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.

It cant be a 1-digit number

2 digit number:

ideally, it would be: (for number xy)

1.5 (10x + y) = 10y + x

15x + 1.5y = 10y + x

14x = 8.5y

Becuase 14x and 8.5y are both WHOLE NUMBERS, y must be even 2,4,6,8 (im not agreeing with Penny on the 000 idea). In addition x and y must both be whole numbers between 0 and 9.

8.5 x 2= 17 REJECT not divisible by 14

8.5 x 4= 34 REJECT not divisible by 14

8.5 x 6= 51 REJECT not divisible by 14

8.5 x 8= 68 REJECT not divisible by 14

It cannot be a 2-digit number

3-digit number:

ideally, it would be: (for number xyz)

1.5 (100x + 10y + z) = 100z + 10x + y

150x + 15y + 1.5z = 100z + 10x + y

140x + 14y = 98.5z

14(10x + y) = 98.5z

Because 14, 10x+y, and 98.5z are all WHOLE NUMBERS, z must be even, so its either 2,4,6,8. (not including 0 coz i disagree with penny with the 000 idea)

98.5 x 2= 197 REJECT not divisible by 14

98.5 x 4= 394 REJECT not divisible by 14

98.5 x 6= 591 REJECT not divisible by 14

98.5 x 8= 788 REJECT not divisible by 14

So, it is not a 3-digit number.

to be continued

*Edited on ***January 25, 2004, 1:45 pm**