The Tower Of Fractions

Algebra Level 5

$\large \frac { \frac { \frac { \frac { \frac { \frac { \frac { \frac { \frac { 9 }{ 8 } }{ 7 } }{ 6 } }{ 5 } }{ 4 } }{ 3 } }{ 2 } }{ 1 } }{ \frac { 2 }{ \frac { 3 }{ \frac { 4 }{ \frac { 5 }{ \frac { 6 }{ \frac { 7 }{ \frac { 8 }{ 9 } } } } } } } } =\dfrac { A }{ B }$

Given that $A$ and $B$ are positive and coprime integers.

Find $A+B$ .

The answer is 16393.

1 solution

Lee Care Gene
Feb 16, 2016

The convention is that the length of the fraction indicates the order in which to evaluate the fraction.

$\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{9}{8}}{7}}{6}}{5}}{4}}{3}}{2}}{1}}{\frac{2}{\frac{3}{\frac{4}{\frac{5}{\frac{6}{\frac{7}{\frac{8}{9}}}}}}}}$

$\mathrm{Divide\:fractions}:\quad \frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a\cdot \:d}{b\cdot \:c}$

$=\frac { \frac { \frac { \frac { \frac { \frac { \frac { \frac { 9 }{ 56 } }{ 6 } }{ 5 } }{ 4 } }{ 3 } }{ 2 } }{ 1 } }{ \frac { 2 }{ \frac { 3 }{ \frac { 4 }{ \frac { 5 }{ \frac { 6 }{ \frac { 63 }{ 8 } } } } } } }$

$=\frac { \frac { \frac { \frac { \frac { \frac { \frac { 9 }{ 336 } }{ 5 } }{ 4 } }{ 3 } }{ 2 } }{ 1 } }{ \frac { 2 }{ \frac { 3 }{ \frac { 4 }{ \frac { 5 }{ \frac { 48 }{ 63 } } } } } }$

$=\frac { \frac { \frac { \frac { \frac { \frac { 9 }{ 1680 } }{ 4 } }{ 3 } }{ 2 } }{ 1 } }{ \frac { 2 }{ \frac { 3 }{ \frac { 4 }{ \frac { 315 }{ 48 } } } } }$

$=\frac { \frac { \frac { \frac { \frac { 9 }{ 6720 } }{ 3 } }{ 2 } }{ 1 } }{ \frac { 2 }{ \frac { 3 }{ \frac { 192 }{ 315 } } } }$

$=\frac { \frac { \frac { \frac { 9 }{ 20160 } }{ 2 } }{ 1 } }{ \frac { 2 }{ \frac { 945 }{ 192 } } }$

$=\frac { \frac { \frac { 9 }{ 40320 } }{ 1 } }{ \frac { 384 }{ 945 } }$

$=\frac{\frac{9}{40320}}{\frac{384}{945}}$

$=\frac{9\cdot \:945}{384\cdot \:40320}$

$=\frac{8505}{15482880}$

$=\frac{9}{16384}$

Please clean up your solution. It doesn't provide a clear explanation of what is happening. It does not properly explain how to deal with the fractions.

I also disagree with the second equal sign in this equation

$\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{9}{8}}{7}}{6}}{5}}{4}}{3}}{2}}{1}=\frac{\frac{\frac{\frac{\frac{\frac{\frac{9}{8}}{7}}{6}}{5}}{4}}{3}}{2}$

$=\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{9}{8}}{7}}{6}}{5}}{4}}{3}}{2}}{\frac{2}{\frac{3}{\frac{4}{\frac{5}{\frac{6}{\frac{7}{\frac{8}{9}}}}}}}}$

Calvin Lin Staff - 5 years, 3 months ago

I have already updated the solution.

Lee Care Gene - 5 years, 3 months ago

Thanks! Looks good now. The convention is that the length of the fraction indicates the order in which to evaluate the fraction.

Calvin Lin Staff - 5 years, 3 months ago

I got the answer as 99225/16384. Sorry for not latexing, I'm tired for it right now :P

Can you tell me where I went wrong?

Mehul Arora - 5 years, 3 months ago

Hmm maybe you did the last step wrong. Can I see your steps?

Lee Care Gene - 5 years, 3 months ago

Hmm, that'd be funny, coz you understand my scribbl-ish handwriting.

I'll tell you what I did

I started from the 9 at the bottom, and worked my way up.

So like, my first step was converting it to $\dfrac {7 \times 9}{8}$ Then as I went up, my answer came out to be $\dfrac {{(3\times 5 \times 7 \times 9)}^2}{(2\times 4 \times 6 \times 8)^2}$

Mehul Arora - 5 years, 3 months ago

@Mehul Arora – Wait let me redo the question...

Lee Care Gene - 5 years, 3 months ago

@Lee Care Gene – You did the upper half of the fraction wrongly. $\frac { \frac { (This\quad one\quad never\quad changes)9 }{ (Just\quad times\quad 8\quad all\quad the\quad way\quad to\quad 1)8! } }{ \frac { 384 }{ 945 } }$ .

Lee Care Gene - 5 years, 3 months ago

@Lee Care Gene – @Mehul Arora Wait how can I respond to myself?!

Lee Care Gene - 5 years, 3 months ago

@Lee Care Gene – Lol, actually you can :P

Mehul Arora - 5 years, 3 months ago

@Lee Care Gene – Oh okayy

Thanks :)

Mehul Arora - 5 years, 3 months ago

The Tower Of Fractions

The answer is 16393.

1 solution

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