# Is there an easy way to multiply two 2-digit numbers that have flipped digits?

I am currently a sophomore in high school competing in UIL academics in Texas. I am competing in the number sense test. Among the questions is the need to quickly multiply 2 two digits numbers, but as with all the questions present in the test, there is always a faster way to do things. One type of question that reoccurs in each test at every competition (I've gone to 3 so far (: ) is the need to multiply two 2-digit numbers together that have their digits flipped when compared to the other number(43*34 and 56*65). I was wondering if there was a faster way to do this than to mentally multiply them. Thanks for the help!

P.S. If there is a way to improve the formatting of the question, it would be greatly appreciated if you would kindly tell me!

• $(10a+b)\cdot (10b+a) = 101ab+10a^2+10b^2$. Is this more convenient? That's up to you to decide. E.g. $43\cdot 34 = 1212+160+90$. I don't personally see this as worth memorizing for everyday use, but perhaps for speed contests its useful enough... Mar 1, 2017 at 2:25
• I would think that directly multiplying them in your head is pretty fast... Mar 1, 2017 at 3:10

I propose the following visual scheme.

Edit: after reading the other answers, I noticed $\color{darkblue}{Arby}$ had the same idea.

To multiply the numbers $\overline{ab}\times\overline{ba}$

Start by laying the product $a\times b=\overline{pq}$ twice : $\overline{pq.pq}$

Then calculate $a^2+b^2=\overline{du}\quad$ call the unit $u$ and the decade $d$

So $\overline{ab}\times\overline{ba}=\overline{(pq+d+\varepsilon).(pq+u0)}$ where $\varepsilon=0,1$ is propagating the addition carry if any.

Examples:

$\begin{array}{l|l|l} 56\times65 & 5^2+6^2=25+36\to 6:1 & \overline{(30+6)(30+10)}=3640\\ 43\times34 & 3^2+4^2=9+16\to 2:5 & \overline{(12+2)(12+50)}=1462\\ 73\times37 & 3^2+7^2=9+49\to 5:8 & \overline{(21+5)(21+80)}=2\color{red}701\quad carry\\ 88\times88 & 8^2+8^2=64+64\to 12:8 & \overline{(64+12)(64+80)}=7\color{red}744\quad carry \end{array}$

Of course this is just a mean of presenting $10a^2+101ab+10b^2$, but I think it is easier to do mentally when proceeding this way.

The addition carrying may appear scary, yet in fact it happens only for $73,77,85,88,94,95,96,98,99$ and their mirrors, it is a simple addition in all other cases.

Let the two numbers be $xy$ and $yx$ and we want to calculate their multiplication. There is a very simple procedure.

$1.$ Multiply both the digits.

$2.$ Multiply both the digit and add $2$ trailing zeros.

$3.$ Find the sum of the square of both the digits and $1$ trailing zero.

Add the numbers obtained in above steps and you will get the answer.

Here is an example. Suppose I want to calculate $37\cdot73$. Calculation is as follows:

$1. \ 7\cdot 3=21$

$2. \ 2100$

$3. \ 7^2 + 3^2 = 49 + 9 = 58$ after adding a zero it becomes $580$.

Answer is $21+2100+580=2701.$

P.S. You need to know the squares of all the digits from $0$ to $9$

Algebraically, you're asking about a shortcut for $(10a+b)(10b+a)$. That works out to: $$10a^{2}+101ab+10b^{2}$$

101 isn't hard to multiply by, especially considering that $a \times b$ will never be more than a 2-digit number in this scenario.

If you can square and slap a 0 on the end, you can handle the first and last parts.

Let's try $56 \times 65$ from your example. That's $250 + 360 + 3030 = 610 + 3030 = 3640$.

Get the idea?

There's only 41 different such 2-digit problems, so memorization might also be considered. 9 of those 41 are really just mentally squaring multiples of 11 up to 99, so that really only leaves 32 such problems.

Similarly from Vedic math I've seen before, the product of $a\times10+b$ and its transpose is $ab/(a^2+b^2)/ab$. So for $73\times37$ you would first multiply 3 and 7 getting and remembering 21, next $3^2+7^2=9+49=58$ which would actually be 580 (why?). You would add the 21 getting 601. Next use $3\times7=21$ again, but this time you get 2100 (again, why?). Adding 601 you get the final answer 2701.

This method generalizes well. $ab\times cd=ac/(ad+cb)/bd$ You can figure out formulas for more digits as well.

Edit: by the way, and another easy to remember trick from Vedic math is mutltiplying two 2-digit numbers with the same first digit and whose last digits sum to 10. In this case $ab\times ac$ is simply the first number times one more than the first number, followed by the product of the last numbers (using a zero in the tens digit if necessary). Easier to show than to explain. $79\times 71=(7\times 8)+09=5609$, $24\times 26=(2\times 3)+24=624$, $65\times 65=(6\times 7)+25=4225$

• "Vedic math" sounds very complicated for the standard way everyone learns to multiply in primary school (at least where I came from). You multiply the ones digits, you multiply the cross terms (appending a 0) and you multiply the tens digits (appending two 0s) and add everything up. It's not really a trick.
– TMM
Mar 1, 2017 at 5:08
• Interesting that you learned that way. I learned a different way, making two rows and adding--this applied to many digit numbers, adding rows for every extra digit on the bottom number.
– Arby
Mar 1, 2017 at 5:33

If you are like me, who knows all the squares of two digit numbers, then you can do it like this: $$36\times 63=36\times(100-36-1) = 3600-1296-36 = 2268.$$

• Subtracting twice seems hardly more convenient than adding twice (which would be what you'd normally do).
– TMM
Mar 1, 2017 at 5:09