Best Answer DeGe, 26 September 2013 - 11:07 AM

Nice puzzle

348 x 28

Here's why:

Lets rewrite the puzzle as:

o1 e1 e2

__ e4 e3__

e8 o2 e6 e5

__e9 o3 e7 __

o5 o4 e10 e5

o1 must be less than 5 because e4 multiplied by oee is 3 digit number

o1 is either 1 or 3

If o1 is 1, then the multiplication of oee with e3 with end in 1xxx

**So, o1 must be 3**

Then **e4 must be 2** in order to keep oee x e4 a 3 digit number

**e9 is 6** then **e8 must be 2 and o5 must be 9**

if e8 is2, then e3 is either 6 or 8

If e3 is 6, the carry over needed from e1xe3 is either 3 or 5

in order to get e8o2 as 21 or 23

Then e1 must be 6 or 8

However, e1 must be either 2 or 4 otherwise e9 will not be even

So e3 can not be 6

**e3 must be 8**

If e3 is 8, e1 can be 2 or 4

Also e2 must be either 6 or 8

because if e2 is 2 or 4, it gives an odd carry over and e6 will then not be even

Now, if e1 is 2, carryover sent to multiplication of 3x8 will be 2 (16 + either 4 or 6 carried over from 6x8 or 8x8) and o2 will not be odd

so **e1 must be 4**

Finally, for e2 which is either 6 or 8

If it is 6, the individual multiplications of oee with e4 and e3 are ok;

However, if you add them up, it results in OEEE instead of OOEE

**With e2 as 8, all the results are ok**