If you moved the 3 at the end of a certain large number to the beginning, it would be the same as multiplying it by 3. What is the number?

Let X be the number in question.

Let Y = FLOOR(LOG(X)).

According to the problem, X and Y...

(X-3)/10 + 3*10^Y = 3*X

Multiply both sides of that equation by 10...

X - 3 + 3*10^(Y+1) = 30*X

Subtract X from both sides.

3*10^(Y+1) - 3 = 29X

So the left side of that equation must be divisible by 29.

Or, stated in another way,

3*10^(Y+1) is congruent to 3 mod 29.

which implies 10^(Y+1) is congruent to 1 mod 29.

Now all we do is find power of 10 that is congruent to 1 mod 29.

Since 29 is a prime number, Fermat's theorem tells us that...

10^28, 10^57, 10^86, etc. are congruent to 1 mod 29.

More generally put, 10^(29*n+28) is congruent to 1 mod 29 for every integer n >= 0.

Which means Y+1 = 29*n+28 where n is an integer. And that in turn tells us that...

29*X = 3*10^(29*n+28) - 3 where n is an integer >= 0

Take n=0 for example.

29*X = 3*10^28 - 3

29*X = 29999999999999999999999999997

X = 1034482758620689655172413793

So, to sum that up... the general formula for the complete set of answers is

X = (10^(29*n+28)-1)*3/29 where n is an integer >= 0

X is guaranteed to be a whole number because we found the set of power of 10 that have the right congruency in mod 29. And thus everything's in order here. Final , complete answer...