Kilometers to Miles Numerology

[kuruman]

I recently had to convert 7,923 km to miles and used the web converter (see right). It turned out that 7,923 km = 4,923 mi ignoring the decimal change. The matching of the last three digits astonished me and I wondered if there is a systematic way to find additional conversion matches like this and if they exist. I tried a few things with little progress. What do the numbers experts say?

 

[PeroK]

Miles to kilometres, coincidentally, has approximately a Fibonacci relationship. One mile is 1.609km, while the limiting factor for a Fibonacci sequence is $\frac{1 + \sqrt 5}{2} \approx 1.618$.

1 mile is approximately 2 km
2 miles are approximately 3 km
3 miles are approximately 5km
5 miles are approximately 8 km
8 miles are approximately 13 km

etc.

 

[kuruman]

Miles to kilometres, coincidentally, has approximately a Fibonacci relationship. One mile is 1.609km, while the limiting factor for a Fibonacci sequence is $\frac{1 + \sqrt 5}{2} \approx 1.618$.

1 mile is approximately 2 km
2 miles are approximately 3 km
3 miles are approximately 5km
5 miles are approximately 8 km
8 miles are approximately 13 km

etc.

It looks like I didn't make my question clear. Here is what I meant to ask:

Find integer values X ,Y, U, W and Z such that
XUWZ km = YUWZ mi (rounded four digit distances)
given the conversion factor 1 km = 0.621371 mi.

I stumbled into one possibility. Are there more 4-digit distances and, if so, how can they be found? My gut feeling is that there are not.

More generally, if the conversion factor is a number between zero and 1, is there a way to find
integer values X ,Y, U, W and Z such that
XUWZ units 1 = YUWZ units 2 (rounded four digit distances)?

 

[jedishrfu]

I think you could work it out by defining your numbers as polynomial sums. Then your relationship becomes

$$\Sigma ( b_n * 10^n ) = constant * \Sigma( a_n * 10^n )$$

Next add the constraint that the first 3 a values are equal to the first 3 b values

The remaining work is left to the OP

 

[kuruman]

I think you could work it out by defining your numbers as polynomial sums. Then your relationship becomes

$$\Sigma ( b_n * 10^n ) = constant * \Sigma( a_n * 10^n )$$

Next add the constraint that the first 3 a values are equal to the first 3 b values

The remaining work is left to the OP

That's the first (brute force) approach I tried without much success. So I thought that there might be a clever shortcut known to experts. If there isn't, I'll try again. Thanks.