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.

1. T(n) = n(n+1)/2
If n is even,n=2k
T(2k)=k(2k+1) where kmod10=0,1,2,3,4
T(2k) mod 10 = 0,3,0,1,6
If n is odd,n=2k+1
T(2k+1)=(2k+1)(k+1) where kmod10=0,1,2,3,4
T(2k+1) mod 10 = 1,6,5,8,5

No. of times 0,1,5,6 occurs is twice that of 3,8.

2. T*= T(m)T(n)
Possible values of T*mod10={0,1,3,5,6,8}*{0,1,3,5,6,8}
=0,1,3,5,6,8,9,4.
So, T* doesn't end in either 2 or 7.

Posted by Praneeth on 2008-02-05 22:53:46