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.
ideally, it would be: (for number abcd)
1.5 (1000a + 100b + 10c + d) = 1000d + 100a + 10b + c
1500a + 150b + 15c + 1.5d = 1000d + 100a + 10b + c
1400a + 140b + 14c = 998.5d
14 (100a + 10b + c)= 998.5d
because 14(100a + 10b + 10c) and 998.5d are both WHOLE NUMBERS, d must be even, so it is 2,4,6,8 (not including 0 look at previous post) also, a, b, c are between 0 and 9.
998.5 x 2= 1997 REJECT not divisible by 14
998.5 x 4= 3994 REJECT not divisible by 14
998.5 x 6= 5991 REJECT not divisible by 14
998.5 x 8= 7988 REJECT not divisible by 14
Ahhh, it's not a 4-digit number either.. -cries-
to be continued
Penny's idea seems more and more plausible lol.
Edited on January 25, 2004, 1:52 pm