1. Multiplier is 9, 99, 999, 9999, ….
17 X 9 = 153, 17 X 999 = 16983
Here, we treat 17 as m (multiplicand) and 9, 999,…. (multiplier) as n
1.1. no of digits of n >= no of digits of m
Answer : (m – 1) | (n - LHS)
eg. 17 X 999
Answer: (17 - 1) | (999 - LHS) >> 16|(999 - 16) >>> 16983
Sly 17 X 99 >> 16|(99-16) >> 1683
1.2. else
Therefore we split m into p|q where no of digits of q = no of 9’s in n
Answer : (m - 1 - p) | (n + 1 - q)
Eg. 177 X 99
Split >> 1 | 77
Answer: (177 – 1 - 1) | (99 + 1 - 77) >> 175|23 >> 17523
Sly, 243 X 9 >> 24 | 3 >> (243-1-24)|(9+1-3) >> 218|7 >>2187
2. Multiplier = ca, multiplicand = cb & a+b = 10, 100, 1000, ….
ie total of last digits is 10, 100, 1000, … & whose prev part is exactly same
Answer: c(c+1) | ab
Eg. 237 X 233
Here c = 23, a = 7, b = 3
Answer: 23 X 24 | 7 X 3 >> 552 | 21 >>55221
Sly, 54 X 56 >> 30|24>>3024
3. Multiplicands & multiplier are near base
Find 2 small multipliers($) with signs
3.1. Base is 10, 100, 1000, ….
Answer: (base+ sum of 2 small multipliers) | (product of 2 small multipliers : Note base)
For, 97 X 105,
Base is 100, and $ are -3, +5
Answer: (100 + (-3+5) ) | -15 >> 102 | -15
To remove –ve, take carry :: 101 | (100 -15) >> 101|85 >> 10185
Sly, 102 X 103 >> 105|06 (Since base is 100, therefore 2 digits after |)
3.2. Else (working base (wb))
Multiply or divide only LHS of answer
(if wb is greater,multiply else divide)
I always prefer to keep wb greater.
Consider 31 X 33
Here wb is 30 (& base is 10), $ are +1, +3
Therefore, (30+1+3) | 3 >> 34 | 3
Since wb greater, multiply by 30/10 = 3
Answer: 34X3 | 3 >> 102 | 3 >> 1023
4. General case : Cross Multiplication
4.1. 2 digit multiplication
ab X cd >>> ac | ad+bc | bd
Eg. 23 X 12 >> 2 | 4+3 | 6 >> 276
4.2. 3 digit multiplication
abc X def >>> ad | ae+bd | af+dc+be | bf+ce | cf
Eg. 103 X 143 >> 1 | 4 | 6 | 12 | 9 >> 14729 (1 carried forward)
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