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I've seen various versions of the above image floating around social media on how to multiply large numbers in your head. I know it doesn't work for all numbers, but why does it work for some numbers and what is the rule for knowing whether it will work or not for a certain pair of numbers? A brief Google search didn't pull anything up for me, but perhaps I missed something. I tried multiplying 97 and 96, 98 and 99, 90 and 90, 89 and 89, 89 and 87, and then I tried 50 and 50, the first for which it didn't work for me, and I could not figure out why. Thanks so much!

amWhy
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米凯乐
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  • $(100-3)(100-4)=10000-7100+12$ as long as your 100-residuals multiply less than 100 the trick works – N74 Aug 07 '18 at 17:58
  • @N74- but 8989 worked, and so did 8987, as long as you just carried over the hundreds digit of the product to the hundreds place in the final number. – 米凯乐 Aug 07 '18 at 18:07
  • $(100−a)(100−b)=(100−a-b)100+ab$. Note that $(100-a-b)$ is at most a two digit number, $ab$ should have at most two digits for concatenation to work. Otherwise you have to worry about the carry. – karakfa Aug 07 '18 at 18:14

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$(A-B)(A-C)\;=\;A^2-AB-AC+BC\;=\;A(A-(B+C))+BC.$

Therefore, with $A=10^2$ and $B=3$ and $C=4$, we have $$(97)(96)=(10^2-3)(10^2-4)=10^2(10^2-(3+4))+3\cdot 4= (100)(93) +12.$$

Another example: $(91)(92)=10^2(10^2-(9+8))+9\cdot8=8300 +72.$

DanielWainfleet
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  • There are many other algebraic short-cuts for arithmetic. E,g $(x+y)(x-y)=x^2-y^2,$ so $(33)(27)=(30+3)(30-3)=30^2-3^2=900-9=891. $ If you want to see some arithmetical "magic" see The Trachtenburg Speed System of Basic Mathematics. – DanielWainfleet Aug 07 '18 at 18:23