FFS. It's parsing my less than sign as HTML. Try this.

It doesn't say that all of a,b,c,d are distinct (i.e. not equal to each other). If 0 is allowed then a=b=c=d is a cheat solution.

B is the tens digit of at most 9x8 so it can be at most seven.

The last digit of D x C is equal to C. So D can only be one of:
3 ? in which case C = 5
6 ? in which case C = 4 OR C = 8
7 - in which case C = 5
9 - in which case C = 5

Just as an example take the first, so assume D = 3, C = 5 and we know B is at most 7. The sum is then

3 x AB5 = BA35

For there to be any carry, A must be at least 4. 3 x B must have a units digit of 2, so B must be 4 (can't be 8 as B is at most 7). So now we have

3 x A45 = 4A35

But this has no solutions, because if the tens digit of 3xA is 4, A must be at least 14 but we know A at most 9 because it's a digit.

You can work through the other cases. I don't think this has solutions unless you allow digits to be zero.

Posted By: Old Man on January 10th 2016 at 13:41:32

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