Multiplication of large numbers, e.g. $97times 96$: why does this work?

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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!




arithmetic




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  • $(100-3)*(100-4)=10000-7*100+12$ as long as your 100-residuals multiply less than 100 the trick works
    – N74
    Aug 7 at 17:58










  • @N74- but 89*89 worked, and so did 89*87, as long as you just carried over the hundreds digit of the product to the hundreds place in the final number.
    – ç±³å‡¯ä¹
    Aug 7 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 7 at 18:14















up vote
1
down vote

favorite












enter image description here



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!




arithmetic




share|cite|improve this question





















  • $(100-3)*(100-4)=10000-7*100+12$ as long as your 100-residuals multiply less than 100 the trick works
    – N74
    Aug 7 at 17:58










  • @N74- but 89*89 worked, and so did 89*87, as long as you just carried over the hundreds digit of the product to the hundreds place in the final number.
    – ç±³å‡¯ä¹
    Aug 7 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 7 at 18:14













up vote
1
down vote

favorite









up vote
1
down vote

favorite











enter image description here



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!




arithmetic




share|cite|improve this question













enter image description here



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!





arithmetic





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edited Aug 7 at 17:59









amWhy

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米凯乐

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  • $(100-3)*(100-4)=10000-7*100+12$ as long as your 100-residuals multiply less than 100 the trick works
    – N74
    Aug 7 at 17:58










  • @N74- but 89*89 worked, and so did 89*87, as long as you just carried over the hundreds digit of the product to the hundreds place in the final number.
    – ç±³å‡¯ä¹
    Aug 7 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 7 at 18:14

















  • $(100-3)*(100-4)=10000-7*100+12$ as long as your 100-residuals multiply less than 100 the trick works
    – N74
    Aug 7 at 17:58










  • @N74- but 89*89 worked, and so did 89*87, as long as you just carried over the hundreds digit of the product to the hundreds place in the final number.
    – ç±³å‡¯ä¹
    Aug 7 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 7 at 18:14
















$(100-3)*(100-4)=10000-7*100+12$ as long as your 100-residuals multiply less than 100 the trick works
– N74
Aug 7 at 17:58




$(100-3)*(100-4)=10000-7*100+12$ as long as your 100-residuals multiply less than 100 the trick works
– N74
Aug 7 at 17:58












@N74- but 89*89 worked, and so did 89*87, as long as you just carried over the hundreds digit of the product to the hundreds place in the final number.
– ç±³å‡¯ä¹
Aug 7 at 18:07




@N74- but 89*89 worked, and so did 89*87, as long as you just carried over the hundreds digit of the product to the hundreds place in the final number.
– ç±³å‡¯ä¹
Aug 7 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 7 at 18:14





$(100−a)(100−b)=(100−a-b)100+ab$. 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 7 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 $$(97)(96)=(10^2-3)(10^2-4)=10^2(10^2-(3+4))+3cdot 4= (100)(93) +12.$$



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






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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 7 at 18:23











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

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up vote
1
down vote



accepted










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



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



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






share|cite|improve this answer





















  • 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 7 at 18:23















up vote
1
down vote



accepted










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



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



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






share|cite|improve this answer





















  • 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 7 at 18:23













up vote
1
down vote



accepted







up vote
1
down vote



accepted






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



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



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






share|cite|improve this answer













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



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



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







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Aug 7 at 18:18









DanielWainfleet

31.8k31644




31.8k31644











  • 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 7 at 18:23

















  • 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 7 at 18:23
















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 7 at 18:23





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 7 at 18:23













 

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