Yet Another Remainder Problem!
∣ ∣ ∣ ∣ ∣ ∣ 2 0 1 4 2 0 1 4 2 0 1 7 2 0 1 7 2 0 2 0 2 0 2 0 2 0 1 5 2 0 1 5 2 0 1 8 2 0 1 8 2 0 2 1 2 0 2 1 2 0 1 6 2 0 1 6 2 0 1 9 2 0 1 9 2 0 2 2 2 0 2 2 ∣ ∣ ∣ ∣ ∣ ∣ Find the remainder when the determinant above is divided by 5.
The answer is 4.
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1 solution
Absolutely correct Approach
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You can use \ begin{vmatrix} \ end{vmatrix} for determinant.
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Thankyou sir. I will remember it.
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@Md Zuhair – I learnt it from here
Select the symbols and the codes will be shown below.
∣ ∣ ∣ ∣ ∣ ∣ 2 0 1 4 2 0 1 4 2 0 1 7 2 0 1 7 2 0 2 0 2 0 2 0 2 0 1 5 2 0 1 5 2 0 1 8 2 0 1 8 2 0 2 1 2 0 2 1 2 0 1 6 2 0 1 6 2 0 1 9 2 0 1 9 2 0 2 2 2 0 2 2 ∣ ∣ ∣ ∣ ∣ ∣ ≡ ∣ ∣ ∣ ∣ ∣ ∣ ( 2 0 1 5 − 1 ) 2 0 1 4 ( 2 0 1 5 + 2 ) 2 0 1 7 2 0 2 0 2 0 2 0 2 0 1 5 2 0 1 5 ( 2 0 2 0 − 2 ) 2 0 1 8 ( 2 0 2 0 + 1 ) 2 0 2 1 ( 2 0 1 5 + 1 ) 2 0 1 6 ( 2 0 2 0 − 1 ) 2 0 1 9 ( 2 0 2 0 + 2 ) 2 0 2 2 ∣ ∣ ∣ ∣ ∣ ∣ (mod 5) ≡ ∣ ∣ ∣ ∣ ∣ ∣ ( − 1 ) 2 0 1 4 2 2 0 1 7 0 0 ( − 2 ) 2 0 1 8 1 2 0 2 1 1 2 0 1 6 ( − 1 ) 2 0 1 9 2 2 0 2 2 ∣ ∣ ∣ ∣ ∣ ∣ (mod 5) ≡ ∣ ∣ ∣ ∣ ∣ ∣ 1 2 2 0 1 7 0 0 2 2 0 1 8 1 1 − 1 2 2 0 2 2 ∣ ∣ ∣ ∣ ∣ ∣ (mod 5) ≡ 2 2 0 4 0 + 2 2 0 1 7 + 1 (mod 5) Since g cd ( 2 , 5 ) = 1 , Euler’s theorem applies. ≡ 2 2 0 4 0 mod ϕ ( 5 ) + 2 2 0 1 7 mod ϕ ( 5 ) + 1 (mod 5) Euler’s totient function ϕ ( 5 ) = 4 ≡ 2 2 0 4 0 mod 4 + 2 2 0 1 7 mod 4 + 1 (mod 5) ≡ 2 0 + 2 1 + 1 (mod 5) ≡ 1 + 2 + 1 (mod 5) ≡ 4 (mod 5)