







More on the last 2 digits of a^k = N 
20200807 

From Jay: I've seen solution methods for finding the last 2 digits
of a^k = N, where a and k are known. My question
is:
What if the number N is known and the last 2
digits are, say, ...62. Is there a way to find, for
any given a, the value of k which will result in a
number where the last 2 digits are ...62? and if
that is possible, then do the same for
a number with the last n digits (say 7 or 53 or
something)? Answered by Harley Weston. 





The last two digits of a phone number 
20160403 

From Joshua: I want to find the last two digits of a phone number whose first eight numbers of a ten digit phone numbers are known. Answered by Penny Nom. 





The last two digits 
20060722 

From Sam: You have answered this question already but you only answered to one digit.
What are the last TWO digits of:
3 to the power of 1994
7 to the power of 1994
3 to the power of 1994 + 7 to the power of 1994
7 to the power of 1994  3 to the power of 1994
Answered by Paul Betts and Penny Nom. 





A palindrome question 
20040928 

From Lilly: 12, 21, 13, and 31 are the only doubledigit numbers whose squares are the palindromes of the squares of the palindromes of doubledigit numbers. I was wondering why it works for these numbers, and if only these numbers work this way. Answered by Chris Fisher and Penny Nom. 

