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The positive integers $A$, $B$, $A-B$ and $A+B$ are all prime numbers. The sum of these primes is?

  1. Even
  2. Divisible by 3
  3. Divisible by 5
  4. Prime

I did try adding them to get $3A + B$ but could not identify if it was a prime or even or anything.

The only thing I was able to realize was that $A$ can not be 3 or 2.

So, how can we do it??

EDIT

Please do consider that this is a question for 9th grade and we have not been taught about advanced mathemeatics like arithmetic progression etc...

I would be thankful to anyone providing the solution in simple terms

Sambhav Khandelwal
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    Hint: $B=2$, which eliminates the first two options – lhf Jun 05 '24 at 13:32
  • @lhf Hmmm. That is useful. Now how can we get rid of one more option? – Sambhav Khandelwal Jun 05 '24 at 13:33
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    You should be able to find $A,B$, though that is not asked for. Note you have three primes in arithmetic progression. – Ross Millikan Jun 05 '24 at 13:34
  • take $A=5$ and $B=2$. Then $3A+B=17$ which eliminate the 3 first option (but I would be surprised that the sum is always prime...) – joshua Jun 05 '24 at 13:35
  • @RossMillikan edited the question – Sambhav Khandelwal Jun 05 '24 at 13:37
  • @joshua trial and error is not something I would like the answer in... – Sambhav Khandelwal Jun 05 '24 at 13:37
  • if the problem is correct, then the only possible answer is "prime". But I wouldn't know how to prove it. – joshua Jun 05 '24 at 13:39
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    We know $B=2$, so $A-B, A, A+B$ is determined. There are many twin primes but only one prime triplet. Should be easy for you to prove that the solution is unique. – lulu Jun 05 '24 at 13:40
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    In the event that $A,B$ are not equal to $2$ it follows that $A,B$ are both odd primes at which point $A-B$ must be even and $A+B$ must be even... at most one of those two can be prime and equal to exactly $2$. Thanks to this observation and that $A-B$ must be positive and prime, we know that $B=2$ is the only possibility... at which point $A=5$ is the only possibility and we are done. – JMoravitz Jun 05 '24 at 13:41
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    @lulu : why $B=2$ ? – joshua Jun 05 '24 at 13:41
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    @joshua Because if $A,B$ are both odd then $A+B$ is even, so not prime. If $A=2$ then $A-B$ can't be prime (can't even be positive). That leaves $B=2$. – lulu Jun 05 '24 at 13:42
  • As for "only one prime triplet", that follows from the fact that $(n,n+2,n+4)$ at least one of these three must be divisible by $3$. For this to be a prime triplet, the only way that can happen is if the term divisible by $3$ must be exactly equal to $3$. – JMoravitz Jun 05 '24 at 13:45

4 Answers4

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If $A,B$ are both odd then both $A+B$ and $A-B$ are both even and cannot be prime. If they are both even all four values are even, so one is even and one is odd. $2$ is the only even prime, so that must be $B$. One of $A-2, A, A+2$ is divisible by $3$, so must be $3$ to be prime. That has to be $A-B$, so $A=5$. $3A+B=17$ is prime but none of the other choices.

Ross Millikan
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  • Your answer justifies the situation where A = 5 and B = 2 as I feel(correct me if I'm wrong). How can we generalize it? – Sambhav Khandelwal Jun 05 '24 at 13:44
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    @SambhavKhandelwal The point is that this is the only possible case, given the assumptions that $A$, $B, A+B$ and $A-B$ are prime. – user1008978 Jun 05 '24 at 13:46
  • @user1008978 : But how can we prove they are the only ones? Sorry if I'm missing something – Sambhav Khandelwal Jun 05 '24 at 13:46
  • @SambhavKhandelwal Ross has proved that $A=5,B=2$ is the only possible solution set. Read his answer carefully. – user Jun 05 '24 at 13:47
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    @SambhavKhandelwal: because of the statement that one of three numbers must be prime and divisible by $3$. There is only one prime like that. If $A=3$ then $A-B=1$, which is not prime. I leave the case $A+B=3$ to you. – Ross Millikan Jun 05 '24 at 13:48
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  1. If $A$ and $B$ are both odd, then $A+B$ and $A-B$ are both even, a contradiction. With the fact that $A-B>0$, we conclude that $A$ is odd and $B=2$.

  2. Now the sum becomes $3A+2$, which is obviously odd and not divisible by $3$. But since $3,5,7$ are the only three consecutive odd natural numbers that are prime(see proof here), the only choice of $A$ is $5$, and since 17 is prime, D is correct.

Harvey Peng
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I'm going to post another answer with the intent of very simplified mathematical language for, as OP states, $9$th graders.

I assume you know how to define the prime numbers, and can likely list the first few readily. You therefore know there is exactly one even prime, $2$, and all other primes are odd.

Now imagine if both $A$ and $B$ are both odd primes. The difference of two odd numbers is even, and the sum of two odd numbers is even. But then $A-B$ and $A+B$ cannot be prime--there is only one even prime!

Therefore either $A$ or $B$ must be equal to $2$. As $A$ is larger, we see $B=2$.

We now see that we have three odd primes: $(A-2, A, A+2)$. Of this set of numbers, exactly one must be divisible by $3$. We'll show using remainders, as if we were doing grade school division with remainders (which turns out to have powerful applications in higher math).

Suppose $A$ is divisible by $3$: That means $A$ has a remainder of $0$ when divided by $3$. $A+2$ must then have a remainder of $2$. And $A-2$ will have a remainder of $1$. It's not obvious, but this stays true no matter what remainder $A$ actually has--try it out for yourself!!

Now we have three primes, one of which must be divisible by $3$. There's only one prime that first the bill: $3$. Since $1$ and $-1$ aren't prime, we have to have $3 = A-2$. Then our other two numbers are $A=5$ and $A+2=7$.

The use of remainders to prove things like this involves modular arithmetic. Link is to Wikipedia, if you want to learn more.

Eric Snyder
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For a straight-forward list of answers (i.e. without much explanation), the value $3A+B$...

  1. ...will be even if and only if $A=2=B$ or $A\neq2\neq B$,

  2. ...will be divisible by $3$ if and only if $B=3$ (as a consequence of Euclid's Lemma),

  3. ...may or may not be divisible by $5$. An example would be $A=3$ and $B=11$. (Holds true if and only if $5C-3A=B$ for some $C\in\mathbb{Z}$.)

  4. ...will not always be prime (see the other responses for an immediate example).

If you have any questions, please comment on this post and I will be happy to reply!

J.G.131
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