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Long Division isn't working for me!

  1. Feb 13, 2009 #1
    The problem is that I am weak at long division and need it for ratio and rate.1. The problem statement, all variables and given/known data



    2. Relevant equations
    Like big questions such as 236 divided by 6.(I know it doesn't go in by using a calculator but if it has a point.



    3. The attempt at a solution
    I just can't do it and thats why I need advice and helpful tips.
    Ward.
     
  2. jcsd
  3. Feb 13, 2009 #2

    mgb_phys

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  4. Feb 13, 2009 #3
    Thanks now how about normal long division?
     
  5. Feb 13, 2009 #4

    symbolipoint

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    Certainly why not!

    You were interested in "236 divided by 6". The dividend is 236 and the divisor is 6.

    The "6" is written at the left of the curve of the long-division structure, and the "236" is written under the segment of the long-division structure.

    Ask: does the 2 contain any 6 ? No. Then include the next digit of the dividend.

    Ask: Does 23 contain any quantity of 6 ? Yes.
    How many? 6*3=18, 6*4=24, so 23 contains 3 of the quantity 6.
    Write the digit, "3" directly over hte "3" of the dividend, and subtract 18 from 23, and then bring down the next digit of the dividend (write it next to the result of subtracting 18 from 23). You are looking at (23-12) with a "6" next to it. This "6" is the one which is part of the dividend.

    Ask: Does this (23-18) with a "6" next to it contain any quantity of 6? If so, how many?
    ....
    Can you see how this works?
     
  6. Feb 14, 2009 #5
    Sort of.... Maybe my brain isn't working:(
     
  7. Feb 14, 2009 #6
    6into 56 10 times but it doesn't go in fully it has a reminder of 4.
     
  8. Feb 14, 2009 #7

    tiny-tim

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    Welcome to PF!

    Hi WadeS! Welcome to PF! :smile:
    Nooo … you have to sneak up on it from below! :wink:

    You have to say "6 into 56 9 times fully, plus it has a remainder of 2" :smile:
     
  9. Feb 14, 2009 #8

    arildno

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    Perhaps this might help:

    We are to perform the division 236 with 6.

    1. We start out optimistically, and say that each of the 6 "guys" gets 50 of the whole.
    Thus, the remainder is 236-6*50=236-300=-64..ooops! (each getting 50 units)

    2. That was a downer. We now let each of the 6 give back to the whole 12 units, thus we get:
    236-300+72=8 (each having 50-10=40 units)

    3. Now, we give each 1 unit, leving 8-6=2 as remainder (each getting 40+1=41 units)

    4. Now, we give each a half-unit, leaving 2-3=-1 as remainder (each getting 41+0.5=41.5 units)

    5. Darn, that was too much! Each then gives away 1/6, whereby there is no remainder, each sitting with 41.5-1/6 units, an expression you may fiddle further with.

    Using powers of ten in the division process is just a clever systematization of the division process, not a necessary, intrinsic feature.
     
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