Math Shortcuts

Squaring of Two numbers

Consider a simple example 9²

Step 1: Consider the nearest base (here 10).

Step 2: As 9 has a deficiency of 1 (10 – 9 = 1), we should decrease it further by 1, and set down our LHS of the Answer as ‘8’.

Step 3: On the RHS put the square of the deficiency (here 1).

 We get 9² = 81.

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Now Consider 102

1) Base is 100

2) Deficiency is ‘-2’ (100 – 102 = -2)

  Therefore we subtract ‘-2’ from 102

102 – (-2) = 104

This is our RHS

3) Our LHS now becomes (-2)² which is 4

 Since the base is 100 we write it as ’04’, so that we get 102² = 10404

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For multiplication also we use the same technique

Consider 28²

1) Let 20 be the Working Base and 10 as the Main Base.

 Therefore x = (Main Base)/ (Working Base) = 10/20 = 1/2

2) Here the deficiency = 20 – 28 = -8

Therefore RHS = 28 – (-8 ) = 36

Divide by x i.e. by (1/2).

We get   36/ (1/2) = 72.   This is the required RHS.

3) LHS = (-8 )² = 64

Since Main Base is 10, we put only ‘4’ on the LHS and carry over ‘6’ to the RHS

Therefore we get

28² = 72+6 | 4 == 784

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Some More Examples

36² = 128 |₁ 6 = 1296                (Base = 40)

522 = 27 | 04 = 2704               (Base = 50)

9972 = 994 | 009 = 994009      (Base = 1000)

10042 = 1008 | 016 = 1008016    (Base =1000)

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This technique exclusively for numbers ending with 5

For numbers ending with 5 the LHS will be 25.

The RHS will be the product of the other digits….

Examples:

25² = (2 * 3) | 25 = 625

35² = (3 * 4) | 25 = 1225

45² = (4 * 5) | 25 = 2025

55² = (5 * 6 ) | 25 = 3035

65² = (6 * 7) | 25 = 4225

75² = (7 * 8 ) | 25 = 5625

105² = (10 * 11) | 25 = 11025

125² = (12 * 13) | 25 = 15625

175² = (17 * 18 ) | 25 = 30625

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  Direct Squaring

‘duplex combination’ can be used for general purpose squaring.

To proceed further we need to know the formula for certain numbers.

D( a ) = a²

D( ab ) = 2ab

D( abc ) = 2ac + b²

D( abcd ) = 2ad + 2bc

D( abcde ) = 2ae + 2bd + c²

D( abcdef ) = 2af + 2be + 2cd      and so on….

As we can see above, D of any number is the sum of square of the middle number and two times the product of the other pairs.

Square of a number is given by

( ab )² = D( a ) | D( ab ) | D( b )

( abc )² = D( a ) | D( ab ) | D(abc) | D( bc ) | D( c )

( abcd )² = D(a) | D(ab) | D(abc) | D(abcd) | D(bcd) | D(bc) | D (c)

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Example : 

(23)² = (ab)² = D(a) | D(ab) | D(b)

                    = 4     |    12    |    9

Since this technique must have only one digit, we carry over ‘1’ of ’12’ to LHS.

Therefore it becomes     4 |₁ 2| 9

Hence the answer is 529

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Example 2:

(527)² = ( abc )²  

           = D( a ) | D( ab ) | D(abc) | D( bc ) | D( c )

           =   25   |     20    |     74   |    28    |   49

           =   25   |₂     0    |₇     4   |₂    8    |₄   9

          =   277729

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 Cubing

 

This method is similar to squaring.  It is just modified a bit, as we shall see in the next few examples.

Consider 13³

   Step 1: Consider nearest base (here 10).

Step 2: As 13 has an excess of ‘3’ (13 – 10 = 3), we double the excess and add the original number (13) to it, and put it on the LHS.

Therefore we get 13 + 6 = 19

Step 3: Now find the new excess. In this case it is 19-10 = 9. Now multiply this with the original excess to get the middle part of the answer.

Therefore we get 9 * 3 = 27

Step 4: Now cube the original excess and put it as the last part

Carry over any big numbers and total to get the answer.

                       19  7  7                     

                             2  2                       

                       21  9  7

Therefore 13³ = 2197

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Now consider 47³

1) Let the main base be 10 and the working base is 50

 Therefore the ratio

 x = (Main Base)/ (Working Base) = 10/50 = 1/5

2) Excess is -3 (47 – 50 = -3).  Double the excess and add the original number (here 47) to it.

We get 47 – 6 = 41.

The Base correction for this part is achieved by dividing by x² .

Therefore we get 41/ (1/25) = 41 * 25 = 1025

3) Excess in the new uncorrected number (41 – 50 = -9) is multiplied by the original excess (-3) to obtain the second part.

Therefore we get -9 * -3 = 27

The Base correction for this part is achieved by dividing by x.

Therefore we get 27 * 5 = 135

4) The third part is obtained by cubing the excess.

    (-3)³ = -27

5) Carry over the extra numbers and total to obtain the final answer

                      1025  0  0                      

                     13  5  0                

                          -2  7                 

  1038  2  3

Therefore the final answer is 103823 

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Direct Cubing 

This method is simple and easy.

If the number is ( ab ) its cube can be calculated as

              a³     a²b     ab²    b³

                      2a²b   2ab²     

Sum them up taking care of the carryovers.

 Consider (16)³

  Writing it as above

         1     6     36    216

              12     72         

       1     18    108   216

 Considering the carryovers                             

                               Therefore we get the answer as 4096…. 

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