Why does conversion from base $b_1$ to $b_2$ give this strange result in Wolfram Alpha?

I was trying to convert the $5$-ary number $44444444_5$ to base $25$ using Wolfram Alpha:

44444444 from base 5 to base 25


Curiously, I got the result $0000_{25}$.

Clearly, $0\times25^{3}+0\times25^{2}+0\times25^{1}+0\times25^{0}$ does not equal $44444444_5$. So why am I getting this result? Is this a bug?

• Note the buttons labeled "Show block form" and "Show digit key". Try them. Commented Jan 23, 2021 at 5:45

It's not a bug, and it's not $0000$, it's $oooo$. $o$ being the symbol $o$ (the one between $n$ and $p$ in the alphabet).

The letter $o$ in base $25$ stands for the twenty-fourth "digit" ( the "digits" in base $25$ are $1,2,3,4,5,6,7,8,9,a,b,c,d,e,f,g,h,i,j,k,l,m,n,o$), which means it equals $24$ in base $10$, which means $$oooo_{25} = (24 + 24\cdot 25 + 24\cdot 25^2 + 24\cdot 25^3)_{10} = 390624_{10}$$

which fits well with the fact that

$$44444444_5 = (4+4\cdot 5 + 4\cdot 5^2+4\cdot 5^3+4\cdot 5^4+4\cdot 5^5+4\cdot 5^6+4\cdot 5^7)_{10} = 390624_{10}$$

Another way of seeing this (without calculating the number) is

\begin{align}44444444_5 &= 4\cdot(1+5+5^2+5^3+5^4+5^5+5^6+5^7) \\& = 4\cdot((1+5)+(5^2+5^3)+(5^4+5^5)+(5^6+5^7)) \\& = 4\cdot(1+5+5^2(1+5) + 5^4(1+5)+5^6(1+5)) \\&= 4\cdot(6+6\cdot25+6\cdot25^2+6\cdot25^3) \\&= 24 + 24\cdot 25 + 24\cdot 25^2 + 24\cdot 25^3 \\&= oooo_{25}\end{align}

• The direct way from base 5 to base 25 is very neat :).
– Surb
Commented Mar 14, 2018 at 13:14