R=2
23*11 base 4 is
023
230
-----
313
+ 2
------
321
If 321 base 4 is divided by 23 base 4 and leaves a remainder 'R' what is the value of 'R' ?
3 answers
or:
321base 4 = 3(4^2) + 2(4) + 1
= 48 + 8 + 1 = 57 in base 10
23base 4 = 2(4) + 3 = 11 in base 10
57/11 = 5 2/11
so the remainder in base 10 is 2
and since 2base 4 = 2base 10
R = 2
Steve's method is more direct, however you have to know how to do arithmetic in base 4.
321base 4 = 3(4^2) + 2(4) + 1
= 48 + 8 + 1 = 57 in base 10
23base 4 = 2(4) + 3 = 11 in base 10
57/11 = 5 2/11
so the remainder in base 10 is 2
and since 2base 4 = 2base 10
R = 2
Steve's method is more direct, however you have to know how to do arithmetic in base 4.
or:
321base 4 = 3(4^2) + 2(4) + 1
= 48 + 8 + 1 = 57 in base 10
23base 4 = 2(4) + 3 = 11 in base 10
57/11 = 5 2/11
so the remainder in base 10 is 2
and since 2base 4 = 2base 4
R = 2
Steve's method is more direct, however you have to know how to do arithmetic in base 4.
321base 4 = 3(4^2) + 2(4) + 1
= 48 + 8 + 1 = 57 in base 10
23base 4 = 2(4) + 3 = 11 in base 10
57/11 = 5 2/11
so the remainder in base 10 is 2
and since 2base 4 = 2base 4
R = 2
Steve's method is more direct, however you have to know how to do arithmetic in base 4.