# 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.

 Number Method 3 If the sum of its digits is divisible by 3 4 If the number formed by the last two digits is divisible by 4 6 If it is divisible by both 2 & 3. 7 If after subtraction of a number consisting of the last three digits from a number consisting of the rest of its digits the result is a number that can be divisible by 7. Ex.: 414141 is divisible 7 as 414-141= 273 is divisible by 7. 8 If the last three digits of the number are divisible by 8 9 If the sum of its digit is divisible by 9. 11 If the difference of the sum of its digits at odd places and sum of its digits at even places, is either 0 or a number divisible by 11. Ex: 4832718 is divisible by 11, since:(Sum of digits at odd places) â€“ (sum of digits at even places)= (8+7+3+4)-(1+2+8) = 11 12 A number is divisible by 12 if it is divisible by both 4 and 3 14 If a number is divisible by 2 as well as 7 15 If a number is divisible by both 3 & 5 16 If the number formed by the last 4 digits is divisible by 16

1. 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.

 Number Method Example 7 Subtract 2 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 7, the original number is also divisible by 7 Check for 945: : 94-(2*5)=84. Since 84 is divisible by 7, the original no. 945 is also divisible 13 Add 4 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 13, the original number is also divisible by 13 Check for 3146:: 314+ (4*6) = 338:: 33+(4*8) = 65. Since 65 is divisible by 13, the original no. 3146 is also divisible 17 Subtract 5 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 17, the original number is also divisible by 17 Check for 2278:: 227-(5*8)=187. Since 187 is divisible by 17, the original number 2278 is also divisible. 19 Add 2 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 19, the original number is also divisible by 19 Check for 11343:: 1134+(2*3)= 1140. (Ignore the 0):: 11+(2*4) = 19. Since 19 is divisible by 19, original no. 11343 is also divisible 23 Add 7 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 23, the original number is also divisible by 23 Check for 53935:: 5393+(7*5) = 5428 :: 542+(7*8)= 598:: 59+ (7*8)=115, which is 5 times 23. Hence 53935 is divisible by 23 29 Add 3 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 29, the original number is also divisible by 29 Check for 12528:: 1252+(3*8)= 1276 :: 127+(3*6)= 145:: 14+ (3*5)=29, which is divisible by 29. So 12528 is divisible by 29 31 Subtract 3 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 31, the original number is also divisible by 31 Check for 49507:: 4950-(3*7)=4929 :: 492-(3*9) :: 465:: 46-(3*5)=31. Hence 49507 is divisible by 31 37 Subtract 11 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 37, the original number is also divisible by 37 Check for 11026:: 1102 – (11*6) =1036. Since 103 – (11*6) =37 is divisible by 37. Hence 11026 is divisible by 37 41 Subtract 4 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 41, the original number is also divisible by 41 Check for 14145:: 1414 – (4*5) =1394. Since 139 – (4*4) =123 is divisible by 41. Hence 14145 is divisible by 41 43 Add 13 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 43, the original number is also divisible by 43. Check for 11739:: 1173+(13*9)= 1290:: 129 is divisible by 43. 0 is ignored. So 11739 is divisible by 43 47 Subtract 14 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 47, the original number is also divisible by 47. Check for 45026:: 4502 – (14*6) =4418. Since 441 – (14*8) =329, which is 7 times 47. Hence 45026 is divisible by 47

1. 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.

1. 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.

1. 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

 4, (7) 4 5 0 2 6 0 9 5 8 0

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.

JPSC Notes brings Prelims and Mains programs for JPSC Prelims and JPSC Mains Exam preparation. Various Programs initiated by JPSC Notes are as follows:- For any doubt, Just leave us a Chat or Fill us a querry––

JPSC Mains Test Series 2022

Subscribe our Test Series program to get access to 20 Quality mock tests for JPSC Preparation.