Let T be the set of triangular numbers and T* be the set of all products of any two triangular numbers. Show that:
1. Among elements of T, each of the digits 0,1,5 and 6 occur in the units place twice as frequently as each of the digits 3 and 8. (More precisely, if MBk is the set of elements of T that are less than B and end in k, then, e.g., MB1/MB8 approaches 2 as B approaches infinity.)
2. None of the elements of T* end in 2 or 7.
Each triangular number T(i) equals the previous triangular number plus i: T(i) = T(i-1)+i.
So:
i T(i)
1 1
2 3
3 6
4 10
5 15
6 21
7 28
8 36
9 45
10 55
11 66
12 78
13 91
14 105
15 120
16 136
17 153
18 171
19 190
20 210
21 231
22 253
23 276
24 300
25 325
26 351
27 378
28 406
29 435
30 465
31 496
32 528
33 561
34 595
35 630
36 666
37 703
38 741
39 780
40 820
Note that there is a cycle of 20 in the last digits. For i itself, there's a cycle of 10 in the last digits, but it is not until i=21 that T(i) also returns to a last digit of 1, as at T(1). From then on, due to the nature of addition, the length-20 cycle repeats.
In the cycle of 20, there are the following counts of last digits:
final
digit count
0 4
1 4
2 0
3 2
4 0
5 4
6 4
7 0
8 2
9 0
Therefore, over the long run, the occurrence of each of 0, 1, 5 and 6 is twice as frequent as either of 3 or 8.
When taking the products of these numbers, which never end in 2, 4, 7 or 9, if either one ends in a zero, the product will end in a zero. If either ends in a 1, the product will end in the same digit as the other triangular number and never be 2, 4, 7 or 9. The other possible pairs of ending digits result in products ending as follows:
1 3 5 6 8
3 9 5 8 4
5 5 5 0 0
6 8 0 6 8
8 4 0 8 4
None of these is a 2 or a 7.
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Posted by Charlie
on 2008-02-05 14:48:22 |
solution
