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What would be the best approach to solve 73 x 42 in my head?

I started with 70 x 40 and then 3 x 40 and combined, but at this point I forgot what I had done and ended up getting lost and not figuring it out.

Is there a good method for solving multiplication as such in my head?

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    Multiplying large numbers always requires keeping at least of couple of numbers in your head. I suggest just practicing over and over. If you can't do two-digit by two-digit numbers in your head yet, start with two-digit by one-digit numbers. Then when you're comfortable with those, try go up to two-digit by two-digit where both numbers are under $20$. Then go up to two-digit by two digit where only one number is under $20$. Keep increasing like this. Basically you're trying to remap your synaptic connections and that takes time. Just keep practicing every day and you'll... –  Oct 05 '15 at 15:56
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    keep getting better and better. Then one day you'll realize that now you're able to multiply very large numbers in your head easily. –  Oct 05 '15 at 15:56

5 Answers5

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Many mental calculators do mutliplications from left to right this way : \begin{array}{r} 73\\ \times\; 42 \end{array}

\begin{array}{c l} \text{'cross' computation} &\text{ partial result}\\ 7\cdot 4=28 & 28\\ 7\cdot 2+3\cdot 4=26 & 306\\ 3\cdot 2=6 &3066 \end{array}

Other example : \begin{array}{r} 237\\ \times\;543 \end{array}

\begin{array}{c l} \text{'cross' computation} &\text{ partial result}\\ 2\cdot 5=10 & 10\\ 2\cdot 4+3\cdot 5=23 & 123\\ 2\cdot 3+3\cdot 4 +7\cdot5=53 &1283\\ 3\cdot 3+7\cdot 4=37 & 12867\\ 7\cdot 3=21 & 128691\\ \end{array}

To compure the square of a number use $a^2=(a+b)(a-b)+b^2$.
For example $$78^2=80\cdot 76+2^2$$

Other methods and examples in this MSE thread.

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I would double $73$ to get $146$, and then double it again to get $292$. I take advantage of the fact that doubling a number in your head is relatively easy since you only have to remember one number at a time. Then I would multiply that result by ten to get $2920$. This is $73\times 40$. Now I just need to add $2\times 73$, which we already calculated as $146$. Adding, we get $2920+146=3066$.

Plutoro
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Your method is good. But keep the idea that (x10 + y)(a10 + b) = xy*100 + (10a)y + (10x)b + by. In other words there will be four things to do.

So 73*42. 1) 70*40 + 2)70*2 + 3)3*40 + 4) 2*3

which is still hard.

but you can do short cuts to simplify. Try to get close to 50 and 25 because those factor into 100 nicely:

73*42 = 73*(50 - 8); 73*50 is 7300/2 is 3650. Put that in the back burner.

Subract 8*72. That's 4*144 = 2*288 = 560 + 16 = 576

Subtract from, what was it, oh yeah, 3650. 3650 - 576 = (3650-500) - 76 = 3150 - 76 = 3100 - 26 = 3066.

There's also:

73*42 = (75 - 2)*42; 75*42 = 150*21 = 3000+ 150 = 3150. 2*42 = 84. 3150 - 84 = 3100 - 34 = 3066.

fleablood
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Another way of doing these is to change the numbers to something close but simpler to calculate, add or subtract the required amounts, and expand brackets, so in this case $$73\times 42=(75-2)(40+2)=75\times 40-2\times 40+2\times 75-4=3000+70-4=3066$$

David Quinn
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No two people's neurons are wired the same way, but if you want to multiply 2-digit numbers together in your head, you can try using the following idea.

Given

$\quad n \,= 10 n_1 + n_0 \quad \;\;\,\text{ with } n_1, n_0 \;\,\in \{1,2,3,4,5,6,7,8,9\}$ $\quad m = 10 m_1 + m_0 \quad \text{ with } m_1, m_0 \in \{1,2,3,4,5,6,7,8,9\}$

To multiply $n \times m$,

1). Multiply the left digits and store in your L-register,

$\tag 1 L = n_1 \times m_1$

Regard the register $L$ as having a length of $2$ (pad on the left with $0$ if necessary).

2). Multiply the right digits and store in your R-register,

$\tag 2 R = n_0 \times m_0$

Regard the register $R$ as having a length of $2$ (pad on the left with $0$ if necessary).

3.) Split the $R$ register string into your $R1$ and $F$ registers,

$\tag 3 R = R1 \, || \, F$

Release the $R$ register to 'clear' memory.

4.) Concatenate the $R_1$ digit to the right of the $L$ and store in your T-register,

$\tag 4 T = L \, || \, R_1$

Release the $R_1$/$L$ registers to clear memory.

5). Add the '-outers' and '-inners' into your T1-register,

$\tag 5 T1 = [n_1 \times m_0] + [n_0 \times m_1]$

The number $T1$ in $\text{base-}10$ has length of at most $3$.

6). Add the two registers $T$ and $T1$ together, storing the results in 3-digit register $G$.

7). Say or think the answer,

$\tag 6 G \, || \, F$


Example 1: $73 \times 42$

Combining steps, you are saying in your head $280$, and, if nobody is looking, using your fingers to store $6$.

You add $14$ (outers) to $12$ (inners), saving $26$.

You add $280 + 26$ giving $306$.

You say the answer, $3,066$.

Example 2: $18 \times 17$

Combining steps, you are saying in your head $015$, and, if nobody is looking, using your fingers to store $6$.

You add $7$ (outers) to $8$ (inners), saving $15$.

You add $015 + 15$ giving $30$.

You say the answer, $306$.

Example 3: $64 \times 43$

Combining steps, you are saying in your head $241$, and, if nobody is looking, using your fingers to store $2$.

You add $18$ (outers) to $16$ (inners), saving $34$.

You add $241 + 34$ giving $275$.

You say the answer, $2,752$.

Example 4: $97 \times 97$

Combining steps, you are saying in your head $814$, and, if nobody is looking, using your fingers to store $9$.

You add $63$ (outers) to $63$ (inners), saving $126$.

You add $814 + 126$ giving $940$.

You say the answer, $9,409$.


The starting point for this technique can be found here:

$\quad$ What is the fastest way to multiply two digit numbers?

CopyPasteIt
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