Asked by Ekingdeft
If 451 in base6 - P in base7 = 301 in base6, find d. value of P
Answers
Answered by
Damon
451 base 6
1*6^0 + 5*6^1 + 4*6^2
= 1 + 30 + 144
= 175 in base 10
301 base 6
1*10^0 + 0*6^1 + 3*6^2
= 1 + 0 + 108
= 109 in base 10
so in base 10
175 - x = 109
x = 66 in base 10
put that in base 7
7^2 = 49
17 left
2*7^1 = 14
3 left
so
1^7^2 + 2*7^1 + 3*7^0
or
P = 123 in base 7
1*6^0 + 5*6^1 + 4*6^2
= 1 + 30 + 144
= 175 in base 10
301 base 6
1*10^0 + 0*6^1 + 3*6^2
= 1 + 0 + 108
= 109 in base 10
so in base 10
175 - x = 109
x = 66 in base 10
put that in base 7
7^2 = 49
17 left
2*7^1 = 14
3 left
so
1^7^2 + 2*7^1 + 3*7^0
or
P = 123 in base 7
Answered by
Damon
By the way, I just did that by common sense. I do not know what routine you are really supposed to know. You may have a better way to shift back and forth.
Answered by
Damon
301 base 6
1*10^0 + 0*6^1 + 3*6^2
should be
301 base 6
1*6^0 + 0*6^1 + 3*6^2
= 109
1*10^0 + 0*6^1 + 3*6^2
should be
301 base 6
1*6^0 + 0*6^1 + 3*6^2
= 109
Answered by
Steve
You could just do the subtraction in base 6 and then convert the answer to base 7:
451-301 = 150
150<sub>6</sub> = 123<sub>7</sub>
451-301 = 150
150<sub>6</sub> = 123<sub>7</sub>
Answered by
Damon
good point :)
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