fill-with-different-natural-numbers-1-19-1-1-1-1-

Question Number 194144 by mr W last updated on 28/Jun/23

f i l l w i t h d i f f e r e n t n a t u r a l n u m b e r s :

\frac{1}{19} = \frac{1}{()} + \frac{1}{()} + \frac{1}{()} + \frac{1}{()}

Answered by York12 last updated on 28/Jun/23

e g y p t i a n f r a c t i o n s

\frac{1}{k} = \frac{1}{k + 1} + \frac{1}{k (k + 1)}

Answered by BaliramKumar last updated on 28/Jun/23

\frac{1}{19} = \frac{1}{(20)} + \frac{1}{(381)} + \frac{1}{(144781)} + \frac{1}{(20961393180)}

Commented by Frix last updated on 28/Jun/23

\frac{1}{21} + \frac{1}{381} + \frac{1}{420} + \frac{1}{144780}

Commented by BaliramKumar last updated on 28/Jun/23

Also

\frac{1}{19} = \frac{1}{19 \times 20} + \frac{1}{20 \times 21} + \frac{1}{21 \times 22} + \frac{1}{22}

\frac{1}{19} = \frac{1}{(22)} + \frac{1}{(380)} + \frac{1}{(420)} + \frac{1}{(462)}

\begin{matrix} \frac{1}{n} = \frac{1}{n (n + 1)} + \frac{1}{(n + 1) (n + 2)} + \frac{1}{(n + 2) (n + 3)} + \frac{1}{(n + 3)} \dots \dots \end{matrix}

Answered by witcher3 last updated on 28/Jun/23

\frac{1}{19} = \frac{1}{19} . 1

1 = \frac{10}{10} = \frac{5 + 2 + 2 + 1}{10} = \frac{1}{2} + \frac{1}{5} + \frac{1}{5} + \frac{1}{10}

Commented by BaliramKumar last updated on 28/Jun/23

Nice approach but \frac{1}{5} & \frac{1}{5} same number

\frac{1}{19} = \frac{1}{19} \times \frac{20}{20} = \frac{1}{19} (\frac{10 + 5 + 4 + 1}{20})

\frac{1}{19} = \frac{1}{19} (\frac{1}{2} + \frac{1}{4} + \frac{1}{5} + \frac{1}{20})

\frac{1}{19} = \frac{1}{2 \times 19} + \frac{1}{4 \times 19} + \frac{1}{5 \times 19} + \frac{1}{20 \times 19}

\frac{1}{19} = \frac{1}{38} + \frac{1}{76} + \frac{1}{95} + \frac{1}{380}

Commented by York12 last updated on 28/Jun/23

t h e r e a r e a l o t a c t u a l l y a s I h a v e m e n t i o n e d

a b o v e i t i s a n e x t e n s i o n t o

t h e e g y p t i a n f r a c t i o n s w h i c h a r e s t i l l

a m y s t r e y t i l l t o d a y

Commented by MM42 last updated on 28/Jun/23

i f d_{i} ∣ m; 1 ⩽ i ⩽ k & Σ d_{i} = n

\Rightarrow \frac{Σ d_{i}}{m} = \frac{1}{m_{1}} + \frac{1}{m_{2}} + \dots + \frac{1}{m_{k}}; m_{i} = \frac{m}{d_{i}}

f o r e x a m p l e :

\frac{1}{24} = \frac{12 + 8 + 3 + 1}{24} = \frac{1}{12} + \frac{1}{3} + \frac{1}{8} + \frac{1}{24}

o r \frac{1}{24} = \frac{8 + 4 + 6 + 3 + 2 + 1}{24} = \frac{1}{3} + \frac{1}{6} + \frac{1}{4} + \frac{1}{8} + \frac{1}{24}

Answered by Spillover last updated on 01/Jul/23

B y u s i n g E g p t i a n f r a c t i o n . w e c a n e x p r e s s

\frac{1}{19} a s s u m o f u n i t f r a c t i o n

\frac{1}{19} = \frac{1}{(3)} + \frac{1}{(37)} + \frac{1}{(228)} + \frac{1}{(342)}

Answered by Spillover last updated on 01/Jul/23