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
ramblings
