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Question Number 6133 by sanusihammed last updated on 15/Jun/16

F i n d a l l p o s i t i v e i n t e g e r s n f o r w h i c h t h e r e e x i s t n o n n e g a t i v e

i n t e g e r a_{1}, a_{2}, a_{3}, \dots . ., a_{n} . S u c h t h a t .

\frac{1}{2^{a_{1}}} + \frac{1}{2^{a_{2}}} + \dots . . + \frac{1}{2^{a_{n}}} = \frac{1}{3^{a_{1}}} + \frac{2}{3^{a_{2}}} + \dots . + \frac{n}{3^{a_{n}}} = 1

Commented by prakash jain last updated on 15/Jun/16

\frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + \dots up to \infty = 1

if all of a_{1}, a_{2}, \dots, a_{n} are non - zero then for

finite n .

so \frac{1}{2^{a_{1}}} + \frac{1}{2^{a_{2}}} + \dots + \frac{1}{2^{a_{n}}} < 1

This implies that for

\frac{1}{2^{a_{1}}} + \frac{1}{2^{a_{2}}} + \dots + \frac{1}{2^{a_{n}}} = 1

(n = 1 and a_{1} = 0)

only one solution exists .

i . e n = 1 and a_{1} = 0