If ABC+DEF+GHI=JJJ, each letter stands for a different digit, and no number starts with zero, what is J?

Here's how I worked this out.

We know that A, D, and G are non-zero; we also know that their sum (combined with the sum of any carries) will be 9 or less.<o:p></o:p>

<o:p> </o:p>

The lowest values for A/D/G (in no particular order) is 1/2/3. The highest possible value for these three is 2/3/4 (and no carry), which would add up to 9.<o:p></o:p>

<o:p> </o:p>

So 6 >= J >= 9. Moreover, we’re pretty limited in what A/D/G can be. The options are:<o:p></o:p>

1/2/3 w/ carry of 0 … J=6<o:p></o:p>

1/2/3 w/ carry of 1 … J=7<o:p></o:p>

1/2/3 w/ carry of 2 … J=8<o:p></o:p>

2/3/4 w/ carry of 0 … J=9<o:p></o:p>

<o:p> </o:p>

Of course, the actual ordering of the #/#/# in each of those cases is variable as well, although I’m pretty sure that’s not a significant fact.<o:p></o:p>

<o:p> </o:p>

Now, in the first option, the only possible fit for B/E/H is 0/4/5, because out of the remaining digits available (0,4,5,7,8,9), those are the only 3 that can be added while retaining a carry of 0 in the first column. Unfortunately, 0/4/5 add up to 9, and furthermore it would be impossible to keep the third column from having a carry and adding a carry that would create a carry for the first column as well; so this option is completely impossible.<o:p></o:p>

<o:p> </o:p>

For the second option, I can never get the sum to be anything but 767. In the last column, I only have two options for numbers that end with 7 from (0,4,5,6,8,9): either 0/9/8, or 8/5/4, both of which add up to 17. Thus one of these must be in the right column; but putting the remaining numbers (4/5/6 or 0/6/9, respectively) in the middle column, and counting the add, always leaves me with a total of 16 for that column. So this option seems to be right out as well.<o:p></o:p>

<o:p> </o:p>

Leading us to the third option. Unfortunately this option requires for the second column to add up to 28. This is impossible to do with single digits unless I had 9+9+9 and a carry of 1, and I can’t reuse 9 like that, so this option is also right out.<o:p></o:p>

<o:p> </o:p>

So by process of elimination it must be the fourth option, right? I feel good about this one. And it works! I recognize that I can put 5/6/8 in the right column, which gives me 19. The only option left for the middle column is 0/1/7, which, carrying the one from the 19, also adds up to 9, giving us the solution:<o:p></o:p>

<o:p> </o:p>

205<o:p></o:p>

316<o:p></o:p>

478<o:p></o:p>

-----<o:p></o:p>

999

The "divisible by 3" property was a good catch, but I didn't catch it, oh well. :)

Posted by IQpierce on 2005-11-11 11:54:41