Math Teachers at Play #5 is up at Let’s Play Math! It’s organized by topic, with pictures and neat quotations thrown in for good measure. And as usual, it is full of interesting posts!

Speaking of Carnivals, apparently the Carnival of Mathematics was just hibernating and it will reappear next week. At least, that’s the rumor, where “rumor” means I read it on jd2718.

And speaking of reading things, after reading Keith Raskin’s comment I headed over to Natural Blogarithms to look for Pythagorean Triple stuff, and I was immediately distracted by the following puzzle:

The four numbers A, B, A+B and A-B are all prime. The sum of these four numbers is

A) Even
B) Divisible by 3
C) Divisible by 5
D) Divisible by 7
E) Prime

(from the 2002 AMC 10/12B #15).

I quickly convinced myself of one answer, then decided there was a misprint and talked to Batman about it at work, and then we he realized that exactly one of the possible answers was correct. At that point the problem seemed a lot more interesting.

If you need something a little more lighthearted this Friday afternoon, here’s the 1-20 roll call from Sesame Street.

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3 Responses to “Math Teachers at Play and some other stuff”

Well, a quick thought says that in order for the sum of two primes to be prime, one of them has to be 2. So we have B=2. An example, then, is A=5, A+B=7, A-B=3, and sum(A, B, A+B, A-B)=17.

Given that, if there’s a correct answer from the list, it has to be “E) Prime”.

Am I missing something? What answer did you quickly convince yourself of? What did you think the misprint was? What makes the problem interesting? (Because unless I’ve missed something basic, I don’t see it as interesting at all.)

I’d initially convinced myself that both primes had to be odd (because if B=2, then one of A-B, A, A+B must be divisible by 3) and so the sum of all four was 3A+B, which would have to be an even number. But on the other hand, as Batman pointed out, you can’t have both primes being odd because otherwise A-B and A+B will both be distinct even primes, which is a contradiction.

So the sneakiness of this problem is that if you combine both things you need B=2 and A-B=3 [in order to be both divisible by 3 and prime]. So the ONLY set of numbers that fits this problem is A=5 and B=2.

[I think I liked the fact that it reads like a general problem, but hiddenly describes a unique pair of numbers so you just need to pick something that describes 17. Plus it’s kind of neat that the sum is also prime.]

Solution (I think) to the prime number problem: (I think it’s also neat that it can be solved by proof, stronger than elimination or example.)

Given A and B are primes.

Either A or B must be 2 in order for A + B to be prime (odd > 2)

It must be B in order for A – B to be prime (assuming primes are positive: 2, 3, 5 …)

The only prime bracketed between primes like this: A – 2, A, A + 2
is 5, since A must be 1 or 2 mod 3, and either A – 2 or A + 2 is 0 mod 3, divisible by 3, which is only okay if it is 3 itself. A can’t be 3, since 1 is not prime; that’s how I know it’s got to be 1 or 2 mod 3.

April 18, 2009 at 10:53 am |

Well, a quick thought says that in order for the sum of two primes to be prime, one of them has to be 2. So we have B=2. An example, then, is A=5, A+B=7, A-B=3, and sum(A, B, A+B, A-B)=17.

Given that, if there’s a correct answer from the list, it has to be “E) Prime”.

Am I missing something? What answer did you quickly convince yourself of? What did you think the misprint was? What makes the problem interesting? (Because unless I’ve missed something basic, I don’t see it as interesting at all.)

April 18, 2009 at 11:05 am |

Barry,

I’d initially convinced myself that both primes had to be odd (because if B=2, then one of A-B, A, A+B must be divisible by 3) and so the sum of all four was 3A+B, which would have to be an even number. But on the other hand, as Batman pointed out, you can’t have both primes being odd because otherwise A-B and A+B will both be distinct even primes, which is a contradiction.

So the sneakiness of this problem is that if you combine both things you need B=2 and A-B=3 [in order to be both divisible by 3 and prime]. So the ONLY set of numbers that fits this problem is A=5 and B=2.

[I think I liked the fact that it reads like a general problem, but hiddenly describes a unique pair of numbers so you just need to pick something that describes 17. Plus it’s kind of neat that the sum is also prime.]

April 19, 2009 at 2:05 pm |

Solution (I think) to the prime number problem: (I think it’s also neat that it can be solved by proof, stronger than elimination or example.)

Given A and B are primes.

Either A or B must be 2 in order for A + B to be prime (odd > 2)

It must be B in order for A – B to be prime (assuming primes are positive: 2, 3, 5 …)

The only prime bracketed between primes like this: A – 2, A, A + 2

is 5, since A must be 1 or 2 mod 3, and either A – 2 or A + 2 is 0 mod 3, divisible by 3, which is only okay if it is 3 itself. A can’t be 3, since 1 is not prime; that’s how I know it’s got to be 1 or 2 mod 3.

So, A = 5, b = 2

A + B + A – B + A + B = 3A + B = 17 prime.