Tips And Tricks For Speedy Calculations Module 4 Division



TIPS AND TRICKS FOR SPEEDY CALCULATIONS – MODULE IV – DIVISION

Division is no more tedious as it used to be! In this module we deal with divisibility rules of various numbers, how to find remainders and general division rules. After reading this module and solving problems by heart you may find division as easier as multiplication.

1. Simple divisibility rules by numbers such as 3, 4, 6, 7, 8, 9, 11, 12, 14, 15, 16

To check whether a number is divisible by the above numbers or not either you can perform the actual division or use the pragmatic short cuts given. We recommend you to use these short cuts as these are much simpler than the long and conventional division methods.

2. Advanced divisibility rules for 7, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47

These are must to be known for beginning and building your grip on division as a full mathematical operation. The following rules explains whether test for a number to be fully divisible by the above mentioned numbers.

3. Calculating remainder when a very large number is divided by a smaller one.

The statement- “When x is divided by z, it leaves y as the remainder.” is represented in modular arithmetic as x=y(mod z)

It can also be interpreted as “x and y leave the same remainder when divided by z.” This is also known as the congruence relation and we can say that “x is congruent to y modulor z.”

There is a property of this relation which is very useful.

Suppose x1=y1(mod z) and x2=y2(mod z), then- x1.x2=y1.y2(mod z)

Now, coming to the original problem, suppose you have been given the seemingly tedious task of finding the remainder when a very large number x^r is divided by z. Let us tackle this problem with an example.

Suppose we want to find 3^48 divided by 11. Do we go about computing 3^48 and then dividing it by 11? Seemingly impossible! So what we do is we find ‘a’ in (3)^(2k) = a mod 11 for every 2k less than or equal to r(=48).

Hence, 3^2 = 9 (mod 11)
3^4 = 81 = 4 (mod 11)
3^8 = 3^4.3^4 = 4^2 (mod 11) = 5 (mod 11) (Realize how easy finding 3^8 mod 11 was using the property stated above!)
Similarly,
3^16 = 5^2 (mod 11) = 3 (mod 11)
3^32 = 3^2 (mod 11) = 9 (mod 11)

Now, note that 48 = 2^6 + 2^5 = 32 + 16.

So to find 3^48=a mod 11, we write a (mod 11) = 6*3 (mod 11). (Again using the property)

3^48 = 3^(16+32)
= 3^16.3^32 = 3*9 (mod 11) = 5

Hence, we can very easily find the remainder, on paper, without using supercomputers, when a large number is divided by some smaller number.

4. General division rule in vedic mathematics.

This might seem to you a little complicated in the beginning. But a little careful analysis will find your doubts melting. Mastery over this method once done will replace the conventional methodology of dividing huge numbers wherein decimal digits are required rather than just finding the remainder.

This method will be explained by means of examples;

1. Divide 716769 by 54;

Reduce the divisor 54 to 5 pushing the remaining digit 4 “on top of the flag”. Corresponding to the number of digits flagged on top (in this case, one), the rightmost part of the number to be divided is split to mark the placeholder of the decimal point or the remainder portion.

The following are the steps to be done;

1. 7 ÷ 5 = 1 remainder 2. Put the quotient 1, the first digit of the solution, in the first box of the bottom row and carry over the remainder 2
2. The product of the flagged number (4) and the previous quotient (1) must be subtracted from the next number (21) before the division can proceed. 21 – 4 x 1 = 1717 ÷ 5 = 3 remainder 2. Put down the 3 and carry over the 2
3. Again subtract the product of the flagged number (4) and the previous quotient (3), 26 – 4 x 3 = 1414 ÷ 5 = 2 remainder 4. Put down the 2 and carry over the 4
4. 47 – 4 x 2 = 3939 ÷ 5 = 7 remainder 4. Put down the 7 and carry over the 4
5. 46 – 4 x 7 = 1818 ÷ 5 = 3 remainder 3. Put down the 3 and carry over the 3
6. 39 – 4 x 3 = 27. Since the decimal point is reached here, 27 is the raw remainder. If decimal places are required, the division can proceed as before, filling the original number with zeros after the decimal point27 ÷ 5 = 5 remainder 2. Put down the 5 (after the decimal point) and carry over the 2
7. 20 – 4 x 5 = 0. There is nothing left to divide, so this cleanly completes the division

Thus the answer is 13273.5 as reflected in the second box.

2. Divide 45026 by 47;

Reduce the divisor 47 to 4 pushing the remaining digit 7 “on top of the flag”. Corresponding to the number of digits flagged on top (in this case, one), the rightmost part of the number to be divided is split to mark the placeholder of the decimal point or the remainder portion.

                                                             4                      9                      7                       5

1. 4 ÷ 4 = 0 remainder 4. Put the quotient 0, the first digit of the solution, in the first box of the bottom row and carry over the remainder 4
2. The product of the flagged number (7) and the previous quotient (0) must be subtracted from the next number (45) before the division can proceed. 45 – 7 x 0 = 4545 ÷ 4 = 9 remainder 9. Put down the quotient 9 and carry over the remainder 9.
3. Again subtract the product of the flagged number (7) and the previous quotient (9), 90 – 7 x 9 = 2727 ÷ 4 = 5 remainder 7. Put down the quotient 5 and carry over the remainder 7.
4. 72 – 7 x 5 = 3737 ÷ 4 = 8 remainder 5. Put down the quotient 8 and carry over the remainder 5.
5. 56 – 7 x 8 = 0there is nothing left to divide, so this cleanly completes the division.

Thus the answer is 958.0