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187. Which of the following CANNOT be the sum of two prime numbers?
(A) 19
(B) 45
(C) 68
(D) 79
(E) 88

Any prime number more than 3 can be expressed as \(p=6n+1\) or\(p=6n+5\) (\(p=6n-1\)), where n is an integer >0 (check this: primality-check-108425.html).

So the sum of two primes more than 3 can yield the following remainders upon division by 6:
0 - if two primes are of a type \(p=6n+1\) and \(p=6n+5\);
2 - if both primes are of a type \(p=6n+1\);
4 - if both primes are of a type \(p=6n+5\);

Now, we are looking for the choice which is not a prime+2, or prime+3 or has a remainder other than 0, 2, or 4 upon division by 6.

(A) 19 --> 19-2=17=prime;
(B) 45 --> 45-2=43=prime;
(C) 68 --> yields a remainder of 2 upon division by 6 so theoretically can be the sum of two primes (and it is 61+7=68);
(D) 79 --> 79-2 is not a prime, 79-3 is not a prime and also 79 yields a remainder of 1 upon division by 6, so it can not be the sum of two primes;
(E) 88 --> 71+17=88.

Answer: D.

Of course the above can be done much easier by just subtracting the primes starting from 2 from the answer choices and seeing whether the result is also a prime.
Awesome explanation based on the crucial theory that any prime number >3 can be expressed in 6n+1 or 6n-1 format. I did not know this hence ended up taking an alternative logic. Where can I find more of such theorems on number theory.
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Bunuel
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187. Which of the following CANNOT be the sum of two prime numbers?
(A) 19
(B) 45
(C) 68
(D) 79
(E) 88

Any prime number more than 3 can be expressed as \(p=6n+1\) or\(p=6n+5\) (\(p=6n-1\)), where n is an integer >0 (check this: primality-check-108425.html).

So the sum of two primes more than 3 can yield the following remainders upon division by 6:
0 - if two primes are of a type \(p=6n+1\) and \(p=6n+5\);
2 - if both primes are of a type \(p=6n+1\);
4 - if both primes are of a type \(p=6n+5\);

Now, we are looking for the choice which is not a prime+2, or prime+3 or has a remainder other than 0, 2, or 4 upon division by 6.

(A) 19 --> 19-2=17=prime;
(B) 45 --> 45-2=43=prime;
(C) 68 --> yields a remainder of 2 upon division by 6 so theoretically can be the sum of two primes (and it is 61+7=68);
(D) 79 --> 79-2 is not a prime, 79-3 is not a prime and also 79 yields a remainder of 1 upon division by 6, so it can not be the sum of two primes;
(E) 88 --> 71+17=88.

Answer: D.

Of course the above can be done much easier by just subtracting the primes starting from 2 from the answer choices and seeing whether the result is also a prime.
Awesome explanation based on the crucial theory that any prime number >3 can be expressed in 6n+1 or 6n-1 format. I did not know this hence ended up taking an alternative logic. Where can I find more of such theorems on number theory.

Theory on Number Properties: math-number-theory-88376.html
Tips on Number Properties: number-properties-tips-and-hints-174996.html

All DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
All PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59
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banksy
Which of the following CANNOT be the sum of two prime numbers?

(A) 19
(B) 45
(C) 68
(D) 79
(E) 88
Sum of two prime numbers are odd only when one of the numbers is 2
A)2+17 possible
B)2+43 possible
D)2+77 is not possible
Hence D is the answer.
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This is a great Question.
Essentially this is testing our knowledge on Even/odd numbers.
If sum of 2 primes is even => 2 must not be either of them.
If sum of 2 primes is odd => 2 must be one of them.

Hence as 79 =2+77 =>79 can never be written as sum of 2 prime numbers.

Hence D
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prime nos. >3 can be written in the form: 6n+1 or 6n-1

Sum of two prime nos.= 6n+1+6n+1= 2(6n+1)

C and E can be eliminated on this basis.

Al the prime nos. are odd except 2.
Odd +Odd= Even
Odd + 2= Odd

The remaining three options are odd.
So subtract 2 from every choice.
D is the number that is always a multiple of some no.

