# Thread: Writing a Fraction in Simplest Form

1. After a week I thought I'd give this another try. I'm using Soroban's formula to reproduce the mathematical joke,

$\displaystyle \frac{1\!\!\!\not{6}}{\not{6}4} = \frac{1}{4}$

While incorrectly solved, it still gives the correct answer. Any hints are appreciated on where I'm going wrong.

Soroban's formula:

We want digits $\displaystyle a, b, c$ so that: $\displaystyle \frac{10a + b}{10b + c} \:= \:\frac{a}{c}$

Solve for $\displaystyle c:\;\;c\:=\:\frac{10ab}{9a + b}$

Step 1

$\displaystyle \frac{10a + b}{10b + c} = \frac{a}{c} = \frac{a}{c}$

$\displaystyle = \frac{10 \cdot 1 + 6}{10 \cdot 4 + 4} = \frac{16}{44}$

Now, $\displaystyle \frac{a = 16}{c = 44}$

Step 2

$\displaystyle c = \frac{10ab}{9a + b} = \frac{10 \cdot 16 \cdot 16}{9 \cdot 16 + 16} = \frac{960}{144}$

$\displaystyle = \frac{\not2 \cdot \not2 \cdot \not2 \cdot \not2 \cdot 2 \cdot 2 \cdot \not3 \cdot 5}{\not2 \cdot \not2 \cdot \not2 \cdot \not2 \cdot \not3 \cdot 3} = \frac{20}{3}$

2. Originally Posted by Euclid Alexandria
After a week I thought I'd give this another try. I'm using Soroban's formula to reproduce the mathematical joke,

$\displaystyle \frac{1\!\!\!\not{6}}{\not{6}4} = \frac{1}{4}$

While incorrectly solved, it still gives the correct answer. Any hints are appreciated on where I'm going wrong.

Soroban's formula:

We want digits $\displaystyle a, b, c$ so that: $\displaystyle \frac{10a + b}{10b + c} \:= \:\frac{a}{c}$

Solve for $\displaystyle c:\;\;c\:=\:\frac{10ab}{9a + b}$

Step 1

$\displaystyle \frac{10a + b}{10b + c} = \frac{a}{c} = \frac{a}{c}$
Error #1
$\displaystyle = \frac{10 \cdot 1 + 6}{10 \cdot 4 + 4} = \frac{16}{44}$
you say b=6 in the numerator, but you change that to b=4 in the denominator!

Now, $\displaystyle \frac{a = 16}{c = 44}$

Step 2

$\displaystyle c = \frac{10ab}{9a + b} = \frac{10 \cdot 16 \cdot 16}{9 \cdot 16 + 16} = \frac{960}{144}$

$\displaystyle = \frac{\not2 \cdot \not2 \cdot \not2 \cdot \not2 \cdot 2 \cdot 2 \cdot \not3 \cdot 5}{\not2 \cdot \not2 \cdot \not2 \cdot \not2 \cdot \not3 \cdot 3} = \frac{20}{3}$
This is not incorrect, however, to produce the mathematical joke you can't have a b or c more than a single-digit number.

3. ## The punchline rears its worn out head

Thanks, Quick. Your tip about the single digit number for c sent me in the right direction. Actually, I see numerous errors now. I also substituted 16 for b in my step 2. And steps 1 and 2 are themselves erroneous. I repeatedly followed Soroban's steps in the order that he presented them in, rather than the order that was intended to be followed.

To solve for c we use

$\displaystyle \frac{10ab}{9a+b}$

where a and b are any single digits we want to test. We want the result of the test to be a whole number, rather than a fraction. For instance, going through various combinations, if

a = 1 and b = 2

$\displaystyle \frac{10 \cdot 1 \cdot 2}{9 \cdot 1 + 2} = \frac{20}{11}$

or if a = 1 and b = 3

$\displaystyle \frac{10 \cdot 1 \cdot 3}{9 \cdot 1 + 3} = \frac{30}{12} = \frac{\not2 \cdot \not3 \cdot 5}{\not2 \cdot 2 \cdot \not3} = \frac{5}{2}$

or if a = 1 and b = 4

$\displaystyle \frac{10 \cdot 1 \cdot 4}{9 \cdot 1 + 4} = \frac{40}{13}$

and so on, then our formula fails the test. However, if a = 1 and b = 6, then

$\displaystyle c = \frac{10 \cdot 1 \cdot 6}{9 \cdot 1 + 6} = \frac{60}{15}$

$\displaystyle = \frac{2 \cdot 2 \cdot \not3 \cdot \not5}{\not3 \cdot \not5} = 4$

and our formula passes the test, because we have a single digit for c.

We then apply our solution for c to the formula

$\displaystyle \frac{10a+b}{10b+c} = \frac{a}{c}$

This formula results in the butt of the joke, because

$\displaystyle \frac{10 \cdot 1 + 6}{10 \cdot 6 + 4} = \frac{16}{64} = \frac{\not2 \cdot \not2 \cdot \not2 \cdot \not2}{\not2 \cdot \not2 \cdot \not2 \cdot \not2 \cdot 2 \cdot 2} = \frac{1}{4}$

Which means that

$\displaystyle \frac{1\not6}{\not64} = \frac{1}{4}$

might be an incorrect method of simplifying the fraction, but it certainly does not give the wrong answer.

Another fraction in Soroban's list,

$\displaystyle \frac{1\!\!\!\not{9}}{\not{9}5} = \frac{1}{5}$

might show an incorrect way of simplifying, but if we solve for c

$\displaystyle \frac{10 \cdot 1 \cdot 9}{9 \cdot 1 + 9} = \frac{90}{18}$

$\displaystyle = \frac{\not2 \cdot 5 \cdot \not3 \cdot \not3}{\not2 \cdot \not3 \cdot \not3} = 5$

and apply the formula

$\displaystyle \frac{10 \cdot 1 + 9}{10 \cdot 9 + 5} = \frac{19}{95}$

$\displaystyle \frac{19}{5 \cdot 19} = \frac{1}{5}$

And so on. Well I really drilled the humor out of the joke, but at least I get the joke now.

Don't all clap at once.

4. good job

(you did do some errors in the begining of your post )

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