What is the remainder when 4100 is divided by 10?
41 = 4, its last (only) digit is 4.
42 = 16, its last digit is 6.
43 = 64, its last digit is 4.
since 4×4 = 16 and 4×6 = 24, the last digits go
4,6,4,6,4,6,...
The odd powers of 4 end in 4 and the even numbered powers of 4
end in 6
The 100th power or 4 will be an even numbered term of that
sequence so it will end in 6.
When a positive integer is divided by 10 the remainder is
always the last digit.
Answer: 6
For how many different positive integer values of d does (dn – 12)² = 0 have integer solutions?
These 6 values of d, since (dn - 12)² is 0 when and only when dn = 12
d n dn (dn - 12) (dn - 12)²
--------------------------------
1 12 12 0 0
2 6 12 0 0
3 4 12 0 0
4 3 12 0 0
6 2 12 0 0
12 1 12 0 0
Answer: 6 values
Edwin
