The question is:Find the last two digits of 9^2013.(No calculators allowed to solve this question.)
Options: (A)01 (B)29 (C)41 (D)81 (E)89
From this point,I will show my working,
9^2=81(Last Digit 1) 9^3=1x9=9 (Now only focusing on ONLY last digit) 9^4=9(Last Digit of the previous answer)x9(The next power)=1 (Last Digit)
Noticing a pattern that as the powers increases,the pattern goes 9,1,9,1,9,1,9,1,9 so the repeated pattern is 1 and 9.So I took
2013÷2=1006R(Remainder)1,where R1 is the number 9.So we know the digit is 9.
This eliminates the option (A),(C) and (D),leaving with (B) and (E)
I attempted to find the 2md last digit number,but to no avail as the number increases,no pattern can be formed.Any ideas how?(The solution doesn't have to use concept of patterns)
Edited working error.