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a + b + c = 34, where a, b and c are primes 08 Oct 2021, 10:26
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E
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95% (hard)

Question Stats:

44% (02:28) correct 56% (02:14) wrong based on 152 sessions

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a + b + c = 34, where a, b and c are primes. How many ordered solution sets of (a, b, c) are possible, for the given equation?

(A) 0

(B) 2

(C) 3

(D) 6

(E) 12
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E
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a + b + c = 34, where a, b and c are primes 08 Oct 2021, 10:39
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a + b + c = 34, where a, b and c are primes. How many ordered solution sets of (a, b, c) are possible, for the given equation?

(A) 0

(B) 2

(C) 3

(D) 6

(E) 12
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We start by finding the possible amount of nonordered solutions first. Then we multiply that by 6 as for each unordered set of solutions, we have 6 ways of ordering it.

First, note we cannot have odd + odd + odd as that results in an odd number. Then one of the numbers must be even, hence equal to 2.

Therefore we can rewrite this as (Two Odd Primes added up = 32). We can try 3 + 29, 5 + 27, 7 + 25, 11 + 21, 13 + 19 and stop there.

We can see 3 + 29 and 13 + 19 are the only options for two primes. Hence we have two nonordered sets of solutions.

Multiply by 6 to get 12 ordered sets of solutions.

Ans: E
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a + b + c = 34, where a, b and c are primes 09 Oct 2021, 05:17
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Can someone please explain the concept of ordered set and unordered set please?
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a + b + c = 34, where a, b and c are primes 11 Oct 2021, 06:33
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a + b + c = 34, where a, b and c are primes. How many ordered solution sets of (a, b, c) are possible, for the given equation?

(A) 0

(B) 2

(C) 3

(D) 6

(E) 12
Show more

Solution:

We are given \(a + b + c = 34\), where a, b and c are primes.

One inference which you should make very early is that one of \(a, b\) or \(c\) has to be even otherwise \(odd+odd+odd\) would never give you \(34 (even)\).

Since \(a, b\) and \(c\) are primes, the only possible even prime number is \(2\). So, one of \(a, b\) or \(c\) is equal to \(2\).

Let us assume \(a=2\).

So, \(b+c=34-2=32\).

We can see that 32 can be broken into 2 prime numbers as \(3+29\) and \(13+19\).

Now, we have 2 triplets with us \((2, 3, 29)\) and \((2, 13, 19)\)

Number of ways in which \(a, b\) and \(c\) can be \((2, 3, 29)= 3!=6\)

Number of ways in which \(a, b\) and \(c\) can be \((2, 13, 19)= 3!=6\)

So, the total number of ordered sets \(= 6+6=12\).

Hence the right answer is Option E.
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a + b + c = 34, where a, b and c are primes 16 May 2024, 00:09
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GMATWhizTeam
a + b + c = 34, where a, b and c are primes. How many ordered solution sets of (a, b, c) are possible, for the given equation?

(A) 0

(B) 2

(C) 3

(D) 6

(E) 12
Show more
­a + b + c = 34 (even) only possible when even + odd + odd = even.

therefore the even prime that we have is 2 and now equation will become,

odd + odd = 34 - 2 = 32.

set of possible values of those two odd primes (3, 29) and (13, 19) from the set of (3,5,7,11,13,17,19,23,29)

hence possible set of values (2,3,29) and (2,13,19) 

and their arrangements for each possibility is 3!.

Therefore total possibilities are 2*3! = 12.

Hence E.
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a + b + c = 34, where a, b and c are primes 16 May 2024, 01:09
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Given: a + b + c = 34, where a, b and c are primes.

Asked: How many ordered solution sets of (a, b, c) are possible, for the given equation?

Primes below 34 = {2,3,5,7,11,13,17,19,23,29,31}

Since sum a+b+c = 34 is even, one number should be even and other 2 odd.

2 is one number (even)

34 - 2 = 32; The sum of 2 odd numbers should be 32

32 = 3 + 29 = 13 + 19

2,3,29 & 2,13,19 are possible solutions which add upto 34.

Number of Ordered pairs (a,b,c) = 3! + 3! = 6 + 6 = 12

IMO E
­
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a + b + c = 34, where a, b and c are primes 12 Jun 2024, 07:51
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a + b + c = 34, where a, b and c are primes. How many ordered solution sets of (a, b, c) are possible, for the given equation?

(A) 0

(B) 2

(C) 3

(D) 6

(E) 12
­a + b + c = 34 (even) only possible when even + odd + odd = even.

therefore the even prime that we have is 2 and now equation will become,

odd + odd = 34 - 2 = 32.

set of possible values of those two odd primes (3, 29) and (13, 19) from the set of (3,5,7,11,13,17,19,23,29)

hence possible set of values (2,3,29) and (2,13,19) 

and their arrangements for each possibility is 3!.

Therefore total possibilities are 2*3! = 12.

Hence E.
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­Hi. If it is an ordered set, can we consider 3! for each set for their arrangements?
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a + b + c = 34, where a, b and c are primes 30 Jun 2024, 00:08
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a + b + c = 34, where a, b and c are primes. How many ordered solution sets of (a, b, c) are possible, for the given equation?

(A) 0

(B) 2

(C) 3

(D) 6

(E) 12
­a + b + c = 34 (even) only possible when even + odd + odd = even.

therefore the even prime that we have is 2 and now equation will become,

odd + odd = 34 - 2 = 32.

set of possible values of those two odd primes (3, 29) and (13, 19) from the set of (3,5,7,11,13,17,19,23,29)

hence possible set of values (2,3,29) and (2,13,19) 

and their arrangements for each possibility is 3!.

Therefore total possibilities are 2*3! = 12.

Hence E.
­Hi. If it is an ordered set, can we consider 3! for each set for their arrangements?
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­Yes 3! for each 
like for 2 3 29
can be arranged in (2 3 29) (2 29 3) (3 29 2) (3 2 29) (29 2 3) (29 3 2)
similary for 2 13 19 you can make 6 possible arrangements.
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