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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  [#permalink]

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New post 19 Jul 2017, 23:50
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  85% (hard)

Question Stats:

44% (02:08) correct 56% (02:25) wrong based on 124 sessions

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How many three-digit numbers contain three primes that sum to an even  [#permalink]

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New post 24 Jul 2017, 20:06
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1
Bunuel wrote:
How many three-digit numbers contain three primes that sum to an even number?

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



Hi,
if you dont want to calculate separately..

Ways the sum will be ODD 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
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Re: How many three-digit numbers contain three primes that sum to an even  [#permalink]

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New post 20 Jul 2017, 00:20
4
Bunuel wrote:
How many three-digit numbers contain three primes that sum to an even number?

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


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
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Re: How many three-digit numbers contain three primes that sum to an even  [#permalink]

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New post 24 Jul 2017, 19:57
1
Bunuel wrote:
How many three-digit numbers contain three primes that sum to an even number?

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


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
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Re: How many three-digit numbers contain three primes that sum to an even  [#permalink]

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New post 26 Jul 2017, 21:45
chetan2u wrote:
Bunuel wrote:
How many three-digit numbers contain three primes that sum to an even number?

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



Hi,
if you dont want to calculate separately..

let the first number be 2, the remaining 2 can be filled with any of the 3,5,7..
so 1*3*3
this 1, that is 2, can be placed in of three position so 1*3*3*3=27..

ofcourse a number without any odd prime 222..

so 27+1=28

B


Much easier approach. Thanks Chetan.
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Re: How many three-digit numbers contain three primes that sum to an even  [#permalink]

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Re: How many three-digit numbers contain three primes that sum to an even &nbs [#permalink] 03 Oct 2018, 00:23
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