CHALLENGE✋: Integers, roots and division

Question

The point is to:
Find all the positive integers "n" that are divisible by all the positive integers not larger than the square root of n.
Any language is welcomed.

Answer 1

Odd. The question made perfect sense to me. At least this is how I interpreted it.

*Remember to take heed for cases 0 and 1. :>

https://code.sololearn.com/cyLCMBAHgdg4/?ref=app

Answer 2

recursion only ☺
//a modification can be made to improve code but'll increase the size of code
https://code.sololearn.com/cBLVKqmodgX9/?ref=app

Answer 3

(x Ɛ Z : 0 < x < ∞) where x % [1, [(√x)]] == 0 , [.] denotes greatest integer function & and % denotes modulus operator(which we use in java 😂)
//in math we state it like this
@blackcat1111

/*it would be wrong to say that 0 is included in set of +ve numbers , u can 0 is included in set of non negative numbers */

Answer 4

guys are u all out of the confusion that why 24 is maximum , if not then u can see this link ☺👍
https://publish.pothi.com/preview/?sku=ebook3477

Answer 5

https://code.sololearn.com/c40R8yE5I1v9/?ref=app

Answer 6

My try
https://code.sololearn.com/cDieqs1c6v5j/?ref=app

Answer 7

@blackcat1111
Absolutely. Unless you haven't understand the challenge properly, or I haven't been clear enough while explaining...

Answer 8

@Martin
An example: 32 is not a solution because sqrt(32)=5.something, so 32 has to be divisible by 1, 2, 3, 4 and 5. But 32%5 or 32%3 is not 0. Did you get it now??

Answer 9

@Guarav Agrawal Sorry, haven't done math in two months 😉

Answer 10

Based on my understanding of your question:
https://code.sololearn.com/cGuulwT0K3Li/?ref=app

Answer 11

https://code.sololearn.com/clf0JFFUjFbT/?ref=app

Answer 12

All? You are sure there is not an infinite quantity of such?

Answer 13

@Ledio Deda I shall rephrase the question; please tell me if I am wrong.
Find all positive integers (x Ɛ Z : 0 ≤ x < ∞) where
x MOD (1.. INT(√x)) == 0.

Answer 14

Well
@Hatsy Rei found them🎉🎉...
Other ways of solution expected...

Answer 15

@blackcat
✅Correct, but in order to avoid time limit exceeded, you can follow a simple maths algorythm to find the larges possible integer "n" who can satisfy the condition...

Answer 16

@VcC
Yes, but that kp must be less or equal than sqrt(n).
Got it???

Answer 17

@yerucham
Absolutely correct!!!👏👏
The only advice is to find the largest possible solution and put the limit to that nr and not 100000.
But that is optional...👍

Answer 18

Well, I would let it turn as an infinite loop, to keep the initial logic ... if the thing has to be used as a checker, one shouldn't stop it unless he knows the answer. But the code playground here doesn't like it.
https://code.sololearn.com/cJmR8F49KA0X/#py

Answer 19

https://code.sololearn.com/crqZak5WxYed/?ref=app

Answer 20

@Ledio Deda Thanks for the tip! It was actually pretty easy. Fixed my code now.
[Spoilers Alert]
Consider the largest number p satisfying the conditions where p MOD (0..q], where q = int(√p).
Following the definition of q, p must be q! as p is the largest number satisfying the conditions.
Therefore: p = q! where q = int(√p).
Hence the largest number is 24.