Concept originated dd.[ December 27, 1999 ]
Find more smallest Squares, Cubes and higher Powers whose digits occur a same number of times ? E.g. 100112 = 100220121 has exactly three zero's, three ones and three two's ! Note that the number of different digits in the decimal expansion may differ from their required frequency. The next tables already give most solutions for frequencies 1 to 6 for the powers 2 to 5. Though the numbers are definitely 'base-related' they surely display a notable beauty !
Some of the first questions I came up with are : Are there always solutions possible for any frequency ? What are the possible shortcuts one can make while programming this puzzle ?
Take a look at the following table and decide if you want to partake to complete it. The task looks formidable but with joined efforts I think we can make a success.
What it means is best explained with an example. Take for instance cell 3.2 of the square table P2. This means we want to find the smallest, the largest and the total of all squares whose decimal expansion contains exactly 3 different digits (from the 10 available) and these 3 digits each occur 2 times (the order is irrelevant - but leading zero's are not allowed). Clearly the first solution for P2(3.2) is 478 which gives square 228484. To indicate it's the smallest solution I'll add a superscript 'A' to the notation, a 'Z' for the largest or greatest one, and a 'T' for the total number of solutions. The final notation for the above example becomes thus P2(3.2)A = 4782 = 228484
JH = Jeff Heleen JES = Jon E. Schoenfield
The smallest solutions
The largest solutions
P2(1.1)Z = 32 = 9 P2(1. >1)Z = nihil P2(2.1)Z = 92 = 81 P2(2.2)Z = 882 = 7744 = P2(2.2)A P2(2. >2)Z = nihil ? P2(3.1)Z = 312 = 961 P2(3.2)Z = 8932 = 797449 P2(3.3)Z = 310862 = 966339396 P2(3.4)Z = nihil P2(3.5)Z = 164315632 = 269996262622969 = P2(3.5)A P2(3.6)Z = 8954391002 = 801811181808810000 P2(3.7)Z = 150085150002 = 225255522505225000000 P2(4.1)Z = 992 = 9801 P2(4.2)Z = 99332 = 98664489 P2(4.3)Z = 9997022 = 999404088804 P2(4.4)Z = 999933332 = 9998666644448889 P2(4.5)Z = 99999333332 = 99998666664444488889 P2(4.6)Z = 9999993333332 = 999998666666444444888889 P2(4.7)Z = 999999933333332 = 9999998666666644444448888889 P2(5.1)Z = 3112 = 96721 P2(5.2)Z = 993302 = 9866448900 P2(5.3)Z = 316204502 = 999852858202500 P2(5.4)Z = 99999415722 = 99998831443413831184 P2(5.5)Z = 31622131674852 = 9999592116615516661225225 (JH) P2(5.6)Z = 9999993644859392 = 999998728972281878121728711721 (JH) P2(6.1)Z = 9682 = 937024 P2(6.2)Z = 9975352 = 995076076225 P2(6.3)Z = 9999377492 = 999875501875187001 P2(6.4)Z = 9999941572002 = 999988314434138311840000 P2(6.5)Z = 9999994032782262 = 999998806556808076875565707076 (JH) P2(7.1)Z = 31422 = 9872164 P2(7.2)Z = 99943022 = 99886072467204 P2(7.3)Z = 316209872042 = 999886831755531737616 P2(8.1)Z = 99162 = 98327056 P2(8.2)Z = 999377892 = 9987561670208521 P2(8.3)Z = 9999443616812 = 999888726457622541145761 P2(9.1)Z = 303842 = 923187456 P2(9.2)Z = 9994166812 = 998833702261055761 P2(9.3)Z = 316210178081822 = 999888767225363175346145124 P2(10.1)Z = 990662 = 9814072356 P2(10.2)Z = 99943634882 = 99887301530267526144 P2(10.3)Z = 9999443871187112 = 999888777330214565264406301521
The total number of solutions
P2(1.1)T = 4 02 = 0 12 = 1 22 = 4 32 = 9
P2(4.1)T = 36 P2(4.2)T = 38 P2(4.3)T = 71 P2(4.4)T = 102 P2(4.5)T = 210 (JH) P2(5.1)T = 66 P2(5.2)T = 165 P2(5.3)T = 992 P2(5.4)T = 5527 (JH) P2(5.5)T = ? P2(6.1)T = 96 P2(6.2)T = 1020 P2(6.3)T = 20700 (JH) P2(7.1)T = 123 P2(7.2)T = 5360 P2(7.3)T = ? P2(8.1)T = 97 P2(8.2)T = 24553 P2(8.3)T = ? P2(9.1)T = 83 P2(9.2)T = 98442 (JH) P2(10.1)T = 87 P2(10.2)T = 468372 (JH)
While working on the data for this page I came up with the following infinite pattern that I like to share with you :
932 = 8649 99332 = 98664489 9993332 = 998666444889 999933332 = 9998666644448889 99999333332 = 99998666664444488889
Each square belongs to the general classification P2(4.n) with n = 1, 2, 3, 4, 5, etc.
This one was sent in by Jeff Heleen where he noticed that the root consists of only 2 digits
3333032 = 111090889809 333330032 = 1111089088998009 33333300032 = 11111088908899980009
Each square belongs to the general classification P2(4.n) with n = 3, 4, 5, etc.
332 = 1089 could belong to the pattern also if the condition of the root having 2 distinct digits is dropped. But since I haven't a solution for P2(4.2) of this kind, the pattern's smoothness is lost!
Another neat arrangement of digits is for this member of P2(4.5): 36001800362 = 1296.1296.2916.1296.1296 Periods used here only to separate out the interesting stuff.
Here is my collection of such numbers with palindromic squareroots. 02 = 0 12 = 1 22 = 4 32 = 9 42 = 16 52 = 25 62 = 36 72 = 49 82 = 64 92 = 81 332 = 1089 442 = 1936 552 = 3025 662 = 4356 882 = 7744 992 = 9801 1812 = 32761 1912 = 36481 2322 = 53824 2522 = 63504 2722 = 73984 2822 = 79524 2922 = 85264 3232 = 104329 3532 = 124609 6162 = 379456 6262 = 391876 6862 = 470596 7172 = 514089 7372 = 543169 7572 = 573049 7772 = 603729 7972 = 635209 9292 = 863041 14412 = 2076481 19912 = 3964081 25522 = 6512704 28822 = 8305924 29922 = 8952064 45542 = 20738916 75572 = 57108249 77772 = 60481729 104012 = 108180801 285822 = 816930724 324232 = 1051250929 358532 = 1285437609 pandigital 401042 = 1608330816 504052 = 2540664025 507052 = 2570997025 846482 = 7165283904 pandigital 977792 = 9560732841 pandigital
Jeff Heleen (email) from New Hampshire, USA.
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