Find the smallest number such that if its rightmost digit is placed at its left end, the new number so formed is precisely 50% larger than the original number.

(In reply to

ramblings 3 by Victor Zapana)

6-digit number:

ideally, it would be: (for number abcdef)

1.5 (100000a + 10000b + 1000c + 100d + 10e + f) = 100000f + 10000a + 1000b + 100c + 10d + e

150000a + 15000b + 1500c + 150d + 15e + 1.5f = 100000f + 10000a + 1000b + 100c + 10d + e

140000a + 14000b + 1400c + 140d + 14e = 99998.5f

14 (10000a + 1000b + 100c + 10d + e) = 99998.5f

Like before, f must be 2, 4, 6, or 8.

99998.5 x 2= 199997 REJECT not divisible by 14

99998.5 x 4= 399994 ACCEPTED divisible by 14.

So, f is 4.

399994 / 14= 28571.

to be continued