Is 7 × 5 × 3 × 2 + 3 a composite number? Justify your answer.

Given number is: 7 × 5 × 3 × 2 + 3 We know that, Composite numbers are numbers that have more than two factors. 7 × 5 × 3 × 2 + 3 = 3 × 7 × 5 × 2 + 1 = 3 × 70 + 1 = 3 × 71 The given expression has 3 a n d 71 has its factors. Hence, it is a composite number.

' a ' and ' b ' are two positive integers such that a b × b a = 800 . Find ' a ' and ' b '.

Given, a b × b a = 800 The number 800 can be factorised as 800 = 2 × 2 × 2 × 2 × 2 × 5 × 5 = 2 5 × 5 2 Hence, a b × b a = 2 5 × 5 2 This implies that a = 2 and b = 5 or a = 5 and b = 2 .

Determine the value of d such that 15 ≡ 3 mod d .

We know that, Two integers a a n d b are congruence modulo n if they differ by an integer multiple of n . That b - a = k n for some integer k . ⇒ a ≡ b m o d n Now, 15 ≡ 3 mod d means 15 - 3 = k d , for some integer k 12 = k d I.e., d divides 12 The divisors of 12 are 1 , 2 , 3 , 4 , 6 , 12 . But d should be larger than 3 and so the possible values for d are 4 , 6 , 12 . Hence, value of d are 4 , 6 , 12

Find the least positive value of x such that 67 + x ≡ 1 mod 4

Given, 67 + x ≡ 1 mod 4 We know that, Two integers a a n d b are congruence modulo n if they differ by an integer multiple of n . That b - a = k n for some integer k . ⇒ a ≡ b m o d n , a - b is divisible by n ⇒ 67 + x - 1 = 4 n , for some integer n ⇒ 66 + x = 4 n ⇒ 66 + x is a multiple of 4 Therefore, the least positive value of x must be 2 , since 68 is the nearest multiple of 4 more than 66 .

Find the least positive value of x such that 98 ≡ x + 4 mod 5

Given, 98 ≡ x + 4 mod 5 Two integers a a n d b are congruence modulo n if they differ by an integer multiple of n . That b - a = k n for some integer k . ⇒ a ≡ b m o d n , a - b is divisible by n ⇒ 98 - x + 4 = 5 n , for some integer n ⇒ 94 - x = 5 n ⇒ 94 - x is a multiple of 5 Therefore, the least positive value of x must be 4 Since 94 - 4 = 90 is the nearest multiple of 5 less than 94 .

Solve 8 x ≡ 1 mod 11

Given, 8 x ≡ 1 mod 11 We know that, Two integers a a n d b are congruence modulo n if they differ by an integer multiple of n . That b - a = k n for some integer k . ⇒ a ≡ b m o d n 8 x ≡ 1 mod 11 can be written as 8 x - 1 = 11 k , for some integer k . ⇒ x = 11 k + 1 8 When we put k = 5 , 13 , 21 , 29 , . . . then 11 k + 1 is divisible by 8 x = 11 × 5 + 1 8 = 7 x = 11 × 13 + 1 8 = 18 x = 11 × 21 + 1 8 = 29 Therefore, the solutions are 7 , 18 , 29 , 40 , . . .