3=9

Odd+Odd+Even=Even—————>3*3*1=9

So 27+1+9+9+9=55

Therefore answer is D

I think you double counted some cases (332, etc.).

28

That’s it!

B.28

That’s right!

Answer is B: 28.

There are 4 possible prime numbers including the only even (E) prime number 2 and the odd (O) prime numbers 3, 5, and 7. Let the 3-digit number be XYZ. X+Y+Z is even in four possible cases:

1) XYZ is EOO, thus there are 1*3*3=9 possible numbers

2) XYZ is EEE, thus it is the number 222

3) XYZ is OOE, thus there are 3*3*1=9 possible numbers

4) XYZ is OEO, thus there are 3*1*3=9 possible numbers

In total, there 9*3+1=28 possible numbers.

Good explanation! That’s the answer ðŸ™‚

28

That’s it!

_ _ _ : Three digit number to be filled by prime numbers such that they add up to an even number.

Case 1: Hundred’s place has 2

The tens place can be filled by any of the four prime numbers and similarly the units place.

So, total three digit nos in this case is 1 X 4 X 4 = 16

Case 2: Hundred’s place has 3

So any of the tens or units has to be occupied by 2 and other place can be occupied by any of the four primes. Sp total nos: 1 X 1 X 4=4

Similarly case 3 has 5 in hundreds place and case 4 has 7 in hundreds place and rest follows similar to case 2.

So, total nos of three digit numbers is 16 + 4 + 4 + 4 = 28

And the answer is B: 28

Great explanation!

28.

The numbers have to be EOO,OEE,OOE, EEE (Even/Odd). With 2,3,5,7 primes, first gives 9 choices, 2nd gives 9 choices, 3rd gives 9 choices and 4th gives 1. Total 28

That’s it ðŸ™‚

Well from the above comments all of them have come up with answer but forgot the same prime number (3,5,7) can repeat twice for the 3 digit number. 2 can repeat only once i.e 222 . So answer will be total possibilities – number possible with two 2s.

64( 4*4*4) – 9 ( 3+3+3 ). = 54. Answer is 54

I think you double counted some cases (332, 552, etc.)

27..

One off…I think you forgot 222.

A.27

Sorry 28 :p

Yep, that’s the right one ðŸ™‚

Hmm. I am not sure whether I understood this question clearly. Here’s my thought:

So among the 10 digits from 0-9 there are 4 digits which are prime (2,3,5,7) which will give us a total of 64 different numbers with these digits (2,3,5,7). However, the question requires the sum of the digits to be even. Therefore, there are only two possibilities: 1) All the numbers must be even or 2) 1 even number must be followed by 2 odds. According to option 1, we can get the number 222 where the individual digits add up to 6. As for option no. 2, 2 must be present in the number along with 2 odd prime numbers. If we plug in 2 at the hundreds place we get a total of 3*3 = 9 possibilities. Since our number is 3 digits the number of possibilities will be 3*9 = 27. Hence, the answer should be 27+1 = 28, i.e. B

Got it–and a great explanation to boot!

28

That’s it!