Question

Let A=0.a1a2a3.. and B=0.b1b2b3... where a1 a2 a3 b1 b2 are integers from 1-9 not necessarily distinct prove that 10989 x A+B is an integer

Answer 1

Correct Question :- Let A = 0.a1a2a3____ and B = 0.b1b2b1b2_____ where a1 a2 a3 b1 b2 are integers from 1 - 9 not necessarily distinct . Prove that 10989(A + B) is an integer

Answer :-

→ A = 0 . a1a2a3a1a2a3_______

multiply by 1000 both sides,

→ 1000A = a1a2a3 . a1a2a3_______

subtracting both we get,

→ 1000A - A = (a1a2a3 . a1a2a3_______) - (0 . a1a2a3a1a2a3_______)

→ 999A = a1a2a3

→ A = (a1a2a3/999) --------- Eqn.(1)

and,

→ B = 0 . b1b2b1b2_________

multiply by 100 both sides,

→ 100B = b1b2 . b1b2______

subtracting both we get,

→ 100B - B = (b1b2 . b1b2______ ) - (0 . b1b2b1b2______ )

→ 99B = b1b2

→ B = (b1/b2/99) ----------- Eqn.(2)

putting values of A and B from Eqn.(1) and Eqn.(2) now,

→ 10989(A + B)

→ 10989[(a1a2a3/999) + (b1/b2/99)]

→ (10989/999) * a1a2a3 + (10989/99) * b1b2

→ 11 * a1a2a3 + 111 * b1b2

since it has been given that, a1, a2, a3, b1, b2 are integers from 1 - 9 .

therefore, we can conclude that,

→ 11 * a1a2a3 + 111 * b1b2 = Is a integer .

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