What is the largest possible integer value of x, so that 30^x divides 345! evenly?
I think that the problem has to do something with finding all possible numbers (and products) that are divisible by 30, however, I can't seem to think of any way to really find them all, without trying to multiply all the numbers and then getting prime factors of it... It seems like a hassle. I don't think that trying to take modulus 30 would help to solve the problem, because we need to find what modulus is it, not what remainder is (sure, we could try to brute force the answer by looking at 30^0, 30^1, 30^2, but that would take way too long and I am not willing to do so). Do anyone have any ideas?
Since \(345! = 345\cdot 344 \cdot 343 \cdot ... \cdot 1\), what if you just find the smallest possible \(k\) such that \(30^k > 345 \). Then any \(k < x\) is a factor of 345!
That should be \( x < k\)
@kmeis002 but smallest k would be 2. however, since 345! include 200 and 135 we can see that 345! is divisible by 30^3 as well.
Or do you mean smallest k that 30^k > 345! ?
No offense given by this post, but the original post states \( 30^k > 345!\)
Well, your answer had a mistake then, sorry. However, how would I find such number?
I apologize, the first post was correct with no typo, but as you stated we haven't collected all the other factors of 30. One moment.
I apologize for the error in logic. Let me rephrase the first post. It should be \(30k > 345\). Then every \(x < k\) must be included in \(345!\)
Thank you for clearing it up! Your idea seems solid, but I can't make it really work. lowest k would be 12. So, x = 11? (Answer of the problem?) Sorry to ask this as it looks quite a bad question, but it doesn't get through me...
Well, that would mean that \(30 \cdot 11, 30 \cdot 10, 30\cdot 9, ... 30 \cdot 0\) are all factors of \(345!\). In fact, they are the only factors of 30 in 345!
Thank you for your help!
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