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#1 2016-01-19 15:44:19

blackcap
Member
Registered: 2015-11-20
Posts: 7

Mysterius semiprime fact in other bases

So you have a semiprime n
n = p*q where p<q

A curious fact about bases is that if a number x ends with a zero in base y, then x is divisible by y. Therefor, if we where to represent n in all bases from 2 to n, then n would only end with a zero in base p and q.

Bellow is n represented in all the bases from n and downwards:
10
...
p0
...
11x
...
1(q-p)0
...

The first entry is n in base n
The second entry is base q. In base q, n would end with a zero and the first digit would be digit number p
The third entry is the point where we go from 2 to 3 digits. I don't know what x is
The forth entry is base p. For some reason, if n is a semiprime then the second digit will be digit number (q-p)
Between the first and the second entry, the first digit is going to increment in a curve that is very flat at the beginning, and very steep towards the end.

Can anyone explain why the second digit of n represented in base p is equal to q-p?

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#2 2016-02-12 16:16:01

anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,049

Re: Mysterius semiprime fact in other bases

Here is a way of looking at it:


If
, with
, and
has a hundreds place equal to 1, that means that
and
. So, for some
,
, i.e.,
, or further,
, and the representation of
in base
will be exactly
, from which the representation of
will be
.

Last edited by anonimnystefy (2016-02-20 21:42:30)


“Here lies the reader who will never open this book. He is forever dead.
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#3 2016-02-20 18:48:35

Grantingriver
Member
Registered: 2016-02-01
Posts: 129

Re: Mysterius semiprime fact in other bases

The answer to this question is very obvious. First there are some mistakes ought to be mentioned. If we represent n in all bases from 2 to n then n would end also in zero in base n along with bases p and q and it is of course divisible by n. The number which represents the point were we go from two to three digits (in all bases) is ,in fact, 100. Now let's return to the problem, since p<q and n=p×q it follows that in base q (where q=10) q must be added p times to reach the number "n" so the representation of this number in base q is "p0" but if we try to do the same proccess in the case of base p we encounter a problem! The permited digits in that base does not inculde the symbol q since q>p. Fore example, in base 3 we have the symbols 0,1 and 2 only and we do not have the symbol 4. Now each place in base p consists of p units of the previous place, so before the digits of the second place reaches the symbol q (which is a digit in a higher base) that place will reasch the symbol "p-1" and the p units in that place will become "1" unit in the next place while the digit in the second place will become "0", so it is obvious that the units in the third place plus the unit in the second place must equal "q" (since n=p×q), but we know that the number of units in the third place are "p" hence the units in the second place must compensate the difference which is "q-p" since p+(q-p)=q. Therefor the digit in the second place must be "q-p".

Q.E.D

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