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

Question

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

A. 27
B. 28
C. 54
D. 55
E. 64

chetan2u · Accepted Answer

Hi,
if you dont want to calculate separately..

Ways the sum will be EVEN is if  
 
# all 3 are even or 
# one is even and remaining 2 are odd
 
I. 1 even and other 2 odd
 let the first number be 2, the remaining 2 can be filled with any of the 3,5,7.. , so 2,_,_ thus 1*3*3 
  Now 2 can be placed in ANY of three position, that is 2,_,_ or _,2_ or _,_,2, so 1*3*3*3=27..  
II. all even
 ofcourse a number without any odd prime 222..

so  27+1=28

B

broall · Answer

There are 4 prime digit numbers: 2, 3, 5, 7.

Since sum of 3 primes is an even number, all of them are even, or one of them is 2 and others isn't 2.

Case 1. Three digit numbers are different.

We must select 2 first. Now, we need to select 2 others from set {3, 5, 7}. We have 3C2 = 3 possible ways to select.

Now, with 3 different digits, we could create 3 * 2 * 1 = 6 different three-digit numbers from each set of 3 different digits.

Hence, we could totally create 6 * 3 = 18 different three-digit numbers.

Case 2. First digit is 2 and two others are identical.

We simply select a number from set {3, 5, 7}. There are 3 possible ways to select it.

Now, from each set of 2 and two other identical numbers, we could create 3 * 2 * 1 / 2 = 3 different three-digit numbers.

Hence, we could totally create 3 * 3 = 9 different three-digit numbers.

Case 3. All of 3 digits are identical.

In this case, we could create only one number 222.

Hence, the result of this question is: 18 + 9 + 1 = 28.

The answer is B

Dkingdom · Answer

The four single digit prime numbers are 2,3,5 and 7.
a) If we take these numbers in groups of 3s (remember the sum should be even) : 
 1) 2,3,5 - 6 possible numbers
 2) 2,5,7 - 6 possible numbers
 3) 2,3,7 - 6 possible numbers 
All the above combinations will give even numbers upon adding the three digits of the number. 
Hence total 6 * 3 = 18 possibilities.

b) Now since we need '2' to be present out of the three digits to keep the sum even, keep '2' and repeat the remaining primes, i.e. 233, 255, 277
Each of these three numbers can be presented in 3 ways. 
Now total possibilities = 18 + 9 = 27

c) Moreover 222 will also give an even number when the digits are summed up. 
Therefore total possibilities = 27+1 = 28

Answer : B

Observer0911 · Answer

Hi Chetan,

This is really helpful. Just one question - Since we're trying to re-arrange 3 integers - why shouldn't this be multiplied by 3! instead of 3?

stne · Answer

1.We can have one digit even and two digits odd. 
 2.we can have all digits even.

1.One digit even and two digits odd:

\hspace{5mm} (i) All digits are different:
 \hspace{5mm}  1*3C2*3!=18 
 \hspace{5mm} Where 1= one way to select number 2 
 \hspace{5mm}  3C2  number of ways to select two numbers from 3,5,7
 \hspace{5mm}  3!  number of ways to arrange all the 3 different digits.

\hspace{5mm} ii) Two odd digits are same: 
 \hspace{5mm}  1*3C1*3=9  
 \hspace{5mm} Where 1= one way to select number 2
 \hspace{5mm}  3C1  number of ways to select one number from 3,5,7 that will be repeated.
 \hspace{5mm}  3  number of ways to arrange 3 numbers where 2 are same  \frac{3!}{2}=3

2)All digits even: 
  \hspace{5mm} 1*1*1=1 
 \hspace{5mm} ( Only one case 222 )

Hence total ways : 18+9+1=28

Ans B

Hope it helps.

abhimanyuarya · Answer

Sum of three primes can be even if there are two odd primes and one even prime
---
Fix 2 as even in 1st digit. 2nd can be selected in 3 ways (from 3,5,7) and similar 3rd digit can be selected in 3 ways
So, total ways are 3*3 = 9 ways ---------- A

the digits can be scuffled in 3!/2! ways i.e. 3 ways ----------- B

Total ways are A*B = 27 ways

One digit is 222 also

therefore, answer is 28.

hadimadi · Answer

Hi,

there is no need for lengthy calculations here. The only way we can construct that number is with 2,3,5,7. For the sum of three of them to be even, we need all even or 2 odd and 1 even. All even is 222, 2 odd we can pick 3^2 combinations. Adding the 2 at the beginning, mid, or end, we get (3^2)*3=3^3=27 possibilities for 2 odd and 1 even, one for all even, so 28

Simone_Rose · Answer

Out of interest, is there a way from determining the total number of 3 digit numbers 
that contain 3 primes first, and then incorporating the 2nd constraint that the primes add up to an even number?

I think 64 3 digit numbers contain 3 primes, so how would one go from there to find primes that add up to an even number?

Krunaal · Answer

You will have to then subtract the primes that add up to odd number from 64. Easier approach is to directly find primes that add up to even number rather than determining the total and subtracting.

Archit3110 · Answer

prime numbers 2,3,5,7
sum of three digit number is even
possible when all three digits are even 222 1 way
or odd odd even this can be arranged in 3*3*1 *3!/2! ways ; 27 ways
total 28
option B

jveitch12 · Answer

This question overlooks double counting of numbers in prime calculation, I think it needs to be a little more precise. For instance the number 238 contains 3 primes. 23, 2, and 3 and has an even sum of primes (2 + 3 + 23 = 28).

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

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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Perfect 805 Score Debrief by Julia

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