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?

(In reply to complete set of answers, mathy reasoning by empraptor)

The general formula should read

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

I made a mistake in doing modular math.  10^(29*n+28) is NOT congruent to 1 mod 29 for all n.  For example, 10^(29+28) = 1-^57 is congruent to 10 mod 29.

Because 10^28 is congruent to 1 mod 29,  10^28 * 10^28 should also be congruent to 1 mod 29.  If you extend this reasoning, 10^(28*n) is congruent to 1 mod 29 for all integer n >= 0.

Of course, while n=0 does satisfy the equation we set up to model the problem statement, it is not a valid solution because X=0 in that case.  So we limit n to >= 1.

Posted by empraptor on 2004-03-04 03:44:36