Asked by jessica
wouldn't 503 to base 5 be 2003 there are 20 sets of 5 with no extra sets of 5 and 3 left over correct
Answers
Answered by
Damon
I do not understand what you are doing.
If 503 is in base 10
that is 3*10^0 + 0*10^1 + 5*10^2
To put 503 in base 5
first say 503 = 5^x
x log 5 = log 503 (any base logs, use 10)
x = 2.7/.7 = 3.85
so it is 5^3 times something almost 5
1 * 5^3 = 125
2 * 5^3 = 250
3 * 5^3 = 375
4 * 5^3 = 500
we only need 3 more so
3*5^0 + 0*5^1 + 0*5^2 + 4*5^3
or
4 0 0 3 is what I get
If 503 is in base 10
that is 3*10^0 + 0*10^1 + 5*10^2
To put 503 in base 5
first say 503 = 5^x
x log 5 = log 503 (any base logs, use 10)
x = 2.7/.7 = 3.85
so it is 5^3 times something almost 5
1 * 5^3 = 125
2 * 5^3 = 250
3 * 5^3 = 375
4 * 5^3 = 500
we only need 3 more so
3*5^0 + 0*5^1 + 0*5^2 + 4*5^3
or
4 0 0 3 is what I get
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.