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We know that,
4 1 = 4
4 2 = 1 6
4 3 = 6 4
4 4 = 2 5 6
Observed that the last digit repeats in every cycle of 2 , that means that we must divide the exponent by 2 . If there is no remainder, the last digit is 6 . If the remainder is 1 the last digit is 4 .
Dividing 3 5 8 4 9 by 2 gives a remainder of 1 , so the last digit is 4 .
2074 = 4 mod 10
Now 2 0 7 4 2 = 6 mod 10
Again 2 0 7 4 3 = 4 mod 10
Hence for odd n last digit is 4
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Method 1:
2 0 7 4 3 5 8 4 9 ≡ ( 2 0 7 0 + 4 ) 3 5 8 4 9 ≡ 4 3 5 8 4 9 (mod 10) .
Now, we note that 4 n ≡ { 4 (mod 10) 6 (mod 10) if n is odd. if n is even. .
Therefore, 2 0 7 4 3 5 8 4 9 ≡ 4 (mod 10) .
Method 2:
We need to find 2 0 7 4 3 5 8 4 9 mod 1 0 . Since 2074 and 10 are not coprime integers, we have to consider 2 0 7 4 3 5 8 4 9 mod 2 and 2 0 7 4 3 5 8 4 9 mod 5 separately.
2 0 7 4 3 5 8 4 9 2 0 7 4 3 5 8 4 9 ⟹ 5 n + 4 ⟹ n ⟹ 2 0 7 4 3 5 8 4 9 ≡ 0 (mod 2) ≡ 2 0 7 4 3 5 8 4 9 mod ϕ ( 5 ) (mod 5) ≡ 2 0 7 4 3 5 8 4 9 mod 4 (mod 5) ≡ 2 0 7 4 1 ≡ 4 (mod 5) ≡ 0 (mod 2) ≡ 0 ≡ 5 ( 0 ) + 4 ≡ 4 (mod 10) Since g cd ( 2 0 7 4 , 5 ) = 1 , Euler’s theorem applies. Euler totient function ϕ ( 5 ) = 4 where n is an integer.