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)
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