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Converting between base 7 to 3

  1. Oct 7, 2009 #1
    1. The problem statement, all variables and given/known data

    convert (352416)_7 to (...)_3

    the _7 means base 7 and similarly, _3 means base 3.

    Idea : convert base 7 to base 10 then base 10 to base 3.

    I know I can use expansion, 3 * 7^6 + 5 * 7^6 ... to convert it to base 10.

    but in my exam, we are not allowed to use calculators, and I really not want to spend time
    on figuring out the power of X. So do you know a faster way to convert this to base 10?
    From then on I could convert it to base 3.
     
  2. jcsd
  3. Oct 7, 2009 #2

    Mark44

    Staff: Mentor

    I can't think of any way other than brute force, which is what you're doing. BTW, it would be 3*75 + 5*74 + ... + 1*71 + 6

    You could make a couple of shortcuts, though. 74 = 492 = (50 - 1)2 = 502 - 2*50 + 1 = 2500 - 100 + 1 = 2401.
     
  4. Oct 7, 2009 #3
    Thanks for the tip.
     
  5. Oct 7, 2009 #4
    Also when say the number turned out to be something like 443267. Then to change this
    number of base 10 to base 3, I could divide by 3 and get the remainder. But is there another way that you can think of.
     
  6. Oct 7, 2009 #5
    And could I have just divided the base 7 number by 3 to turn it into base 3?
     
  7. Oct 7, 2009 #6

    Mark44

    Staff: Mentor

    Once you have converted to a base-10 number, subtract the highest power of 3 that is <= your number. It's probably easier to just show an example than to explain the process.

    Suppose you want to convert 4325 to its base-4 equivalent. 4325 = 11710.

    The powers of 4 are 1, 4, 16, 64, 256, and so on.

    117 = 64 + 53, so my base-4 representation is going to be 1xxx.
    53 = 3*16 + 5, so now I have 13xx
    5 = 4 + 1, so I have 131x.
    My final remainder is 1, so the number we want is 13114.

    It might be possible to convert directly from base 7 to base 3, but that requires you to know arithmetic in both bases; e.g. to be able to know that 6*4 = 21 in base 7, that sort of thing.
     
  8. Oct 7, 2009 #7
    Yes, it's pretty simple in this case but be sure to express everything in base 7:

    [tex](352416)_7=3 \times (115352)_7+0[/tex]

    [tex](115352)_7=3 \times (26340)_7 +2[/tex]

    [tex](26340)_7=3 \times (6560)_7 +0[/tex]

    [tex](6560)_7=3 \times (2164)_7 +2[/tex]

    [tex](2164)_7=3 \times (521)_7 +1[/tex]

    [tex](521)_7=3 \times (152)_7 +2[/tex]

    [tex](152)_7=3 \times (40)_7 +2[/tex]

    [tex](40)_7=3 \times (12)_7 +1[/tex]

    [tex](12)_7=3 \times (3)_7 +0[/tex]

    [tex](3)_7=3 \times (1)_7 +0[/tex]

    [tex]\Rightarrow (352416)_7=(10012212020)_3[/tex]
     
  9. Oct 7, 2009 #8
    I think that [tex]6\times 4=33[/tex] in that base.
     
  10. Oct 7, 2009 #9
    Yea, that what I do with power of 2's.

    Thanks everyone.
     
  11. Oct 7, 2009 #10

    Mark44

    Staff: Mentor

    My error.
     
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