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

  • Thread starter Thread starter anemone
  • Start date Start date
AI Thread Summary
The discussion focuses on finding the first three figures after the decimal point in the expression of the fraction formed by two decimal numbers, specifically $$\frac{0.12345678910\cdots495051}{0.515049\cdots987654321}$$. Participants attempted various approaches, including creating tables of smaller fractions to identify patterns, but struggled to find a clear solution. One user noted that their calculations stabilized around 0.2396, suggesting that further approximations would not significantly alter the first three decimal places. Additionally, a calculus-based argument was presented to support the stability of the quotient's decimal representation. Overall, the thread highlights the complexity of the problem and the collaborative effort to derive a solution.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
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.
cleardot.gif
 
Mathematics news on Phys.org
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.
 
Last edited:
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!
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top