What six digit number added to itself five times will give a sum each time having the same 6 digits as the original number?

(It's not 000000.)

Let be N the six-digit number that we are searching.

The derived numbers are 2*N, 3*N, 4*N, 5*N and 6*N.

Since 6*N is a six-digit number, the leftmost digit of N must be 1.

If N were even, its righmost digit would be (2, or 4 or 6 or 8). All the successive products would end in (2 or 4 or 6 or 8 or 0). And the digits of N would be (0, 1, 2, 4, 6 and 8). But, no matter where the digit 0 is placed, since 3 times the even digits 4 or 6 would carry 1, when multiplying 3 by the next even digit, would result an odd number. Impossible.

So, N is odd. 

Since N is odd, its rightmost digit is 3 or 5 or 7 or 9.

1 * N would end in (3.....5......7.....9).

2 * N would end in (6.....0......4.....8).

3 * N would end in (9.....5......1.....7).

4 * N would end in (2.....0......8.....6).

5 * N would end in (5.....5......5.....5).

6 * N would end in (8.....0......2.....4).

The only "column" where appear the already knew digit "1" is the third above, and so the digits of N are (7, 4, 1, 8, 5, 2).

To arrange them in order is easy.

 

 

 

The other five digits adds to

Posted by pcbouhid on 2005-07-06 01:12:59