Question

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Let A = 0. a1a2a3aa1a2a3..... And B = 0. b1b2b1b2..... Where a1, a2, a3, b1, b2, are integers from. 1 to 9 not necessarily distinct. Prove that 10989* (A+B) is an integer.

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Answer 1

Answer:

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A= 0.a1a2a3a1a2a3.....

1000A= a1a2a3.a1a2a3......

Subtracting A from 1000A:-

999a= a1a2a3

B= b1b2b1b2b1b2.....

100B= b1b2.b1b2.b1b2....

Subtracting B from 100B:-

99B= b1b2

So,A= $\frac{a1a2a3}{999} and \: </em></strong></p><p><strong><em>b = \frac{b1b2}{99}$

Putting these values in the question:-

$10989( \frac{a1a2a3}{999} + \frac{b1b2}{99} )$

$= 11 \times a1a2a3 + 111 \times b1b2$

THIS IS AN INTEGER BECAUSE a1,a2,a3,b1 and b2 ARE ALL INTEGERS...

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Answer 2

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