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For each of the numbering systems below, can you determine the rule for representing numbers in that system?

Decimal Number System 1 Number System 2 Number System 3
25 9 231 200
392 $, 6040 2003
5,236 W+ 83440 1011002
146,004 00t 1380300 300000012
25,897,872 >3V1 C6345030 123014
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  • $\begingroup$ That's a lot of text... $\endgroup$
    – warspyking
    Commented Oct 23, 2015 at 16:08

2 Answers 2

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Number system 3:

This system is based on prime factors. The rightmost digit is the number of $2$s in the factorization, the next is the number of $3$s, and so on.

$$\small\begin{align}25_{10} &=& 5^2 \centerdot 3^0 \centerdot 2^0 &=& 200_{pf} &\;\;\;\text{(look at the exponents)}\\392_{10} &=& 7^2 \centerdot 5^0 \centerdot 3^0 \centerdot 2^3 &=& 2003_{pf}\\5236_{10} &=& 17^1 \centerdot 13^0 \centerdot 11^1 \centerdot 7^1 \centerdot 5^0 \centerdot 3^0 \centerdot 2^2 &=& 1011002_{pf}\\146004_{10} &=& 23^3 \centerdot 19^0 \centerdot 17^0 \centerdot 13^0 \centerdot 11^0 \centerdot 7^0 \centerdot 5^0 \centerdot 3^1 \centerdot 2^2 &=& 300000013_{pf}\\25,897,872_{10} &=& 13^1 \centerdot 11^2 \centerdot 7^3 \centerdot 5^0 \centerdot 3^1 \centerdot 2^4 &=& 123014_{pf}\\\end{align}$$

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    $\begingroup$ If we used this number system all the time, it would totally destroy modern cryptography, since it's based on the difficulty of factoring large numbers. If the number is represented as the product of its fractions, factorization becomes trivial, and the cryptography is easily broken. $\endgroup$
    – GentlePurpleRain
    Commented Oct 23, 2015 at 21:19
  • $\begingroup$ can you prove that? it seems to me you're saying that if I use this number system for just a moment I can break cryptography. $\endgroup$
    – Octopus
    Commented Oct 23, 2015 at 21:50
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    $\begingroup$ The problem is with the 'if'. Try counting in that system. Try sorting lists of numbers. Simple addition would be impossible. $\endgroup$
    – chasly - supports Monica
    Commented Oct 23, 2015 at 22:35
  • $\begingroup$ @Octopus I meant if everyone used it all the time. Just using it momentarily obviously does nothing. $\endgroup$
    – GentlePurpleRain
    Commented Oct 24, 2015 at 4:26
  • $\begingroup$ @chasly Absolutely. I'm not saying it would be practical or feasible; it was just an interesting observation. $\endgroup$
    – GentlePurpleRain
    Commented Oct 24, 2015 at 4:27
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Partial answer ...

Number System 1:

This is basically a base 95 number system where the value of each digit is the corresponding ASCII value minus 32.
$$\small\begin{array}{rrrrr}\\ &9 \rightarrow &&&&(57-32) \cdot 95^0=&25\\&\$, \rightarrow &&&(36-32) \cdot 95^1 + &(44-32) \cdot 95^0=&392\\&W+ \rightarrow &&&(87-32) \cdot 95^1 + &(43-32) \cdot 95^0=&523600\\&t \rightarrow &&(48-32) \cdot 95^2 + &(48-32) \cdot 95^1 + &(116-32) \cdot 95^0=&146004\\&>3V1 \rightarrow &(62-32) \cdot 95^3 + &(51-32) \cdot 95^2 + &(86-32) \cdot 95^1 + &(49-32) \cdot 95^0=&25897872\\\end{array}$$

Number System 2:

In this system the digits have the following values from least to most significant:

$1^0 \space 2^1 \space 3^2 \space 4^3 \space ...$

Interpreting $C$ as $12$ similar to hexadecimal numbers this gives:

$$\small\begin{array}{rrrrrrrrrr}\\&&&&&&2 \cdot 3^2 + &3 \cdot 2^1 + &1 \cdot 1^0 = &25\\&&&&&6 \cdot 4^3 + &0 \cdot 3^2 + &4 \cdot 2^1 + &0 \cdot 1^0 = &392\\&&&&8 \cdot 5^4 + &3 \cdot 4^3 + &4 \cdot 3^2 + &4 \cdot 2^1 + &0 \cdot 1^0 = &5236\\&&1 \cdot 7^6 + &3 \cdot 6^5 + &8 \cdot 5^4 + &0 \cdot 4^3 + &3 \cdot 3^2 + &0 \cdot 2^1 + &0 \cdot 1^0 = &146004\\&12 \cdot 8^7 + &6 \cdot 7^6 + &3 \cdot 6^5 + &4 \cdot 5^4 + &5 \cdot 4^3 + &0 \cdot 3^2 + &3 \cdot 2^1 + &0 \cdot 1^0 = &25897872\end{array}$$

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