Problem of the Month (January 2005)

A computer was programmed to print out products of integers, but for some reason the spaces for multiplication didn't print. Most of the equations can be recovered uniquely: 946 = 414 can only be 9 46 = 414. This month we are interested in the other cases, the ambiguous products.

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:

16	=	1
64	=	4

19	=	1
95	=	5

26	=	2
65	=	5

49	=	4
98	=	8

What are some other small ambiguous products? What products can be ambiguous? Are there arbitrarily long strings which are ambiguous? What is the shortest product that can be written in 3 different ways? Are there products which can be written in arbitrarily large numbers of ways?

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?

ANSWERS

Here are the shortest strings which are 3-ambiguous or 4-ambiguous.


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.

1999^...9 × 5 = 1 × 999^...95 555^...565000^...04 × 5 = 555^...56 × 5000^...045 666^...674000^...02 × 4 = 666^...67 × 4000^...024

He also showed that if AB × C = A × BC is an ambiguous product, so is A[B]^k × [B]^lC = A[B]^l × [B]^kC 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.

1 × 333^...325 = 1333^...3 × 25

He also found a very general infinite family. let N=10ⁿ, 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:

1 × 666^...64 = 1666^...6 × 4 2 × 666^...65 = 2666^...6 × 5 4 × 999^...98 = 4999^...9 × 8 2 × 777^...756 = 2777^...7 × 56 3 × 999^...975 = 3999^...9 × 75 4 × 324324^...3243 = 432432^...4324 × 3 4 × 8484^...847 = 4848^...484 × 8 6 × 5454^...545 = 65454^...54 × 5 7 × 4242^...424 = 74242^...42 × 4 8 × 333^...32 = 8333^...3 × 32 8 × 648648^...6486 = 864864^...8648 × 6

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:

19 × 75 × 395 = 1975 × 3 × 95 = 562875 1 × 975 × 3 × 95 = 19 × 75 × 39 × 5 = 277875

Philippe Fondanaiche also sent these examples of 3-ambiguous and 4-ambiguous 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

Philippe Fondanaiche also investigated ambiguous products where the equal sign comes before the multiplication signs.

If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 1/31/05.