First 3 decimal places of 0.123...495051 divided by 0.515049...321

[anemone]

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 $$\frac{0.12345678910\cdots495051}{0.515049\cdots987654321}$$).

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.

$$\frac{0.12}{0.21}=0.571...$$

$$\frac{0.123}{0.321}=0.383...$$

$$\frac{0.1234}{0.4321}=0.285...$$

$$\frac{0.12345}{0.54321}=0.227...$$

$$\frac{0.123456}{0.654321}=0.188...$$

$$\frac{0.1234567}{0.7654321}=0.161...$$

$$\frac{0.12345678}{0.87654321}=0.140...$$

$$\frac{0.123456789}{0.987654321}=0.124...$$

$$\frac{0.12345678910}{0.10987654321}=0.123...$$

$$\frac{0.1234567891011}{0.1110987654321}=0.111...$$

$$\frac{0.123456789101112}{0.121110987654321}=0.019...$$

$$\frac{0.12345678910111213}{0.13121110987654321}=0.940...$$

$$\frac{0.1234567891011121314}{0.1413121110987654321}=0.873...$$

$$\frac{0.123456789101112131415}{0.151413121110987654321}=0.815...$$

$$\frac{0.12345678910111213141516}{0.16151413121110987654321}=0.764...$$

$$\frac{0.1234567891011121314151617}{0.1716151413121110987654321}=0.719...$$

$$\frac{0.123456789101112131415161718}{0.181716151413121110987654321}=0.679...$$

$$\frac{0.12345678910111213141516171819}{0.19181716151413121110987654321}=0.643...$$

$$\frac{0.1234567891011121314151617181920}{0.2019181716151413121110987654321}=0.611...$$

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]

I also approached it by making a table, but I started by using the most significant figures in the given denominator:

$$\frac{0.1}{0.5} = 0.2$$,

$$\frac{0.12}{0.51} = 0.2352\ldots$$,

$$\frac{0.123}{0.515} = 0.2388\ldots$$,

$$\frac{0.1234}{0.5150} = 0.2396\ldots$$,

$$\frac{0.12345}{0.51504} = 0.2396\ldots$$,

$$\frac{0.123456}{0.515049} = 0.2396\ldots$$,

$$\frac{0.1234567}{0.5150494} = 0.2396\ldots$$,

$$\frac{0.12345678}{0.51504948} = 0.2396\ldots$$.

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.

 

[anemone]

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!