find remainders when 2^50 and 41^65 are divided by 7. this is a mod problem but I don't know how to work it...
By Fermat's little theorem we have:
a^6 = 1
(equality means equality mod 7)
if GD(a,7) = 1.
2^48 is thus equal to 1, therefore
2^50 = 2^2 = 4
To compute 41^65, we can use that
41 = -1, and taking the 65th power then gives:
(-1)^65 = -1 = 6
To find the remainders when 2^50 and 41^65 are divided by 7, we can use modular arithmetic.
Let's start with 2^50. We know that any number raised to a power can be expressed in terms of remainders when divided by another number. In this case, we are dividing by 7.
We need to find a pattern to solve this problem. We can start by calculating some small powers of 2 modulo 7:
2^1 ≡ 2 (mod 7)
2^2 ≡ 4 (mod 7)
2^3 ≡ 1 (mod 7)
2^4 ≡ 2 (mod 7)
2^5 ≡ 4 (mod 7)
You can see that the remainders are repeating after every 3 powers of 2. This pattern will continue for all higher powers as well.
So, to find the remainder when 2^50 is divided by 7, we need to find the remainder when 50 is divided by 3. Dividing 50 by 3 gives a quotient of 16 and a remainder of 2.
Now, we can find the remainder of 2^50 as follows:
2^50 ≡ 2^(3*16 + 2) (mod 7)
≡ (2^3)^16 * 2^2 (mod 7)
≡ 1^16 * 4 (mod 7)
≡ 4 (mod 7)
Therefore, when 2^50 is divided by 7, the remainder is 4.
Now let's move on to 41^65.
Similarly, we need to find a pattern with powers of 41 modulo 7:
41^1 ≡ 6 (mod 7)
41^2 ≡ 1 (mod 7)
41^3 ≡ 6 (mod 7)
41^4 ≡ 1 (mod 7)
41^5 ≡ 6 (mod 7)
Again, we notice a pattern. The remainders repeat every 2 powers of 41.
To find the remainder when 41^65 is divided by 7, we need to find the remainder when 65 is divided by 2. Dividing 65 by 2 gives a quotient of 32 and a remainder of 1.
Now, we can find the remainder of 41^65 as follows:
41^65 ≡ 41^(2*32 + 1) (mod 7)
≡ (41^2)^32 * 41^1 (mod 7)
≡ 1^32 * 6 (mod 7)
≡ 6 (mod 7)
Therefore, when 41^65 is divided by 7, the remainder is 6.
In conclusion, the remainders when 2^50 and 41^65 are divided by 7 are 4 and 6, respectively.