If \[ 34 + 4 = 43\] \[ 34 \times 4 = ?\]
base 5
What are these numbers? base 5.. base 12... base 10?
Base 5 is like 1,2,3,4,5,10
We use 0-9 digits normally it's base 10 if a number system has digits from 0-3 only, it's base 4 0, 1, 2, 3, 10
@ParthKohli base 5 will have digits from 0-4 so 0, 1, 2, 3, 4, 10
34 x 4 = 34 + 34 + 34 + 34 base 5?
How do we calculate \(34\) in base 5?
yeah:) to easily do this, let's convert the number into base 10. Multiply it and then convert back to base 5
How can we convert?
If we have 25 it's actually \[25=5\times 10^0+2\times 10^1\] Isn't it?
Yeah! :)
34 in base 10 \[34=4*5^0+3*5^1\] what will you get?
So \(34 = 4 \times 5^{0} \times 3 \times 5^{1}\)?
Oops. I meant a + in between.
4 + 15 = 19
4 will be 4
so \[19*4=??\]
\(19 \times 4 = 76\)
\(76 = 7 \times 10^1 + 6 \times 10^0\)
now we have to convert into base 5 divide 76 by 5 15 ______________________ 5|76 75 first division result in quotient of 15 and remainder 1, so 1 is your first digit Now divide 15 by 5 and then tell me the remainder
0 is the remainder.
I get 301. Is that correct?
Correct:D
What? I just guessed it :P
another way is you can do it directly in base 5 without converting
you got remainder 1 then 0 and then 3 so it's 301
\[\large{(10a+b)+b=10b+a}\] \[\large{10a+2b=10b+a}\] \[\large{9a=8b}\] \[\large{a=\frac{8b}{9}}\] \[\large{(10a+b)*b=10ab+b^2=10(\frac{8b}{9})+b^2=\frac{80b+9b^2}{9}}\] b = 4: \[\large{\frac{320+144}{9}=\frac{464}{9}}\]
:d is that correct way @ash2326 ?
How do you get 3? I just guessed it.
as @rsadhvika indicated, you can also multiply directly in base 5
How to do so?
when you divided by 15 you got 3 as quotient and remainder 0 next step is to divide 3 by 5 and then find the remainder
let me show you...
\[3*5^2+0*5^1+1*5^0=76\]
|dw:1343571289892:dw|
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