i PLEASE NEED HELP WITH 4010 base five minus 1234 base 5!! please and thank you I know that the numbers add to more than five so they need to be different
I'd just use the usual method taught in schools, 'borrowing', but it's a little difficult to illustrate on here.
Instead you could use 'equal addition' as follows.
4010-1234
adding 1 to both numbers
=4011-1240
adding 10 to both numbers
=4021-1300
adding 200 to both numbers
=4221-2000
=2221
It's rather less efficient though.
Hello, cathyharris4!
Have you never ever done a subtraction in other bases?
$\displaystyle 4010_5 - 1234_5$
$\displaystyle \begin{array}{ccccc} & ^{C1} & ^{C2} & ^{C3} & ^{C4} \\ & 4 & 0 & 1 & 0 \\ - & 1 & 2 & 3 & 4 \\ \hline \end{array}$
In C4, we can't subtract 4 from 0, so we "borrow" from C3.
$\displaystyle \begin{array}{ccccc} & ^{C1} & ^{C2} & ^{C3} & ^{C4} \\ & 4 & 0 & ^0\rlap{/}1\; & ^10\; \\ - & 1 & 2 & 3 & 4 \\ \hline \end{array}$
The "1" we borrowed represents "5", so we have:
$\displaystyle \begin{array}{ccccc} & ^{C1} & ^{C2} & ^{C3} & ^{C4} \\ & 4 & 0 & 0 & 5 \\ - & 1 & 2 & 3 & 4 \\ \hline \end{array}$
Subtract in C4 and we have:
$\displaystyle \begin{array}{ccccc} & ^{C1} & ^{C2} & ^{C3} & ^{C4} \\ & 4 & 0 & 0 & 5 \\ - & 1 & 2 & 3 & 4 \\ \hline &&&&1 \end{array}$
In C3, we can't subtract 3 from 0, so we "borrow" from C2.
But C2 has "0"; we can't borrow.
So we must borrow from C1.
$\displaystyle \begin{array}{ccccc} & ^{C1} & ^{C2} & ^{C3} & ^{C4} \\ & ^3\rlap{/}4\;\; & ^10\;\; & 0 & 5 \\ - & 1 & 2 & 3 & 4 \\ \hline &&&&1 \end{array} \quad\Rightarrow\quad \begin{array}{ccccc} & ^{C1} & ^{C2} & ^{C3} & ^{C4} \\ & 3 & 5 & 0 & 5 \\ - & 1 & 2 & 3 & 4 \\ \hline &&&&1 \end{array}$
Now we can borrow from C2:
$\displaystyle \begin{array}{ccccc} & ^{C1} & ^{C2} & ^{C3} & ^{C4} \\ & 3 & ^4\rlap{/}5\;\; & ^10\;\; & 5 \\ - & 1 & 2 & 3 & 4 \\ \hline &&&&1 \end{array} \quad\Rightarrow\quad \begin{array}{ccccc} & ^{C1} & ^{C2} & ^{C3} & ^{C4} \\ & 3 & 4 & 5 & 5 \\ - & 1 & 2 & 3 & 4 \\ \hline &&&&1 \end{array}$
And we can complete the subtraction:
$\displaystyle \begin{array}{ccccc} & ^{C1} & ^{C2} & ^{C3} & ^{C4} \\ & 3 & 4 & 5 & 5 \\ - & 1 & 2 & 3 & 4 \\ \hline &2&2&2&1 \end{array}$