# Problem of the Month (December 2000)

Consider the following identities:

43 = 26 . 15

85 . 17 = 23 . 46

27 . 43 . 95 = 610 . 81

127 . 310 = 67 . 81 . 24 . 95

12 . 813 . 914 = 1211 . 67 . 310 . 45

1513 . 311 . 48 . 161 = 214 . 912 . 57 . 106

18 . 413 . 187 . 1517 = 56 . 1011 . 914 . 123 . 216

Each contains the first few integers exactly once, and only products of the form ab are allowed. Is there such an equation using the numbers from 1 to 20? How about 1 to 22? Can this always be done with the numbers from 1 to 2n (for n≥3) ? How many solutions are there?

Brendan Owen found one for n=10:

114 . 311 . 513 . 612 . 819 . 92 . 157 = 1020 . 1817 . 416

Thanks to Philippe Fondanaiche for finding a correct equation (see above) for n=8, since mine was in error. He also found these for n=10, 11, 12, and 20:

1519 . 1011 . 167 . 186 = 517 . 2013 . 914 . 84 . 123 . 21

1422 . 316 . 1513 . 811 . 106 = 2019 . 717 .1812 . 29 . 215 . 41

1424 . 1522 . 1019 . 816 . 36 = 2023 . 518 . 1217 . 713 . 2111 . 29 . 41

416 . 185 . 93 . 2019 .1536 . 3527 . 1437 . 2231 . 337 .1340 = 16 . 324 . 128 . 2529 .1024 . 2830 . 2134 . 1138 . 2623 . 3917

He believes that it is possible to do this for every even n, and gives a technique to do so.

Tom Sundquist found an equation for n≤15 and n=22:

17 . 211 . 1214 . 1610 . 1819 = 35 . 420 . 613 . 815 . 917

13 . 62 . 917 . 1513 . 167 . 2022 = 514 .84 . 1021 . 1211 . 1819

21 . 311 . 817 . 1023 . 128 . 1522 = 414 . 524 . 613 . 919 . 167 . 2021

118 . 211 . 312 . 914 . 1613 . 2022 . 2524 = 47 . 526 . 619 . 817 . 1023 . 1521

112 . 317 . 1618 . 2021 . 2526 . 2719 = 214 . 47 . 522 . 624 . 813 . 911 . 1023 . 1528

17 . 321 . 411 . 529 . 817 . 922 . 1528 . 2026 = 214 . 612 . 1023 . 1613 . 2518 . 2719 . 3024

111 . 524 . 744 . 812 . 2031 . 2138 . 2526 . 2722 . 2839 . 3034 . 3217 = 213 . 319 . 46 . 923 . 1033 . 1441 . 1536 . 1618 . 3543 . 4029 . 4237

He conjectures that there are plenty of these for larger values of n.

John Hoffman wrote a JAVA program to count the number of solutions. Here are his results:

 n Number of Solutions 3 4 5 6 7 8 9 2 4 4 154 428 11938 173476

If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 12/14/00.