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I know this might seem silly, but I am having trouble performing step-by-step long division on $5555 \div 55$.

My main problem is that I don't know when or what the rule is about putting the zeroes. I end up with eleven the way I was taught to calculate it.

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1  
"Bring down" the digits one at a time. –  André Nicolas Feb 26 '13 at 20:05

2 Answers 2

up vote 3 down vote accepted

The traditional grade-school algorithm:

              101  
             ----  
          55)5555  
             55  
             ---  
               5  
               0  
               --  
               55  
               55  
               --
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At what point did you put the zero in? –  Imray Feb 26 '13 at 20:15
    
Oh I see, 55 goes into 0 no times. Then you brought down the next 5... –  Imray Feb 26 '13 at 20:16
2  
@Imray: That’s right. I just did it one digit at a time. –  Brian M. Scott Feb 26 '13 at 20:20

Perhaps seeing it this way would give you an idea:

$$5555 = 5500 + 55 = 55*???$$

Further hint

$$5555 = 5500 + 55 = 55 \cdot 100 + 55 \cdot 1 = 55 \cdot ???$$

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huh? I don't understand what you are getting at –  Imray Feb 26 '13 at 20:15
    
@Imray see the edited version with another hint. –  gt6989b Feb 26 '13 at 20:30
    
I know what the answer is, I do own a calculator. My question was about the process. –  Imray Feb 27 '13 at 1:30
1  
It has nothing to do with a calculator. Note that the decomposition itself implies (because 5500 has 2 zeroes on the right) that you need an extra 0 in division. That comes from bringing things down 1 at a time as was suggested by a couple of people. I tried to explain to you not mechanically where the 0 comes from, but really what happens when you do division - and this is the where the extra 0 really comes from... –  gt6989b Feb 27 '13 at 13:48

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