Tessellate S.T.E.M.S. (2019) - Mathematics - School - Set 1 - Objective Problem 1
⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ a n = 1 − ( 3 n ) + ( 6 n ) − ⋯ b n = − ( 1 n ) + ( 4 n ) − ( 7 n ) + ⋯ c n = ( 2 n ) − ( 5 n ) + ( 8 n ) − ⋯
Consider the sequences above for a natural number n ≥ 2 . Compute the remainder of a 2 0 1 8 2 + b 2 0 1 8 2 + c 2 0 1 8 2 − a 2 0 1 8 b 2 0 1 8 − b 2 0 1 8 c 2 0 1 8 − c 2 0 1 8 a 2 0 1 8 when divided by 7 .
This problem is a part of Tessellate S.T.E.M.S. (2018)
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2 solutions
I figured that the system would generate the same remainder when divided by 7 for each number of the same parity (same for all even numbers). So I took 2018 as just 2. This gave me a = 1, b = -2, c = 1, the required expression as 9 and hence the remainder as 2 . Please let me know the actual solution.
Well your idea is correct (namely looking for a periodicity in the remainders) but your observation is not. As you'll observe 6 is the smallest positive integer m which satisfies 7 ∣ ( 3 m − 1 ) (also known as the "order" of 3 modulo 7 ), so that the remainders are actually periodic with a period of 6 and as 2 0 1 8 leaves a remainder of 2 modulo 6 (which is where you have gotten lucky to get the correct answer nonetheless), it suffices to find the remainder of the second term modulo 7 .
Let ω denote a non-real cube root of unity. Substitute − 1 , − ω , − ω 2 in the binomial expansion of ( 1 + x ) n (for a positive integer n ) to obtain a n , b n , c n and thus show that a n 2 + b n 2 + c n 2 − a n b n − b n c n − c n a n = 3 n and observe that 3 6 ≡ 1 ( m o d 7 ) , 2 0 1 8 ≡ 2 ( m o d 6 ) .