🔥🔥🔥 hard challenge()

Question

Find the smallest Number such that if it's RIGHTMOST digit is placed at the BEGINNING, The new Number so formed is precisely 50% larger than the original Number?

Hint: use a C# or any other program to find the correct answer
or the longer trial and error method.

...good luck!

Answer 1

https://code.sololearn.com/cpmuEsS2VfMY/#py
Run the code to find the answer :D

Answer 2

I'm late 😞. Here's a JS solution 🙋.
https://code.sololearn.com/WO3Co0TYK73y/?ref=app

Answer 3

That's awesome @Visph.

Answer 4

@Sheena Singh:

Much efficient way to do:

n = 10
m = 0
while int(m) != 1.5 * n:
n += 1
m = str(n%10)+str(n//10)
print(n)

Answer 5

@Nomeh Uchenna Gabriel wrote:

<< That's awesome @Visph. >>

Awesome but not 'best answer'? ;P

Answer 6

@Krishna ~Wow 👏👏👏, this is too great! I never thought of the harder JavaScript version.

Answer 7

@Baptiste:

But more verbose ^^ (even if mine should probably been improved ;))

Answer 8

@visph
Even more efficient (no str/int)
n=10
m=0
dec=10
while (n%10*dec+n//10)!=1.5*n:
n+=1
if n>=dec*10:
dec*=10
print(n)

Answer 9

@visph Yes of course more verbose, but as we do not convert integer to strings each time, it stays more efficient ^^
I did not manage to make it less verbose without passing the number to string as you did