What Will People Do For A 40 Le Giveaway?

The first person to win two rounds gets 40 le. I'm guessing this will work better than the counting threads =)

Round One: Math
​

-Each person must multiply the person above's answer by an integer between 2 and 9, inclusive.
-The winner is the first person whose number meets the following criteria:
-Must be a perfect square
-Must be divisible by 3*11*(3!^2!-1!) plus 441^(1/(3-sine(41.9)^2-cosine(41.9)^2))
-Must have the first digit be 1, have an odd second digit, and the last digit cannot be 2, 4, 6, or 8
-And with that... 1!

 

..................
kthen
must be divisible by 1188 aaand.... ummm
9

 

u know... i want the LE... but im on vacations so I refuse to do this... xD what am i Sayin... 121!!!

 

968
has to be perfect square and divisible by 1188 and 21

 

968 * 2 = 1936
perfect square
first digit is 1
odd second digit
last digit is a 6 so it's wrong
not divisible by 1176 (the long equation thing) so it's still wrong

 

Hold on a second....
Someone multiplied 9 * 121/9 = 121.
Correcting this mistake (i. e. dividing by 121/9), the 1936 should be 144.
By the way, kudos to memethyl on the 1176.
Continue with 144.

 

oh by the way, here's a little python 3 script i wrote up to test your number:

Spoiler: the code

Code:

import math
num = int(input("Number: "))

def number_check(i):
  if math.sqrt(i) == round(math.sqrt(i)):                        # If i is a perfect square...
      if str(i)[0] == '1':                                       # ...and the first digit is 1...
          if int(str(i)[1]) % 2 == 1:                            # ...and the second digit is odd...
              if int(str(i)[-1]) % 2 == 1 or str(i)[-1] == '0':  # ...and the last digit isn't 2, 4, 6, or 8...
                  if i % 1176 == 0:                              # ...and it's divisible by 1176,
                      return True                                # return True.
  return False                                                   # If any of those conditions fail, return False.

print(number_check(num))

you can run this at https://repl.it/languages/python3

Spoiler: list of solutions between 1 and 1,000,000,000 if you don't feel like checking

176,400
1,587,600
11,289,600
17,640,000
110,250,000
119,246,400
138,297,600
158,760,000
192,099,600

if you somehow get any of these, you win part 1

 

danm you math

 

144 x 9 = 1,296
Let's go.

 

1296 x 9 = 11,664

Lets get to higher numbers

 

1296 x 2=2592 I don't even know what i'm doing

 

Yay bigger numbers
11664 * 5 = 58320
Let's never have to worry about requirement 3 again

 

58320 * (176400/58320) =176400

the number of the division is rounded to this

3.02469

 

It must be an "integer" cheater.

 

xD i think the guy isn't paying attention to the thread so... thas why i kinda did that xD

 

He wasn't paying attention so do we do his number now?

-RogueHatter

________________________________

Wait, now I'm confused . . .
What number do we do?

-RogueHatter

 

my number is illegal so dont use it

 

So what number do we use now?

-RogueHatter

 

58320 * 5 = 291600

 

291600 * 7 = 2,041,200