find the remainder of 3^250 is divided by 7
Answers
3^250, when divided by 7, gives a remainder of 4.
Given: 3^250 is divided by 7
To find: The remainder when 3^250 is divided by 7
Solution:
We solve the question using the properties of congruence which states;
a ≡ b (mod n)
which means that we get a remainder of 'b' when a number 'a' is divided by 'n'.
For the pattern of 3, we can say that
3^1 ≡ 3 ( mod 7 )
3^2 ≡ 2 ( mod 7 )
3^3 ≡ 6 ( mod 7 )
3^3 ≡ 6 ( mod 7 )
3^4 ≡ 4 ( mod 7 )
3^5 ≡ 5 ( mod 7 )
3^6 ≡ 1 ( mod 7 )
This assumes a cycle for the next consecutive values like,
3^7 ≡ 3 ( mod 7 )
3^8 ≡ 2 ( mod 7 )
3^9 ≡ 6 ( mod 7 )
3^10 ≡ 4 ( mod 7 )
3^5 ≡ 5 ( mod 7 )
3^12 ≡ 1 ( mod 7 ) and so on.
So according to the given question, we have 3^250,
∴ we can write,
3^250 = 3^(6×41 + 4)
= (3^6)^41 × 3^4
= (1)^41 × 3^4 [ as 3^6 ≡ 1 ( mod 7 ) ]
= 1 × 4 [ as 3^4 ≡ 4 ( mod 7 ) ]
= 4
Hence, 3^250 when divided by 7 gives a remainder of 4.
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Answer: 3^250, when divided by 7, gives a remainder of 4.
Step-by-step explanation:
Given: 3^250 is divided by 7
To find: The remainder when 3^250 is divided by 7
Using the properties of congruence which states a ≡ b (mod n)
here remainder of 'b' when a number 'a' is divided by 'n'.
to find the pattern of 3, we can say that
3^1 ≡ 3 ( mod 7 )
3^2 ≡ 2 ( mod 7 )
3^3 ≡ 6 ( mod 7 )
3^3 ≡ 6 ( mod 7 )
3^4 ≡ 4 ( mod 7 )
3^5 ≡ 5 ( mod 7 )
3^6 ≡ 1 ( mod 7 )
Assuming a cycle for the next consecutive values as,
3^7 ≡ 3 ( mod 7 )
3^8 ≡ 2 ( mod 7 )
3^9 ≡ 6 ( mod 7 )
3^10 ≡ 4 ( mod 7 )
3^5 ≡ 5 ( mod 7 )
3^12 ≡ 1 ( mod 7 ) and so on.
Thus 3^250,
hence, 3^250 = 3^(6×41 + 4)
= (3^6)^41 × 3^
= (1)^41 × 3^4 [ as 3^6 ≡ 1 ( mod 7 ) ]
= 1 × 4 [ as 3^4 ≡ 4 ( mod 7 ) ]
= 4
#SPJ2