Multiplication

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Decimal multiplication tricks for quick maths

Bracket notation

Let the square brackets represent in-place decimal expansion such that $[1][2][3]=123=1\ast 10^2+2\ast 10^1+3\ast 10^0$ For simplicity, let this notation allow any number greater than 9 in a brackets, and define it to mean the familiar: “leave the remainder in-place, carry the quotient to the left. $1[72][33]=[1+7][2+3][3]853$ Note: this extends seamlessly to quotients of an number of decimal digits, in this example we are adding a quotiont of size $10$ to the $[43]$ $[43][99][99]=[43][99+9][9]=43[108][9]=[43+10][8][9]=5389$

Sub 12 multiplier

Times 4

Double then double again. Or if its easy do it directly times 4.

$23234\ast 4=46468\ast 2=92936$

Times 5

Double then times 10 $5\ast 9348=4674\ast 10=46,740$

Times 6

When you multiply 6 times an even number, they will end in the same number. $534\ast 6=3000+180+24=3204$

Times 9

Single digit: You don’t even need to say the number outloud. $a\in [0-9]\ast 9=[a-1][10-a]$

Times 11

Multiplying any two digit number times 11 is easy:

$93\ast 11=[9+Q(9+3)][R(9+3)]3$ Where:

• $Q(a,b)$ is the quotient of $\frac{a+b}{10}$
• $R(a,b)$ is the remained of $\frac{a+b}{10}$

Multiply times a three digit number is also the same idea: $793\ast 11=[7][7+9][9+3][3]=8723$ Where we keep the remainder in place and shift the quotient one place to the left.

Times 12

$a\ast (12)=a\ast 10+a\ast 2$

Two digit numbers multiplied

This actually extends to multiplying any two digit numbers together:

$[ab]\ast [cd]=[a\ast c][a\ast d+b\ast c][b\ast d]$

As for $0<a,b,c,d\le 9$:

$(a\ast 10+b)(c\ast 10+d)=a\ast c\ast 10^2+[a\ast d+b\ast c]\ast 10+c\ast d$

Examples

• $37\ast 76=21[3\ast 6+7\ast 7][7\ast 6]=21[67][42]=28[1][2]=2812$
• $53\ast 93=45[27+15][9]=49[2][9]=4929$
• $44\ast 46=16[24+16][24]=16[42][4]=2024$

Multiplying Teens

$13\ast 15=[13+5][3\ast 5]=18[15]=195$ We have $[ab]\ast [cd]=[ab+d][b\ast d{]}_{10}$ We will add the last digit of the secondary number to the primary number for the first two position. For the third position, multiply the last two digits and take remainder 10.

Examples

$19\ast 18=[1][9+8][9\ast 8]=27[72]=342$ $15\ast 13=[1][5+3][5\ast 3]=[1][8][15]=[1][9][5]=195$

This is a direct result of the following expansion: When we multiply two teens we have the following for $3\le a,b\le 9$ $(10+a)\ast (10+b)=10^2+(a+b)\ast 10+a\ast b$ This immediatly tells us our answer will have the following form: $[1][a+b{]}_{10}[a\ast b{]}_{10}$

And of course, we add the quotients. mod ten to the position to the left.

Warning

Note that anything multiplied by 19 is more easily computed as follows: $a\ast 19=a\ast 20-a$
Examples:

• $4324363\ast 19=86,487,260-4,324,363=82,162,897$
• $15\ast 19=300-15=285$

Special format [a][b]*[a][10-b]

In english: Matching first numbers and second numbers adding up to 10.

The trick:

$a(a+1)\times 10^2+b(10-b)\times 10$

This implies:

$[a][b]\ast [a][10-b]=[a\ast (a+1)][][b\ast (10-b)]$

The [] in the middle is to stay consistent with our made up notation, there are no carry forwards, just write down the exact numbers at their corresponding positions

Proof:

$$\begin{array}{rl}(a\ast 10+b)\ast (a\ast 10+(10-b))& =(10a+b)(10(a+1)-b)\\ & =a(a+1)\times 10^2+b(a+1)\times 10-ab\times 10-b^2\\ & =a(a+1)\times 10^2+b\times 10-b^2\\ & =a(a+1)\times 10^2+b(10-b)\times 10\end{array}$$

Examples

$95\ast 95=9025=9025$ $74\ast 76=[7\ast 8][][24]=[56][2][4]5624$

Squaring

pass

Three digit numbers multiplied

We are going to follow the following pattern, resulting in five of our inplace brackets [] Exemplified as follows:

Examples

$348\ast 579=[3\ast 5][3\ast 7+5\ast 4][3\ast 9+4\ast 7+8\ast 5][7\ast 8+4\ast 9][8\ast 9]=[15][41][95][92][72]=[20][1][4][9][2]=201,492$

Warning

Since we are dealing with inplace numbers that can exceed 100 in our bracket notation , the quotient has two digits, we pass it to the second remainder to the left of the current place. Or just pass is as is, as a 10 to the first remainder to the left and this will force you to pass it as a 1 to the next remainder.