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Multiplication By Short Cut-Methods

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Multiplication  of  a  Number  By  99, 999, 9999 etc.  :   Place as many zeros to the right of  the multiplicand as is the number of  nines and from the number so formed, subtract the multiplicand to get the answer.

e.g. Multiplying  596378  by  99999.

Soln. Using the above rule, we have:

596378  99999 = (59637800000 – 596378) = 59637233622.

Multiplication  Of  a  Number  By  5n  :   Put  ‘n’  zeros to the right of  the multiplicand  and  divide the number so formed by  2n.

e.g. Multiply 975436 by 625.

Soln. We may write, 625  =  54.

\therefore 975436    625  =  975436  54  = \frac{9754360000}{16}  =  609647500.

Multiplication  By  Distributive  Law  :   We use the laws :

  1. a \times b  +  a  \times  c   =   a  (b + c)                   II.      a \times b  –  a \times c   =   a  (b – c)

Example Evaluate : 972\times 214 + 972 \times  786.

Solution 972 \times  214 + 972 \times  786  =  972 \times  (214 + 786)  =  972 \times  1000   =   972000.

Example Evaluate : 512 \times  318 – 512 \times  218.

Solution  512 \times  318  –  512 \times  218  =  512 \times (318 – 218)  =  512 \times  100  =  51200.

  1. (a + b)2 = a2  +  b2  +  2ab
  2. (a – b)2 = a2 + b2 – 2ab
  3. (a + b)2 – (a – b)2 = 4ab
  4. (a + b)2 + (a – b)2 = 2(a2 + b2)
  5. a2 – b2 = (a + b) (a – b)
  6. ( a + b )3 =   a3 + b3   +  3 a b (a + b)
  7.  7. ( a – b )3 =  a3 – b3 – 3 a b (a – b )
  8. a3 + b3 =  (a + b) (a2 – ab + b2)

=  ( a + b )3  –  3 a b (a + b)

  1. a3 – b3 = (a – b) (a2 + ab + b2)

=  ( a – b )3 + 3 a b (a – b)

  1. (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
  2. a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2  –  ab  –  bc  –  ca)
  3. If a + b + c =  0,   then   a3  +  b3  +  c3    =   3abc.

Examples :

1 .Evaluate : [(1856)2 – (144)2].

 Solution. [(1856)2 – (144)2] = (1856 + 144) (1856 – 144)

= (2000) (1712)

= 342400

2 .Find the value of  : \left [\frac{\left (905+298 \right )^{2}+\left (905-298 \right )^{2}}{\left (905 \right )^{2}+\left (298 \right )^{2}} \right ]

Solution.  Given Exp.  = \frac{\left (a+b \right )^{2}+\left (a-b \right )^{2}}{\left (a^{2}+b^{2} \right )} ,  where  a = 905  &  b = 298

=   \frac{2\left (a^{2}+b^{2} \right )}{\left (a^{2}+b^{2} \right )}

=        2          [ (a + b)2  +  (a – b)2  =  2(a2 + b2) ]

3. Evaluate : \left (\frac{675\times675\times675-396\times396\times396}{675\times675+675\times 396+396\times396} \right )

Solution.  Given Exp. =  \frac{\left (a^{3}+b^{3} \right )}{\left (a^{2}+ab+b^{2} \right )} ,   where  a = 675  &  b = 396

=    ( a – b )

=     675 – 396

=     279

4. Evaluate : \left (\frac{512\times512\times512+173\times173\times173}{512\times512-512\times 173+173\times173} \right )

Solution.  Given Exp. = \frac{\left (a^{3}+b^{3} \right )}{\left (a^{2}-ab+b^{2} \right )}  ,    where   a  =  512   &   b  =  173

=    ( a + b )

=    ( 512 + 173 )

=          685

5. Evaluate :  (286  286 + 214  214 + 2  286  214).

 Soln. Given Exp. =    (a2 + b2 + 2ab),   where  a = 286  &  b = 212.

= (a + b)2

= (286 + 214)2

= (500)2

= 250000

6. Evaluate : (i)   (1207)2                                            (ii)     (1398)2

 (i) (1207)2 =    (1200 + 7)2

=    (1200)2  +  (7)2  +  2 \times 1200 \times 7   [(a + b)2 = a2 + b2 + 2ab]

=    (1440000 + 49 + 16800)

=    1456849

 (ii) (1398)2 =    (1400 – 2)2

=    (1400)2  +  (2)2  –  2 \times 1400 \times 2     [(a – b)2 = a2 + b2 – 2ab]

=    (1960000 + 4 – 5600)

=    1954404

7. Find the total number of  prime factors in  ( 648 \times 352 \times 118 \times  657 ).

 Solution. Given Exp. = (2 \times 3)48 \times  352 \times  118 \times  (5 \times 13)7

= 248  3100  57 118  137

\therefore  Total number of  prime factors  = (48 + 100 + 7 + 8 + 13)  = 176.

 

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