Find-all-positive-integers-n-for-which-there-exist-non-negative-integer-a-1-a-2-a-3-a-n-Such-that-1-2-a-1-1-2-a-2-1-2-a-3-1-2-a-n-1-3-a-1-2-3-a-

Question Number 5834 by sanusihammed last updated on 31/May/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 \dots . a_{n} . S u c h t h a t

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

P l e a s e h e l p .

Commented by Yozzii last updated on 31/May/16

n = 1 \Rightarrow \frac{1}{2^{a_{1}}} = \frac{1}{3^{a_{1}}} \Rightarrow a_{1} = 0 .

\underset{i = 1}{\sum^{n}} \frac{1}{2^{a_{i}}} = \underset{i = 1}{\sum^{n}} \frac{i}{3^{a_{o}}}

\underset{i = 1}{\sum^{n}} (\frac{1}{2^{a_{i}}} - \frac{i}{3^{a_{i}}}) = 0

\underset{i = 1}{\sum^{n}} (\frac{3^{a_{i}} - i 2^{a_{i}}}{6^{a_{i}}}) = 0

6^{\underset{l = 1}{\sum^{n}} a_{l}} (\underset{i = 1}{\sum^{n}} \frac{3^{a_{i}} - i 2^{a_{i}}}{6 a^{i}}) = 0

\underset{i = 1}{\sum^{n}} {6^{\underset{l = 1}{\sum^{n}} a_{l} - a_{i}} (3^{a_{i}} - i 2^{a_{i}})} = 0

Commented by sanusihammed last updated on 31/May/16

I a p p r e c i a t e