1. 27/13 = 2 with a remainder of 1
27^2/13 = 56 with a remainder of 1
27^3/13 = 1514 with a remainder of 1
27^4/13 = 40,880 with a remainder of 1
...
So 27 to any integer power, divided by 13, has a remainder of 1.
I have three questions:
1. What is the remainder when 27 to the power of 1001 is divided by 13?
2. What is the remainder of when 38 to the power of 101 is divided by 13?
3. How do you show that
70 x 27 (to the power of 1001) + 31 x 38(to the power of 101)
can be divisible by 13?
Thanks
2 answers
Define x Mod(13) as the remainder of x if you divide x by 13. You can iterchange taking Mod with taking powers.
Since:
27 Mod(13) = 1 --------->
27^n Mod(13) = 1^n = 1
2)
Let's omit writing Mod(13) every time, equality simply means both sides are the same Mod (13).
We can use that 38 = 12 = -1
So 38^101 = (-1)^101 = -1 = 12
70 x 27^(1001) + 31 x 38^(101) =
70 - 31 = 39 = 0
Since:
27 Mod(13) = 1 --------->
27^n Mod(13) = 1^n = 1
2)
Let's omit writing Mod(13) every time, equality simply means both sides are the same Mod (13).
We can use that 38 = 12 = -1
So 38^101 = (-1)^101 = -1 = 12
70 x 27^(1001) + 31 x 38^(101) =
70 - 31 = 39 = 0
13 answers