Usually in division we generate 1 digit at a time.
However, there is no rule - if the situation presents itself - that you cannot do 2 digits a a time.
In this example, I start to work with 25 as the starting divisor.
Because 25X40 = 1000.
Because 40 is bigger than 37, the divisor is at least 25.
So I work with that and deal with the remainder in the next step.
Second step is that I calculate the remainder to be 75:
In other words:
(\frac{1000}{37}=25R75)
75 is bigger than 37, so 25 is too small.
Let’s correct this. Divide the remainder again by 37:
( \frac{75}{37} = 2R1 )
We add the 2 to the 25 to get 27:
( \frac{1000}{37} = 27R1 )
When dividing by 37, then 25R75 is the same as 27R1.
Stepwise: 25R75 = 26R38 = 27R1.
In the analogy with apples, this is what we do:
We have a pile of 1000 apples, that need to be divided over 37 baskets.
Step 1: We take 40 baskets and put 25 apples in each. The pile is now empty.
Step 2: We realize we only need 37 baskets so we remove 3 baskets, take the apples out and put them in a pile again. The pile now contains 75 apples.
Step3: We put 2 apples from the pile in each of the 37 baskets, leaving one apple left on the pile.
Now there are 27 apples in each basket and one left in the pile:
( \frac{1000}{37} = 27R1 )
Apart from Willem Bouman and and maybe Nodas, I know of nobody who memorized all the multiplication tables up to 100.
They would immediately take 27 as the first two digits.
Also because from experience they know that 37 X 27 = 32^2 - 5^2= 1024 - 25 = 999.
If you do this the traditional way you get:
Start with 100: 100\37. Guess 2 times. 2X37 = 74. 100 - 74 = 26.
260\37. Guess 7. 7x37 = 210+49 = 259.
260 - 259 = 1
You will end up with the same results. However, starting with 25 gives you much smaller numbers to work with.
The last part of the post is devoted to the fact that we started with 1 (if you move the decimal point) and ended with 1, so we work with a repeating string of digits and a factor 1000.
You ask for another example.
Let me think.
How about:
( \frac{1000}{47} )
Start with increasing the quotient to 50:
1000\50 = 20. Remainder is zero.
Lower the quotient from 50 to 47:
( \frac{1000}{50} = 20)
Therefore:
( \frac{1000}{47} = 20R60 )
(=3 baskets containing 20 apples each)
Divide 60 by 47:
( \frac{60}{47} = 1R13)
Add to 20R60=21R13:
( \frac{1000}{47} = 20R60 = 21R13 )
Let’s do another 2 digits:
Move the decimal point 2 places:
13 => 1300:
( \frac{1300}{50} = 26 )
( \frac{1300}{47} = 26R78 )
(3*26=78)
( \frac{78}{47} = 1R31 )
( \frac{1300}{47} = 26R72 = 27R31)
Another two:
( \frac{3100}{50} = 62 )
( \frac{3100}{47} = 62R(62*3) = 62R186 )
In my mind, I focus now in the 186 and a quotient of 50. 3 times. I subtract 150 from 186 to get 36 and add 3*3=9 to 36 = 45.
( \frac{3100}{47} = 62R186 = 65R45 )
( \frac{4500}{50} = 90 )
( \frac{4500}{47} = 90R270 =95R(270-250+3*5) = 95R35)
35 is more than half of 47, so round the 95 up to 96:
( \frac{1000}{47} \approx 21,276596 )
With 4 imho easy steps we have calculated 1000\47 to 8 significant places.
Observe:
We never multiplied by 47. We divided by 50 which is more like doubling the remainder and used the difference - 3 - as a factor to multiply by.
