How will you show that ( 17 × 11 × 2 ) + ( 17 × 11 × 5 ) is a composite number? Explain.

Composite numbers have factors other than 1 and itself. It can be observed that 17 × 11 × 2 + 17 × 11 × 5 = 17 11 × 2 + 11 × 5 = 17 × 11 × 2 + 5 = 17 × 11 × 7 The given expression has 17 , 11 and 7 as its factors. ∴ It is a composite number. Hence, 17 × 11 × 2 + 17 × 11 × 5 is a composite number because its has factors other than 1 and itself.

Solve 5 x ≡ 4 ( mod 6 ) .

We know that, Two integers a and b are congruence modulo n , if they differ by an integer multiple of n . This can also be written as a ≡ b ( mod n ) . Given, 5 x ≡ 4 ( mod 6 ) 5 x - 4 = 6 n , for some integer n 5 x = 6 n + 4 x = 6 n + 4 5 When we put n = 1 , 6 , 11 , … . then 6 n + 4 is divisible by 5 . x = 6 1 + 4 5 = 2 x = 6 6 + 4 5 = 8 Therefore, the solutions are 2 , 8 , 14 , …

Solve 3 x - 2 ≡ 0 ( mod 11 ) .

We know that, Two integers a and b are congruence modulo n , if they differ by an integer multiple of n . This can also be written as a ≡ b ( mod n ) . Given, 3 x - 2 ≡ 0 ( mod 11 ) Then, 3 x - 2 ≡ 0 ( mod 11 ) ⇒ 3 x - 2 = 11 n , for some integer n ⇒ 3 x = 11 n + 2 ⇒ x = 11 n + 2 3 , where n = 2 , 5 , 8 , … ⇒ x = 11 × 2 + 2 3 = 8 ∴ x = 11 × 5 + 2 3 = 55 + 2 3 = 57 3 = 19 Also, x = 11 × 8 + 2 3 = 88 + 2 3 = 90 3 = 30 ∴ x = 8 , 19 , 30 , …