10 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.first 10 primes are 2,3,5,7,11,13,17,19,23,29
P = 2x3, 2x5, ..., 2x29
3x5, ..., 3x29
5x7, ..., 5x29,
11x13, ..., 11x29
13x17, ..., 13x29
17x19, ..., 17x29
19x23, ..., 19x29
23x29
my logic was that
if x = 2, there are 10 possibilities (2x3, ..., 2x29)
if x = 3, there are again 10 possibilities (2x3, 3x5, ..., 3x29)
and so on till x=29
this makes a total of 100
also since y belongs to A, not P, we have to include the original element itself
eg. 2 divides 2, 3 divides 3 etc
that adds 10 more
why isnt the answer 110?
this is their explanation, i'm not entirely sure what they're doing after
y=lambda*x
Basically, P is not just set of all (p_i)(p_j)
P is set of all products of distinct primes taken two at a time, three at a time, four at a time and so on.
But I don't think there should be a 9C0
Shouldn't there? Because X=S union P so y=x case to be counted no?
Oh yeah nvm. Yup that's also there.
i think there would be 9. 2x2 isn't counted, so the multiplier can only be chosen from 9 distinct numbers, not 10, while the multiplicand is 2.
9 for each multiplicand, 10 total multiplicands, so 90 ordered pairs from that, and then the numbers' pairs with themselves like you said at the end, so the total number of ordered pairs would be 90+10.
i didn't realize the thing that was later pointed out by opt either :').
ah yeah u right
the amount of overcounting i do in this chapter jesus 💀
forgot to reply, understood thank you 👌
+solved @Opt
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