Let $P_1$ be the set of all primes {2,3,5,7,....} and for each integer n, let $P_n$ be the set of all prime multiples of n {2n, 3n, 5n, 7n,....}. Which of the following intersections is nonempty?
A)$P_1\cap P_{23}$
B) $P_7\cap P_{21}$
C)$P_{12}\cap P_{20}$
D)$P_{20}\cap P_{24}$
E) $P_5\cap P_{25}$
Please, help

Question

Let $P_1$ be the set of all primes {2,3,5,7,....} and for each integer n, let $P_n$ be the set of all prime multiples of n {2n, 3n, 5n, 7n,....}. Which of the following intersections is nonempty?
A)$P_1\cap P_{23}$
B) $P_7\cap P_{21}$
C)$P_{12}\cap P_{20}$
D)$P_{20}\cap P_{24}$
E) $P_5\cap P_{25}$
Please, help

loser66 · Answer

By try and error, we have 60 = 12*5 = 20*3, hence C is the correct answer. But I would like to know the logic rather than using try and error method.

kainui · Answer

I guess the way I came to my answer was really this just boils down to taking the two integers and seeing if you can multiply both of them by a single prime to get the other one.

So if we have $P_{n}\cap P_{m}$ and two primes $p$ and $q$ we can really set up this equation and if it's satisfied, the intersection is not empty.

$pn = qm$

This sounds so formal, but really it's quite straight forward, going down the list we see:

n=1 and m=23 well they're both being multiplied by single primes so since 23 is prime we know that n is a product of 1 prime and m is a product of 2 primes. Not a chance!

7 and 21 seem like they might be likely, but even if we multiply 7 by 3 to get 21, we have to multiply 21 by some prime which is greater than 1 by definition, so that's thrown out.

Next one we see 2*2*3 and 2*2*5 Hey they differ by a single prime, so when the first one's multiplied by 5 and the second is multiplied by 3 it works out. We can keep going to check ourselves, but that's the reasoning I used. There's probably another way to do it too, I think I've seen this sorta problem before.

loser66 · Answer

Thanks a lot. I got what you mean.
Can we use LCM or GCD here?
like if 1, 23, LCM = 23, but 23 is not in P_23
          7, 21, LCM = 21 and 21 = 7 *3 in P_7 but not in P_21 because the smallest number in P_21 is 42
          12,20, LCM = 60 and as shown above
    last one. 5,25, LCM = 25 and 25 is in P_5 but not in P_25 since the smallest one whose last digit is 5 in P_25 is 125 = 25 *5 
Is it valid?

ganeshie8 · Answer

The way kai put it looks simpler right
$$ab = cd ~~ \implies~~  a=c,b=d~~\lor~~a=d,b=c$$
whenever $a,b,c,d$ are primes

kainui · Answer

I think we can say, but don't know how to explain this I can only tell you that I know it's true since I see that the LCM will give us the "overflow" amount and the GCD will divide out what's in common leaving just that overflow amount which is the product of our primes: $$\frac{LCM(m,n)}{GCD(m,n)} = p*q$$
So in examples we can see:
$$\frac{LCM(1,23)}{GCD(1,23)}=\frac{23}{1} =23= p*q$$
$$\frac{LCM(7,21)}{GCD(7,21)}=\frac{21}{7} =3= p*q$$
$$\frac{LCM(12,20)}{GCD(12,20)}=\frac{60}{4} =15= p*q$$
etc...

kainui · Answer

The most compact and general way that I can write this is with the divisor counting function as: $$ \tau \left(\frac{LCM(m,n)}{GCD(m,n)} \right) = 2$$

If that statement is true then $P_n\cap P_m$ is nonempty.

kainui · Answer

Actually.. that last statement isn't entirely true since it could be true if we have say m= 5 and n=20. Oh well I think this is taking it too far anyways haha.

loser66 · Answer

Yes, I am supposed to solve this problem in 2minutes.