Loser66
  • Loser66
Please, guide me how to convert 623489base 10 to base 16
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Find the biggest power of 16 smaller than 623489 to determine the number of digits and just repeatedly subtract... it's like Euclidean division algorithm
Loser66
  • Loser66
you mean I have to go backward from biggest power of 16, not start from 16^0?
Loser66
  • Loser66
so 16^4 = 65536 then?

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anonymous
  • anonymous
Yes
Loser66
  • Loser66
then?
phi
  • phi
you need a bigger number
phi
  • phi
623489 is much bigger than 16^4
Loser66
  • Loser66
yes
Loser66
  • Loser66
I need more,please, next step?
phi
  • phi
find the n so that 16^n is the first power of 16 bigger than your number then find 16^(n-1) divide 16^(n-1) into your number. the quotient will be the nth digit in your base 16 number. (use A for 10, B for 11, etc if you have to) now do the same thing for the remainder
Loser66
  • Loser66
I got 136*(16^3) +4096 is the next number after 1*(16^4) but how to express 136(16^3)?
phi
  • phi
16^5 is bigger than your number, so 16^4 is what you divide into your number
phi
  • phi
I was under the impression that 623489 was bigger than 16^5, but it is not...
anonymous
  • anonymous
Yeah phi it's most certainly not
Loser66
  • Loser66
|dw:1371506371646:dw|then? how to express it?
eSpeX
  • eSpeX
\(16^0=1\) \(16^1=16\) \(16^2=256\) \(16^3=4096\) \(16^4=65536\) \(16^5=1048576\)
phi
  • phi
how are you getting 1 for the highest digit. what do you get for 623489/16^4
Loser66
  • Loser66
I don't understand your question
phi
  • phi
how many times does 16^4 go into 623489 ?
e.mccormick
  • e.mccormick
1 would be the lowest didget. Sounds like you did remainder method backwards.
Loser66
  • Loser66
9
phi
  • phi
the idea is your number is a*16^4 + b* 16^3 + c*16^2 + d*16^1 + e*16^0 a is 9 now subtract 9*16^4 from your number to get the remainder
eSpeX
  • eSpeX
So 623489-(9*16^4) give you the remainder that you move to your next power.
Loser66
  • Loser66
got it, thank you
Loser66
  • Loser66
9(16^4) + 33665 and repeat the step with 33665, right?
eSpeX
  • eSpeX
Right
Loser66
  • Loser66
8(16^3) +... so, if the number before 16^ ... is bigger than 10, for example if it is 13 *(16^2) use A,B ... right?
phi
  • phi
yes,
Loser66
  • Loser66
got it, thanks everybody
phi
  • phi
but you should not be getting any digits > 10 for this problem
e.mccormick
  • e.mccormick
Want to see the remainder method for doing this?
Loser66
  • Loser66
yes
Loser66
  • Loser66
I searched but they give me available answer , not method.
Loser66
  • Loser66
my prof taught me calculate from base^0 and up. so I don't know this method
e.mccormick
  • e.mccormick
\(\cfrac{623489}{16}=38968\) with a remainder of 1 \(\cfrac{38968}{16}=2435\) with a remainder of 8 \(\cfrac{2435}{16}=152\) with a remainder of 3 \(\cfrac{152}{16}=9\) with a remainder of 8 \(\cfrac{9}{16}=0\) with a remainder of 9 If any remainder was over 9, you would need to convert to the proper letter. Take the remaioners in reverse order: 98381
Loser66
  • Loser66
ok, I know this method, mod(16) and consider the remainder only.
Loser66
  • Loser66
quite easier than go from 16^...
e.mccormick
  • e.mccormick
Mathematically, they are related. The remainder method works in reverse, so some people like the power method better because it works the same way they would write out the number.
eSpeX
  • eSpeX
It is also easier to use a calculator with. :)
Loser66
  • Loser66
@eSpeX no calculator, need method.XD
e.mccormick
  • e.mccormick
They are pretty close in calculator utility. You have to keep track of fractional values and make them into a useful number either way.
Loser66
  • Loser66
98381 is the answer. I gooooot it hehehe
Loser66
  • Loser66
@e.mccormick you explain me more about the remainder you talked. please
Loser66
  • Loser66
for the first one (remainder 1) the quotient is not an exponent of 16, right? how to link to the problem?
e.mccormick
  • e.mccormick
You are dividing out all possible values of 16, the SECOND digit. The remainder of this process is 1, the first digit. So you get 38968.0625 or 38968 with a remainder of 1. That makes 1 the last digit. Now, from 38968 you divide again. This divides out all possible instances of the third digit and leaves a remainder of the second. So on ans do forth. If I used % to mean get the mod, this becomes: \(\cfrac{623489}{16}=38968\) & \(623489\% 16=1\) \(\cfrac{38968}{16}=2435\) & \(38968\% 16=8\) \(\cfrac{2435}{16}=152\) & \(2435\% 16=3\) \(\cfrac{152}{16}=9\) & \(152\% 16=8\) \(\cfrac{9}{16}=0\) & \(9\% 16=9\)
Loser66
  • Loser66
you are wicked!! I "hate" the way you do it!pal
Loser66
  • Loser66
and if the remainder is 10 use A, 11 use B, right?
e.mccormick
  • e.mccormick
All I am doing is keepting track of the whole part and the remainder. The whole part goes to the next digit. The remainder goes to the answer. Yes. A=10, B=11, C=12, D=13, E=14, and F=15.
e.mccormick
  • e.mccormick
This system also works for base 8, 4, 2, and so on.
Loser66
  • Loser66
thank you

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