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what is321 base 4 + 123 base 4 equal?
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Anonymous said:what is321 base 4 + 123 base 4 equal?
321+123 = 1110
Hello, Guest!
Here it is in baby-steps . . .
\(\displaystyle \;\;\;3\;2\;1\)
\(\displaystyle \;\;\;1\;2\;3\)
\(\displaystyle \;\;----\)
In column \(\displaystyle c\), we have: \(\displaystyle \,1\,+\,3\;=\;10\;\;('4')\)
We "put down the 0, carry the 1".
\(\displaystyle \;\;\;a\;b\;c\)
\(\displaystyle \;\;\;3\;2^{^1}\,1\)
\(\displaystyle \;\;\;1\;2\;3\)
\(\displaystyle \;\;----\)
\(\displaystyle \;\;\;\;\;\;\;0\)
In column \(\displaystyle b\), we have: \(\displaystyle \,2\,+\,2\,+\,1\:=\:11\;\;('5')\)
"Put down the 1, carry the 1".
\(\displaystyle \;\;\;a\;b\;c\)
\(\displaystyle \;\;\;3^{^1}\,2\;1\)
\(\displaystyle \;\;\;1\;2\;3\)
\(\displaystyle \;\;----\)
\(\displaystyle \;\;\;\;\;1\;0\)
In column \(\displaystyle a\), we have: \(\displaystyle \,3\,+\,1\,+\,1\:=\:11\;\;('5')\)
Write down the entire \(\displaystyle 11\).
\(\displaystyle \;\;\;a\;b\;c\)
\(\displaystyle \;\;\;3\;2\;1\)
\(\displaystyle \;\;\;1\;2\;3\)
\(\displaystyle \;\;----\)
\(\displaystyle \;1\;1\;1\;0\)
Therefore: \(\displaystyle \,321_4\,+\,m123_4\:=\:1110_4\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
If you're desperate, change the problem to base-10,
\(\displaystyle \;\;\)do the addition, then change back to base-4.
\(\displaystyle \;\;\;321_4\;=\;57\)
\(\displaystyle \;\;\;123_4\;=\;27\)
. . . . . . . . . .\(\displaystyle \,--\)
. . . . . . . . . . .\(\displaystyle 84\;=\;1110_4\)
Here it is in baby-steps . . .
\(\displaystyle \;\;\;a\;b\;c\)What does \(\displaystyle 321_4\,+\,123_4\) equal?
\(\displaystyle \;\;\;3\;2\;1\)
\(\displaystyle \;\;\;1\;2\;3\)
\(\displaystyle \;\;----\)
In column \(\displaystyle c\), we have: \(\displaystyle \,1\,+\,3\;=\;10\;\;('4')\)
We "put down the 0, carry the 1".
\(\displaystyle \;\;\;a\;b\;c\)
\(\displaystyle \;\;\;3\;2^{^1}\,1\)
\(\displaystyle \;\;\;1\;2\;3\)
\(\displaystyle \;\;----\)
\(\displaystyle \;\;\;\;\;\;\;0\)
In column \(\displaystyle b\), we have: \(\displaystyle \,2\,+\,2\,+\,1\:=\:11\;\;('5')\)
"Put down the 1, carry the 1".
\(\displaystyle \;\;\;a\;b\;c\)
\(\displaystyle \;\;\;3^{^1}\,2\;1\)
\(\displaystyle \;\;\;1\;2\;3\)
\(\displaystyle \;\;----\)
\(\displaystyle \;\;\;\;\;1\;0\)
In column \(\displaystyle a\), we have: \(\displaystyle \,3\,+\,1\,+\,1\:=\:11\;\;('5')\)
Write down the entire \(\displaystyle 11\).
\(\displaystyle \;\;\;a\;b\;c\)
\(\displaystyle \;\;\;3\;2\;1\)
\(\displaystyle \;\;\;1\;2\;3\)
\(\displaystyle \;\;----\)
\(\displaystyle \;1\;1\;1\;0\)
Therefore: \(\displaystyle \,321_4\,+\,m123_4\:=\:1110_4\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
If you're desperate, change the problem to base-10,
\(\displaystyle \;\;\)do the addition, then change back to base-4.
\(\displaystyle \;\;\;321_4\;=\;57\)
\(\displaystyle \;\;\;123_4\;=\;27\)
. . . . . . . . . .\(\displaystyle \,--\)
. . . . . . . . . . .\(\displaystyle 84\;=\;1110_4\)