The ten's digit of 11^n will always be the unit's digit of n
We can see the pattern as follows,
- 11^1 = 11 ......... Ten's digit = 1
11^2 = 121 ........ Ten's digit = 2
11^3 = 1331 ....... Ten's digit = 3
11^4 = 14641 ...... Ten's digit = 4
...
Or we can use binomial expansion as follows...
11^n = (10 + 1)^n = (Terms multiple of 10^2 + (n
C1)*(10^1)*(1^(n - 1)) + 1)
Hence, ten's digit of 11^n will depend upon (n
C1)*(10^1)*(1^(n - 1)) + 1 = 10(n
C1) + 1 = 10n + 1
Hence, 11^n = (Some number multiple of 100 + 10n + 1) = (100K + 10n + 1)
Hence, ten's digit of 11^n will always be n.
Hundred's digit of 5^n will be 6 only when n is an even integer greater than 2
5^3 = 125
5^4 = 625
5^5 = 3125
Now, note that last three digits 5*(number ending with 125) is 625 and last three digits of 5*(number ending with 625) is 125. Hence, for powers greater than 2, the last three digits of 5^n will be 125 for odd powers and 625 for even powers.
