Find all the positive integers n , such that n · 2 n + 1 is divisible by 3 .

We have to find n ∈ N such that 3 | n · 2 n + 1 . Let n · 2 n + 1 is divisible by 3 , therefore So, By definition of congruence modulo n · 2 n + 1 ≡ 0 ( mod 3 ) ⇒ n · 2 n ≡ - 1 ( mod 3 ) And, - 1 ≡ 2 ( mod 3 ) ⇒ n · 2 n ≡ 2 ( mod 3 ) … 1 ∵ a ≡ b mod n & b ≡ c mod n ⇒ a ≡ c mod n Thus, we get if 3 | n · 2 n + 1 , then n · 2 n leaves remainder 2 when divided by 3 . i.e., we have to find all n ∈ N such that n · 2 n leaves remainder 2 when divided by 3 . Now, By Euclid lemma, only n ∈ N (set of natural no.) n = 3 k or n = 3 k + 1 , or n = 3 k + 2 , where k = 0 , 1 , 2 , . . . . , Case I : If n = 3 k , then n · 2 n = 3 k · 2 3 k Hence, on dividing by 3 leaves remainder 0 . So, this case is not possible. Case II : If n = 3 k + 1 , then n · 2 n = ( 3 k + 1 ) 2 3 k + 1 then 3 k + 1 2 3 k + 1 mod 3 = 3 k · 2 3 k + 1 mod 3 + 2 3 k + 1 mod 3 ⇒ 3 k + 1 2 3 k + 1 mod 3 = 0 mod 3 + 2 3 k + 1 mod 3 ⇒ 3 k + 1 2 3 k + 1 mod 3 = 2 · 2 3 k mod 3 For k = 0 , 2 , 4 , 6 , . . . . . , on dividing 2 · 2 3 k by 3 , leaves remainder 2 since 2 · 2 3 k = 2 · 8 k and 8 ≡ - 1 mod 3 ⇒ 8 k ≡ - 1 k mod 3 ⇒ 8 k ≡ 1 mod 3 And, 2 ≡ 2 mod 3 therefore, 2 · 8 k ≡ 2 mod 3 . So, n = 3 k + 1 , where k = 0 , 2 , 4 , 6 , . . . Case 3 : If n = 3 k + 2 n · 2 n = 3 k + 2 2 3 k + 2 Therefore, 3 k + 2 2 3 k + 2 mod 3 = 3 k · 2 3 k + 2 mod 3 + 2 · 2 3 k + 2 mod 3 ⇒ 3 k + 2 2 3 k + 2 mod 3 = 0 mod 3 + 2 · 2 3 k + 2 mod 3 ⇒ 3 k + 2 2 3 k + 2 mod 3 = 2 · 2 3 k + 2 ( mod 3 ) For k = 0 , 2 , 4 , 6 , . . . . On dividing 2 · 2 3 k + 2 by 3 leaves remainder 2 since, 2 · 2 3 k + 2 = 8 · 8 k = 8 k + 1 and 8 ≡ 2 mod 3 ⇒ 8 k + 1 ≡ 2 k + 1 mod 3 ⇒ 8 k + 1 ≡ 2 mod 3 So, n = 3 k + 2 , k = 0 , 2 , 4 , 6 , . . . Hence, 3 | n · 2 n + 1 if n = 3 k + 1 or n = 3 k + 2 where k = 0 , 2 , 4 , 6 , . . .

Prove that when a b c ¯ is a multiple of 37 , then so is the number b c a ¯ .

Since a b c ¯ is a multiple of 37 , then 100 a + 10 b + c = 37 k Therefore, b c a ¯ = 100 b + 10 c + a ⇒ b c a ¯ = 1000 a + 100 b + 10 c - 999 a ⇒ b c a ¯ = 10 100 a + 10 b + c - 37 · 27 a ⇒ b c a ¯ = 10 · 37 k - 37 · 27 a ⇒ b c a ¯ = 37 10 k - 27 a Which is divisible by 37 .