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When 0.123...495051/0.515049...321

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,689
Hello MHB,

Recently I've come across a problem that jumped off the page at me (Find the first 3 figures after the decimal point in the decimal expression of the number \(\displaystyle \frac{0.12345678910\cdots495051}{0.515049\cdots987654321}\)).

I tried to approach it by making a table where I started to divide some smaller numbers but stick to the same pattern that is required by the aforementioned problem, i.e.

\(\displaystyle \frac{0.12}{0.21}=0.571...\)

\(\displaystyle \frac{0.123}{0.321}=0.383...\)

\(\displaystyle \frac{0.1234}{0.4321}=0.285...\)

\(\displaystyle \frac{0.12345}{0.54321}=0.227...\)

\(\displaystyle \frac{0.123456}{0.654321}=0.188...\)

\(\displaystyle \frac{0.1234567}{0.7654321}=0.161...\)

\(\displaystyle \frac{0.12345678}{0.87654321}=0.140...\)

\(\displaystyle \frac{0.123456789}{0.987654321}=0.124...\)

\(\displaystyle \frac{0.12345678910}{0.10987654321}=0.123...\)

\(\displaystyle \frac{0.1234567891011}{0.1110987654321}=0.111...\)

\(\displaystyle \frac{0.123456789101112}{0.121110987654321}=0.019...\)

\(\displaystyle \frac{0.12345678910111213}{0.13121110987654321}=0.940...\)

\(\displaystyle \frac{0.1234567891011121314}{0.1413121110987654321}=0.873...\)

\(\displaystyle \frac{0.123456789101112131415}{0.151413121110987654321}=0.815...\)

\(\displaystyle \frac{0.12345678910111213141516}{0.16151413121110987654321}=0.764...\)

\(\displaystyle \frac{0.1234567891011121314151617}{0.1716151413121110987654321}=0.719...\)

\(\displaystyle \frac{0.123456789101112131415161718}{0.181716151413121110987654321}=0.679...\)

\(\displaystyle \frac{0.12345678910111213141516171819}{0.19181716151413121110987654321}=0.643...\)

\(\displaystyle \frac{0.1234567891011121314151617181920}{0.2019181716151413121110987654321}=0.611...\)

and so on and so forth

but I failed to observe any pattern that's worth mentioning to help me to crack the problem.


Could anyone help me with this particular problem? Thanks in advance.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
I also approached it by making a table, but I started by using the most significant figures in the given denominator:

\(\displaystyle \frac{0.1}{0.5} = 0.2\),

\(\displaystyle \frac{0.12}{0.51} = 0.2352\ldots\),

\(\displaystyle \frac{0.123}{0.515} = 0.2388\ldots\),

\(\displaystyle \frac{0.1234}{0.5150} = 0.2396\ldots\),

\(\displaystyle \frac{0.12345}{0.51504} = 0.2396\ldots\),

\(\displaystyle \frac{0.123456}{0.515049} = 0.2396\ldots\),

\(\displaystyle \frac{0.1234567}{0.5150494} = 0.2396\ldots\),

\(\displaystyle \frac{0.12345678}{0.51504948} = 0.2396\ldots\).

By this time, the fraction had stabilised (within the limits of my 8-digit calculator) to 0.2396987. I doubt whether further approximations would affect the first three digits after the decimal point.

Edit. You can confirm that by using a bit of calculus. If $f(x,y) = x/y$ then $\delta f \approx (1/y)\delta x - (x/y^2)\delta y$. With $x\approx 0.123$, $y\approx 0.515$ and $\delta x = \delta y = 10^{-n}$, you find that $\delta f < 2*10^{-n}$. So the change in the fraction after the eighth decimal places in numerator and denominator is never going to be sufficient to affect the first three decimal places in the quotient.
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,689
Hi Opalg,

I'm very thankful to you for showing me something that I had never thought about before and your solution and the proof work so beautifully...

Thank you so much!