sir pls answer

Given A = 0.a1a2a3a1a2a3.......

and  B = 0.b1b2b1b2.......

To Prove: 10989 × (a + b) is an integer.

Proof: First represent A and B in the form of rational number.

We have, A = 0.a1a2a3 a1a2a3.......(1)

Here, we have three repeating digits after the decimal point. So, we multiply both sides of (1) by 1000 to get

1000 A = a1a2a3·a1a2a3  a1a2a3....... (2)

Now, (2)–(1), will give

999A = a1a2a3

Again, we have, B =0.b1b2b1b2........  (3)

Here, we have two repeating digits after the decimal point. So, we multiply both sides of (3) by 100 to get

100 B = b1b2·b1b2  b1b2 ........  (4)

Now, (4)–(3), will give,

99B = b1b2

It is given that a1,a2,a3,b1 and b2 are integers between 1 to 9.

∴ From (5), it is clear that, 10989 (A + B) is an integer.

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