challenge problem

[galactus]

Here is a fun challenge problem. It involves a little number theory, but nothing nasty and obscure.

Find the smallest positive integer such that when the first digit is shifted to the end of the number, the result is 3/2 times the original number

Ex: 1284 becomes 2841, but this is not 3/2 times 1284. The number is rather huge, so trial and error is not a good idea.

Another problem I think is fun is:

Find the volume of the unit cube using spherical coordinates in a triple integral.

Of course, we know the volume is 1 but using spherical is kind of wacky way to go about it.

I will leave them up for a week. I hope someone finds them interesting and fun.

 

[Deleted member 4993]

Right after I posted it - I realized that - But I couldn't delete my post - you were too fast.

Excel failed me - I got to find some other way.

 

[Denis]

Here's 5 of them (16digits):
1 176 470 588 235 294 * (3/2) = 1 764 705 882 352 941 ; similarly:
2 352 941 176 470 588
3 529 411 764 705 882
4 705 882 352 941 176
5 882 352 941 176 470

Never was exposed to number theory, so:
a = the moving digit ; b = remainder of number

b = a(3*10^x - 2) / 17 where x = length(b)

A li'l double looper (a and x) gets them quick.
I noticed repetition at 31 digits, but not getting involved with that!

 

[galactus]

Right after I posted it - I realized that - But I couldn't delete my post - you were too fast.

Excel failed me - I got to find some other way.

I deleted it for you.

 

[galactus]

That's it, Denis.