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.)

Since we know that the number starts with a "1", and if we admit (without proof) that one of the 6 numbers will end in "1", this "1" was obtained by multiplying 3 x 7.

So, N = 1abcd7, and if we make another "wild guess" that when multiplied by 3, the new number is abcd71,

3*N = (N - 100,000)*10 + 1

3*N = 10*N - 999,999

7*N = 999,999 =========> N = 142,857.

Posted by pcbouhid on 2005-07-05 23:12:43