The most effective way of conveying the reason for handling the division operation with the digit having highest place value, is allowing children to solve the problems of type 'Divide 1525 by 25' in as many ways as possible.
The example above may be solved starting division from the digit with lowest place value in following ways:
In the first method, we start with 5 in Ones' place. Since we cannot make any group of 25, we bring down 2 Tens. We cannot make any group of 25 from 2 Tens. However if we exchange 2 Tens for 20 Ones we have in all 25 Ones, and we have now 1 group of 25. We record 1Ones as the quotient and subtract 25 from 25. This leaves nothing and we record 0 as the result.
Next, we move to 5 Hundreds. We cannot make any group of 25. Hence we find out the next lower place value of 5 Hundreds, which is 50 Tens. Now we get 2 Tens as a quotient. (Remember that 50 tens are 500 Ones, which when made into groups of 25, we get 20 groups).
Similarly, when we move to 1 Thousand, we get 10 Hundred as equivalent lower place value, which cannot give any group of 25. Hence, we get the equivalent of next lower place value of 10 Hundred, which is 100 Tens. When 100 Tens are made into groups of 25 we get 4 Tens, as a quotient.
Thus we complete division of the number 1525.
The total quotient is thus 1 Ones + 2 Tens + 4 Tens, which equals 61.
In the second method, we find out equivalent for each digit in terms of Ones. Wherever necessary we group the Ones conveniently and then carry out partial division. For example, we have grouped 20 Ones and 5 Ones to carry out the first division. After carrying out partial divisions for all the digits in the dividend we add all the quotients obtained to get the final quotient.
Explaining the above methods is very easy using play-money.
Whatever methods we may use for carrying out division, we are basically taking advantage of the distribution of division over addition, as shown below:
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