Thousands of numbers can be multiplied by a computer in less than a second, but we human being can not. Easy availability of calculators, computers and smart phones have made us dependent on these gadgets - not allowing our minds to think how could we multiply faster. Here in this note a few useful multiplication tricks have been listed and we hope that it will help and engage students to figure out new tricks of multiplication. Throughout this article multiplication of two integers $x$ and $y$ are represented as $x.y$, whereas $xy$ is considered as a two digit number with $y$ being the one's digit and $x$ being the ten's digit. Same notation is followed when more than two integer are involved.

**7-trick**

For any integer $x$, $7x = (10x/2) + 2x$. So, double the integer $x$ (which is to be multiplied by $7$) and then add it with half of $10x$ to get the desired result. For example, $$132 . 7 = (1320/2) + 264 = 660 + 264 = 924.$$

**11-trick**

What is the result of multiplying any two digit number $xy$ by $11$?

$$11 . xy = x(x+y)y \hspace{0.2in} if \hspace{0.2in} x + y \leq 9 $$

$$11 . xy = x(x+y)y \hspace{0.2in} if \hspace{0.2in} x + y \leq 9 $$

$$ \hspace{0.7in} = (x+a)by \hspace{0.2in} if \hspace{0.2in} x + y = ab \geq 10 $$

For example, $$(i) \hspace{0.25in} 14.11 = 1(1+4)4 = 154$$

For example, $$(i) \hspace{0.25in} 14.11 = 1(1+4)4 = 154$$

$$(ii) \hspace{0.25in} 86.11 = (8+1)46 = 946$$

This trick can be generalized for any number easily. $$11.X_1X_2X_3X_4X_5 = X_1(X_1 + X_2)(X_2 + X_3)(X_3 + X_4)(X_4 + X_5)X_5,$$ if all of $(X_1 + X_2)$, $(X_2 + X_3)$, $(X_3 + X_4)$, and $(X_4 + X_5)$ are less than equal to $9.$ When the sums $(X_1 + X_2)$, $(X_2 + X_3)$, $(X_3 + X_4)$, $\ldots$, etc are greater or equal to $10,$ then let us consider the following.

$$X_4 + X_5 = a_1b_1$$ $$X_3 + X_4 = a_2b_2 + a_1 = a_2{'}b_2{'}$$ $$X_2 + X_3 = a_3b_3 + a_2{'} = a_3{'}b_3{'}$$ $$X_1 + X_2 = a_4b_4 + a_3{'} = a_4{'}b_4{'}$$

Then the result of multiplication is $= (X_1 + a_4{'})b_4{'}b_3{'}b_2{'}b_1X_5$. For example, $83664.11 = 920304.$

**13-trick**

This trick can be generalized for any number easily. $$11.X_1X_2X_3X_4X_5 = X_1(X_1 + X_2)(X_2 + X_3)(X_3 + X_4)(X_4 + X_5)X_5,$$ if all of $(X_1 + X_2)$, $(X_2 + X_3)$, $(X_3 + X_4)$, and $(X_4 + X_5)$ are less than equal to $9.$ When the sums $(X_1 + X_2)$, $(X_2 + X_3)$, $(X_3 + X_4)$, $\ldots$, etc are greater or equal to $10,$ then let us consider the following.

$$X_4 + X_5 = a_1b_1$$ $$X_3 + X_4 = a_2b_2 + a_1 = a_2{'}b_2{'}$$ $$X_2 + X_3 = a_3b_3 + a_2{'} = a_3{'}b_3{'}$$ $$X_1 + X_2 = a_4b_4 + a_3{'} = a_4{'}b_4{'}$$

Then the result of multiplication is $= (X_1 + a_4{'})b_4{'}b_3{'}b_2{'}b_1X_5$. For example, $83664.11 = 920304.$

It is now easy to devise a trick for $13$ as well. Note that, $$13.X_1X_2X_3X_4X_5 = 11.X_1X_2X_3X_4X_5 + 2.X_1X_2X_3X_4X_5$$.So, double the number $X_1X_2X_3X_4X_5$ is added with the result of multiplying the digit by $11.$

Multiply any three digit number $xyz$ with the numbers $7$, $11$ and $13$ respectively. Te result of the multiplication is the number $xyzxyz$. For example, $432.7.11.13 = 432432$, and this is essentially because $7.11.13 = 1001.$

**3367-trick**

Multiply any two digit number $xy$ with $3367$. The result of the multiplication is $\frac{1}{3}xyxyxy$, i.e., the original number written three times consecutively and the resulting number is then divided by $3.$

For example, $83.3367 = \frac{1}{3}.838383 = 279461$. This is because, $xy.10101 = xyxyxy$.

For example, $83.3367 = \frac{1}{3}.838383 = 279461$. This is because, $xy.10101 = xyxyxy$.

$$\hspace{0.17in}=> xy.(3.3367) = xyxyxy$$ $$\hspace{0.05in}=> xy.3367 = \frac{1}{3}xyxyxy$$

**Squaring tricks**

(A) $(X5)^2 = YZ25, YZ = X.(X+1)$, where $X$ is any digit between $1$ to $9$. For example, $(65)^2 = 4225$, note $YZ = 42 = 6.7$.

(B) $(5X)^2 = abcd$, where $ab = 25 + X$ and $cd = X^2$, where $X$ is any digit between $1$ to $9$. For example, $(54)^2 = 2916$ and note that $ab = 25 + 4 = 29$ and $cd = 4^2 = 16$.

(C) Squaring a number $N$ made up of $6$'s only. Let $c$ be the number of $6$'s in the number $N$. Then $N^2$ is equal to $(c-1)$ number of consecutive $4$'s followed by a $3$, which is again followed by $(c-1)$ number of consecutive $5$'s, followed by a $6$ at the end. For example, $$6^2 = 36$$ $$(66)^2 = 4356$$ $$(666)^2 = 443556$$ $$(6666)^2 = 44435556$$

(D) Lastly, we will see a general trick to square quickly any number $N$. Note that, $N^2 = (N - y) (N + y) + y^2.$ We are now free to choose this $y$ to help us square quickly. For example, $$33^2 = 30.36 + 3^2 = 1089$$ $$44^2 = 40.48 + 4^2 = 1936$$ $$123^2 = 100.146 + 23^2 = 14600 + 20.26 + 3^2 = 15129$$

(B) $(5X)^2 = abcd$, where $ab = 25 + X$ and $cd = X^2$, where $X$ is any digit between $1$ to $9$. For example, $(54)^2 = 2916$ and note that $ab = 25 + 4 = 29$ and $cd = 4^2 = 16$.

(C) Squaring a number $N$ made up of $6$'s only. Let $c$ be the number of $6$'s in the number $N$. Then $N^2$ is equal to $(c-1)$ number of consecutive $4$'s followed by a $3$, which is again followed by $(c-1)$ number of consecutive $5$'s, followed by a $6$ at the end. For example, $$6^2 = 36$$ $$(66)^2 = 4356$$ $$(666)^2 = 443556$$ $$(6666)^2 = 44435556$$

(D) Lastly, we will see a general trick to square quickly any number $N$. Note that, $N^2 = (N - y) (N + y) + y^2.$ We are now free to choose this $y$ to help us square quickly. For example, $$33^2 = 30.36 + 3^2 = 1089$$ $$44^2 = 40.48 + 4^2 = 1936$$ $$123^2 = 100.146 + 23^2 = 14600 + 20.26 + 3^2 = 15129$$

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