We exclude trivial cases where the product is 0, or the product contains the same factors in a different order, or a factor is added to a smaller solution. The smallest ambiguous products are:
1 64 = 64 16 4 = 64 1 95 = 95 19 5 = 95 2 65 = 130 26 5 = 130 4 98 = 392 49 8 = 392
In fact, these are often seen in recreational mathematics as fractions where illegal canceling gives the correct result:




If the equal signs also don't print, we get a new class of ambiguous products. This time we also exclude cases where leading 1's become multiple factors, like 1 18 = 18 and 1 1 8 1 = 8.
The shortest strings of digits that are ambiguous in this fashion are:
2 27 1 = 54 22 7 = 154 2 6 2 1 = 24 2 62 = 124 2 9 5 1 = 90 2 95 = 190 3 1 8 2 = 48 31 8 = 248 3 5 5 1 = 75 35 5 = 175 4 4 5 1 = 80 4 45 = 180
What are the strings of 7 digits that are ambiguous? Can a string give 3 different products? What else can you prove about ambiguous products?
1 4 48 35 = 1 448 3 5 = 14 4 8 3 5 1 4 4 8 63 = 1 448 6 3 = 14 4 8 6 3 1 48 6 35 = 1 4 8 63 5 = 14 8 6 3 5 1 5 5775 = 1 55 7 75 = 15 5 77 5 3 5 1 448 = 35 1 4 48 = 3 5 14 4 8 4 48 6 35 = 4 4 8 63 5 = 448 6 3 5 6 3 1 448 = 6 3 14 4 8 = 63 1 4 4 8 6 35 1 48 = 6 3 5 14 8 = 63 5 1 4 8 6 3 5 448 = 6 35 4 48 = 63 5 4 4 8
1 4 48 6 35 = 1 4 4 8 63 5 = 1 448 6 3 5 = 14 4 8 6 3 5 6 3 5 1 448 = 6 35 1 4 48 = 6 3 5 14 4 8 = 63 5 1 4 4 8
Claudio Baiocchi found several infinite families of arbitrary products.
He also showed that if AB × C = A × BC is an ambiguous product, so is A[B]^{k} × [B]^{l}C = A[B]^{l} × [B]^{k}C for all k and l. He sent me a paper he wrote on the subject.
He gives another interesting example, where A = 333^{...}34, C = 666^{...}67, and B=3AC, then AB × C = A × BC is ambiguous.
Emilio Schiavi found this one.
He also found a very general infinite family. let N=10^{n}, and let d be a divisor of N. Let A = N/d  1, B = N  d, and c=999...9. Then (Ac) × B = A × (cB) is an ambiguous product.
He also analyzed AB × C = A × BC, and gave a way to construct interesting examples.
Philippe Fondanaiche found these:
He notes that the first three (and Claudio Baiocchi notes the next two as well) are ambiguous in more than one way. He also provides this additional example:
Philippe Fondanaiche also sent these examples of 3ambiguous and 4ambiguous strings:
130 3 25 6 24 = 1 30 325 6 24 = 1 30 3 25 624 1 50 2 40 735 = 150 2 40 7 35 = 1 50 240 7 35 1 80 4 50 648 = 180 450 6 48 = 180 4 50 6 48 1 975 3 95 395 = 19 75 395 39 5 = 1975 39 5 3 95 4 80 6 75 972 = 4 80 675 9 72 = 480 6 75 9 72 8 33 83 3 3332 = 833 83 3 33 32 = 833 8 33 3 332 1 75 3 50 4 48 945 = 1 75 3 50 448 9 45 = 1 75 350 4 48 9 45 = 175 3 50 4 48 9 45
If you can extend any of these results, please email me. Click here to go back to Math Magic. Last updated 1/31/05.