Wikipedia:Reference desk/Archives/Mathematics/2021 December 6
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December 6
[edit]Pattern in last 2 digits of polygonal numbers
[edit]Legend: X=any digit; E=any even digit; Q=any odd digit (to distinguish from 0 which O can be confused with); X=0 or 5; X=1 or 6; X=2 or 7; X=3 or 8; X=4 or 9
- Triangular numbers end in X0, X1, X3, X5, X6, or X8.
- Square numbers end in 00, E1, E4, 25, Q6, or E9
- Pentagonal numbers end in X0, X1, X2, X5, X6, or X7
- Hexagonal numbers end in X0, X1, X3, X5, X6, or X8.
- Heptagonal numbers can end in any final 2 digits!
- Octagonal numbers end in E0, E1, 33, E5, Q6, or 08.
- Enneagonal numbers end in X0, X1, X4, X5, X6, or X9.
- Decagonal numbers end in X0, X1, X2, X5, X6, or X7.
- Hendecagonal numbers end in X0, X1, X3, X5, X6, or X8.
- Dodecagonal numbers end in E0, E1, Q2, Q3, E4, E5, Q6, Q7, E8, or E9.
Is there any pattern in this sequence?? What final 2-digit endings can a triskaidecagonal number have?? (Please answer using the legend at the top of this section.) Georgia guy (talk) 14:55, 6 December 2021 (UTC)
- Considering the properties of modular arithmetic, the final 2-digit endings of the (n+100)-gonal numbers are the same as those of the n-gonal numbers; they only depend on the value of n modulo 100. So in the end it repeats, and there are at most 100 different ending sets. In fact, there are precisely 16. There are some patterns: all (10k+7)-gonal numbers have the same ending set of any final 2 digits, as do the (20k+2)-gonal numbers. The 5k-gonal numbers also share their sets, so the entries above for pentagonal and decagonal are identical. The (10k+9)-gonal numbers and the (20k+14)-gonal numbers also have their final 2-digit sets in common. I see no obvious regularity. Transforming this into your legendary code would be a tedious and unrewarding undertaking that I prefer to skip. --Lambiam 17:12, 6 December 2021 (UTC)
- Grouping the remainders by the resulting sets, I get 3 groups of 15, 3 groups of 5, 5 groups of 6 and 5 groups of 2; I assume this matches up with your results. All sets have either 22, 44, 50 or 100 elements. The fact that the modulus is composite seems to make the results rather messy. It would probably be instructive to first look at the remainder sets mod p where p is an odd prime. In that case the sets have size p or (p+1)/2, and it appears that the groups have size 1 or 2. --RDBury (talk) 18:05, 7 December 2021 (UTC)
- Our numbers match up. --Lambiam 20:39, 7 December 2021 (UTC)
- Grouping the remainders by the resulting sets, I get 3 groups of 15, 3 groups of 5, 5 groups of 6 and 5 groups of 2; I assume this matches up with your results. All sets have either 22, 44, 50 or 100 elements. The fact that the modulus is composite seems to make the results rather messy. It would probably be instructive to first look at the remainder sets mod p where p is an odd prime. In that case the sets have size p or (p+1)/2, and it appears that the groups have size 1 or 2. --RDBury (talk) 18:05, 7 December 2021 (UTC)