May 21
The number 192 doubled is 384 and tripled is 576. Notice that these three numbers, 192, 384, 576, contain all of the digits from 1 to 9 once each. Determine the only three-digit number ending in 7 with the same property, that is: the number itself, its double and its triple contain all of the digits from 1 to 9 once each.
Comment:
I started the same way rokapoke did, but went a bit further. Once you realize the hundreds digit can only be 2 or 3.
If the hundred's digit is 2, then 2xx doubled must produce a number between 400 and 599 and tripled between 600 and 899. Since 4 is already used, doubled must be in the 500s and tripled cannot be in the 600s since you cannot carry when doubling unless you also do when tripling. 7 has already been used, so tripled must be in the 800s.
So, you have 2x7, 5x4, and 8x1, leaving 3, 6, and 9 still to place. Since tripled is 8x1, we need to carry 2 from the middle digit and only 9 accomplishes that, so the only possibility is 297, but that gives us 594, which doesn't work.
Which means the hundred's digit must be 3 and the doubled and tripled numbers cannot carry the hundred's digit (or the tripled value would exceed 1000), so we have 3x7, 6x4, and 9x1 with 2, 5, and 8 still to place. Since the 10s digit cannot carry when doubled or tripled, our only posibility is 327, which gives us the correct results of 327, 654, and 981.
It's still effectively trial and error, but with enough information to indicate that there are no other possible solutions.
