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

Question

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!

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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!

Answer 1

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.

Answer 2

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. $79\times 71=(7\times 8)+09=5609$, $24\times 26=(2\times 3)+24=624$, $65\times 65=(6\times 7)+25=4225$

Answer 3

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

Answer 4

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.

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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.

Answer 5

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 $9$