Assumption 1: The following logic assumes that O is an odd digit and E is an even digit, and they always represent the same digit wherever they appear in a calculation.
OEO×OE can be broken down into two multiplications and a sum.
OE=OZ+E so OEO(OZ+E)=OEO(O0+E).
Rules:
odd×odd=odd; odd×even=even×odd=even; even×even=even;
odd+odd=even; odd+even=even+odd=odd; even+even=even.
OEO×O=O×O+10(O×E)+100(O×O).
O×O=xO, where x could be even, odd or zero.
If O=1 or 3, x=0 because O×O=1 or 9.
If O=5, x=2, because O×O=25.
If O=7, x=4.
If O=9, x=8.
Therefore, O×O=O or EO, and 100(O×O)=OZZ or EOZZ.
O×E={ (1×2=2) (1×4=4) (1×6=6) (1×8=8) (3×2=6) (3×4=12) (3×6=18) (3×8=24)
(5×2=10) (5×4=20) (5×6=30) (5×8=40) (7×2=14) (7×4=28) (7×6=42) (7×8=56)
(9×2=18) (9×4=36) (9×6=54) (9×8=72) }
Therefore O×E=E⇒O=1; O×E=OZ⇒O=5. In no cases is E reproduced in any product. So O=1 or 5. E can be any even digit. The carryover can only be 1, 2, 3 or 4. Since O can only be 1 or 5, all other values can be eliminated. So we can write 1E1×1E or 5E5×5E as the arithmetic expression.
E can only be 2, 4, 6, 8 so there are 8 possible expressions:
121×12=1452, 141×14=1974, 161×16=2576, 181×18=3258, 525×52=27300, 545×54=29430, 565×56=31640, 585×58=33930.
None of these 5-digit solutions produce OOZOE (1101E or 5505E), so the original assumptions are incorrect, and we now have to assume that O is any odd number and E any even number, Z continuing to represent 0, of course. The important conclusion is that the product must have zero hundreds.
Having eliminated assumption 1, we can now turn to assumption 2: O represents any odd digit, E any even digit apart from zero, Z=0.
There are many solutions in this case. We can work through all 20 combinations of 2-digit OEs:
12, 14, 16, 18, 32, 34, 36, 38, 52, 54, 56, 58, 72, 74, 76, 78, 92, 94, 96, 98. These are all possible multipliers.
The 5-digit product limits it to between 11012 and 99098 (pattern has to be OOZOE). Also the multiplicand must be only 3 digits long: between 121 and 989.
If the multiplier is 12 then the multiplicand is between 917 and 989, because 12×917=11004 and 12×989=11868. The product must conform to OOZOE between 11004 and 11868.
For other multipliers the range of values for the multiplicand and product will shift. The table below is based on these constraints.
Here are at least some solutions:
OEO (multiplicand) OE (multiplier) OOZOE (product)
921 12 11052
923 12 11076
787 14 11018
941 16 15056
723 18 13014
343 38 13034
923 38 35074
981 52 51012
947 56 53032
983 58 57014
181 72 13032
987 74 73038
961 76 73036
987 76 75012
381 92 35052
181 94 17014
761 96 73056
521 98 51058
541 98 53018
The method is best illustrated by example.
- Pick a potential OE from the list of 20. In this example I've picked 38.
- Multiply by 989 (the highest possible OEO)=37582.
- This doesn't conform to the pattern OOZOE, so find the nearest lower OOZOE candidate: 5 must be reduced to 0, OE=98 as the highest combination, so we get 37098.
- Integer part of 37098/38=976.
- This doesn't conform to the pattern OEO, so find the nearest lower candidate: 7 must be reduced to 6 and 6 changed to 9=969.
- 969×38=36822. Nearest OOZOE is 35098 (see step 3).
- Integer 35098/38=923.
- This conforms to OEO, hence 923×38=35074 is a solution (see table).
- Subtract 76 (that is, 2×38): 34998. This is the next product to work with.
- Nearest OOZOE to 34998=33098 (see step 3).
- 33098/38=871→789 (see steps 4 and 5).
- 789×38=29982→19098; integer 19098/38=502→389; 389×38=14782→13098; integer 13098/38=344→343; 343×38=13034. Hence another solution.
Searching stops when the 5-digit product reaches 11012 or lower. Continuing with this method produced no more solutions using 38. Work through the list to find more solutions using different OEs.
