[Guest]

Find the smallest integer number x that its triple is equal to the double of the number obtained when you move the digit on the left to the right side of the number.

In other words, solve the equation 3x=2y where, let's say, x=abcdef and y=bcdefa (if the number had 6 digits).

No tricks here, or trivial solutions like 0, just a bit of algebra... If doing things right, it shouldn't take more than 15 minutes to solve it.

[akaslickster]

3

[Guest]

4356

as the smallest.

EDIT... sorry, misread the question, was completely reversing the number.

Edited September 4, 2008 by pjbno4

[k-man]

It took me about 5 minutes, but I'm not sure if this is the smallest number. This is what I came up with...

3 * 1176470588235294 = 2 * 1764705882352941

[Guest]

It took me about 5 minutes, but I'm not sure if this is the smallest number. This is what I came up with...

3 * 1176470588235294 = 2 * 1764705882352941

That's the one! But it should be easy for you to demonstrate that it's the smallest, how did you get to the answer?

[Guest]

Could someone please post what the best method for solving this problem is.

Thanks

[Guest]

Could someone please post what the best method for solving this problem is.

Thanks

OK here is the solution:

Let x be the number we look for, with n digits, and 'a' a one-digit number.

Let z be the number x taking out the most significant digit.

Thus,

x=a*10^n+z

So we have to solve:

3(a*10^n+z)=2(10z+a)

rearranging,

z=a(3*10^n-2)/17

so we have to find the smalles number like this 299999...9998 that is divisible by 17.

Or, easier, find a number like this 30000... whose reminder is 2 when we divide by 17.

So we just do the division the normal way, until we get a reminder 2:

3000000000... |17___

130 ............... 176...

..110

....80

........

To get a reminder of two we find that the number is 176470588235294 (pretty long...) so n=16.

We set a=1 to get the minimum number z, so:

x=1176470588235294, which actually has 16 digits (n), so everything's fine, and the relationship holds.

If z is the minimum, and a=1, x cannot be smaller, so it is the minimum.

q.e.d.

Edited September 6, 2008 by despistado

[k-man]

That's the one! But it should be easy for you to demonstrate that it's the smallest, how did you get to the answer?

I did it pretty much how you described, but there is a digit a there that is a part of the equation. I wasn't sure it's 1. I found this number assuming a=1, but I wasn't sure that for some other a there isn't a smaller number.

Edited September 8, 2008 by k-man