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 2018 to 2019 (Posted on 2019-03-01)
Is there a number in the form 201820182018...2018 with 2018 repeating n times that is a multiple of 2019?

 No Solution Yet Submitted by Danish Ahmed Khan Rating: 5.0000 (1 votes)

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 Solution - other way | Comment 2 of 3 |
0.111...*9=1
0.222...*9=2

0.2525252...*99=25
0.2018201820...*9999=2018

0.2018201820...*99999999=20182018

0.2018201820...* (9999... n ....9999)=(2018... n ... 2018)

Call p=(2018... n ... 2018)

2019=673*3 (no other divisors) => p will be multiple of 2019 if it is multiple of 673 and is multiple of 3

Let see when p is multiple of number prime 673

1/673=0,001485884... as a not-regular fraction the decimal part is repeated after 672 digits or less. Concretely for number 673 the decimal part has a period of 224 digits. (Big online numbers calculator for this)

As we have seen in the beginning of the post:

0,001485884... (follows digits since 224, and repeated infinite times) can be express as:

001485884.../9999...-224 nines-...999 (where both numbers are integers).

Then I can choose number p (=2018... n ....2018) with n=56 groups (2018) and with a number of digits 56*4=224 digits.

p/673=p*0,01485884...
=p*(001485884.../9999...-56 groups-..9999)
=p*001485884...*(0,20282028.../p)
So:
p=673*001485884...*0,20282028....
This is an integer as p is an integer an, we can see it, is a multiple of 673. In addition 673 is prime so 001485884...*0,20282028.... is an integer.

Let see when p is not just multiple of 673 but of 2019 (673*3).

Here we will need to work with three full decimal periods of 1/673.

(0.001485884....001485884....001485884)...001485...=
=001485... 001485... 001485 -integer with 224*3 digits- *
* 9999... 224*3 nines ...9999

See that now: =001485... 001485... 001485 -integer with 224*3 digits- is obviously a multiple of 3.

Repeating what we did before, the new p composed of groups of number 2018 will be:
p=673*001485... 001485... 001485...*0,20282028...

The product of the last two terms will be an integer again. Then:

p will be now multiple of 2019, will have now 224*3 digits=672 digits, so that it will be composed by 168 groups of 2018.

Edited on March 4, 2019, 3:04 pm
 Posted by armando on 2019-03-04 06:30:07

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