D it is.
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Bunuel
banksy
187. Which of the following CANNOT be the sum of two prime numbers?
(A) 19
(B) 45
(C) 68
(D) 79
(E) 88

Any prime number more than 3 can be expressed as \(p=6n+1\) or\(p=6n+5\) (\(p=6n-1\)), where n is an integer >0 (check this: https://gmatclub.com/forum/primality-check-108425.html).

So the sum of two primes more than 3 can yield the following remainders upon division by 6:
0 - if two primes are of a type \(p=6n+1\) and \(p=6n+5\);
2 - if both primes are of a type \(p=6n+1\);
4 - if both primes are of a type \(p=6n+5\);

Now, we are looking for the choice which is not a prime+2, or prime+3 or has a remainder other than 0, 2, or 4 upon division by 6.

(A) 19 --> 19-2=17=prime;
(B) 45 --> 45-2=43=prime;
(C) 68 --> yields a remainder of 2 upon division by 6 so theoretically can be the sum of two primes (and it is 61+7=68);
(D) 79 --> 79-2 is not a prime, 79-3 is not a prime and also 79 yields a remainder of 1 upon division by 6, so it can not be the sum of two primes;
(E) 88 --> 71+17=88.

Answer: D.

Of course the above can be done much easier by just subtracting the primes starting from 2 from the answer choices and seeing whether the result is also a prime.

Bunuel when you say


p=6n+1p=6n+1 orp=6n+5p=6n+5 (p=6n−1p=6n−1),

do you mean

p=6n+1p=6n+1 "or" p=6n+5p=6n+5 (p=6n−1p=6n−1), ?
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Nunuboy1994
Bunuel
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187. Which of the following CANNOT be the sum of two prime numbers?
(A) 19
(B) 45
(C) 68
(D) 79
(E) 88

Any prime number more than 3 can be expressed as \(p=6n+1\) or\(p=6n+5\) (\(p=6n-1\)), where n is an integer >0 (check this: https://gmatclub.com/forum/primality-check-108425.html).

So the sum of two primes more than 3 can yield the following remainders upon division by 6:
0 - if two primes are of a type \(p=6n+1\) and \(p=6n+5\);
2 - if both primes are of a type \(p=6n+1\);
4 - if both primes are of a type \(p=6n+5\);

Now, we are looking for the choice which is not a prime+2, or prime+3 or has a remainder other than 0, 2, or 4 upon division by 6.

(A) 19 --> 19-2=17=prime;
(B) 45 --> 45-2=43=prime;
(C) 68 --> yields a remainder of 2 upon division by 6 so theoretically can be the sum of two primes (and it is 61+7=68);
(D) 79 --> 79-2 is not a prime, 79-3 is not a prime and also 79 yields a remainder of 1 upon division by 6, so it can not be the sum of two primes;
(E) 88 --> 71+17=88.

Answer: D.

Of course the above can be done much easier by just subtracting the primes starting from 2 from the answer choices and seeing whether the result is also a prime.

Bunuel when you say


p=6n+1p=6n+1 orp=6n+5p=6n+5 (p=6n−1p=6n−1),

do you mean

p=6n+1p=6n+1 "or" p=6n+5p=6n+5 (p=6n−1p=6n−1), ?

You mean space was omitted?

It's \(p=6n+1\) or \(p=6n+5\) (\(p=6n-1\))...
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banksy
Which of the following CANNOT be the sum of two prime numbers?

(A) 19
(B) 45
(C) 68
(D) 79
(E) 88

Recall that odd + odd = even and odd + even = odd. Also recall that 2 is the only even prime number. If the sum of two distinct prime numbers is odd, then one of them must be 2 since the other one will be odd, and the sum of 2 and an odd number is odd. So let’s check the answer choices that are odd numbers first.

A) 19

19 = 2 + 17

Since 17 is prime, A is not the correct choice.

B) 45

45 = 2 + 43

Since 43 is prime, B is not the correct choice either.

D) 79

79 = 2 + 77

Since 77 is not a prime, D is the correct choice.

Answer: D
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banksy
Which of the following CANNOT be the sum of two prime numbers?

(A) 19
(B) 45
(C) 68
(D) 79
(E) 88

A prime number always ends with 1 2,3,5,7,9.
So if we add any 2 prime numbers we cannot obtain a last digit 9.

So Ans: D
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