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How can I calculate the following things in my head?

  • $_9C_4$ I know this is $\frac{9\cdot 8\cdot 7\cdot 6}{4\cdot 3\cdot 2\cdot 1}$ and then $3\cdot 2\cdot 3\cdot 7$ but I can't immediately come up with that calculation. Are there some quicker ways do this?

  • $6^5$

  • $68\cdot 27$

I don't want to rely on my calculator anymore. Can anyone give me some mental tricks?

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    Powers of small integers ($2$ in particular) come up frequently enough to be just memorized. A younger version of yours truly would have done your last example (mental arithmetic only) as: $68\cdot27=68\cdot30-68\cdot3=2040-204=1836.$ Multiplying a double digit integer by three is simple enough, and can be reused here.2012-08-27

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There are books on the subject, but I’m not familiar with them, having developed techniques of my own that I find adequate for my needs. You might look into the Trachtenberg system for high-speed arithmetic if speed of computation is high on your list of goals; I’ve always been more interested simply in being able to do a reasonably broad range of mental arithmetic. Most of my techniques involve intelligent rearrangement of calculations, which depends greatly on the specific numbers involved, so I can’t easily give you general principles. Here, for what it’s worth, is how I might perform these three calculations mentally.

  1. $\binom94=\frac{9\cdot8\cdot7\cdot6}{4\cdot3\cdot2}=\frac{9\cdot7\cdot6}3=3\cdot7\cdot6=3\cdot42=126$.

  2. $6^5=216\cdot6\cdot6=\left(200\cdot6+16\cdot6\right)\cdot6=(1200+96)\cdot6=1296\cdot66\cdot1300-6\cdot4=$ $7800-24=7776$. (I might actually remember that $6^4=1296$, which would save some time.)

  3. $68\cdot27=1200+420+160+56=1620+160+56=1780+56=1836$, or
    $68\cdot27=70\cdot27-2\cdot27=70\cdot30-3\cdot70-54=2100-210-54=1900-10-54$ $=1900-64=1836$.

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    The Trachtenberg system shares many similarities to [Vedic mathematics](http://en.wikipedia.org/wiki/Bharati_Krishna_Tirtha's_Vedic_mathematics). One difference I noted was that in Vedic math, everything is practiced so that you can derive the most significant digit first, allowing you to have the appearance of just reading off the result much faster. Accuracy is more important than speed, IMHO.2012-08-27
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Similar to Brian M. Scott's suggestion, there are many tricks. The relative usefulness depends on what calculations you want to do. Are you specialized to integer calculations with exact results? In that case some of the below won't help.

The more arithmetic facts you know, the more likely you can find one to help. I find knowing all the perfect powers up to 1000, powers of 2 up to 2^16, factorizations of numbers up to 100 (especially which are prime), factorials to 8,and divisibility tests useful. Great facility with $a^2-b^2=(a+b)(a-b)$ is essential. Common trig values and Pythagorean triangles come in handy. When doing multidigit multiplies I find it easier to start from the most significant digits. You can quit when you have enough accuracy.

Approximations: $(1+x)^n \approx 1+nx$ for $x \ll 1$ is the biggest hitter. Depending on the accuracy you need, $\pi = \sqrt {10}$ and both might equal $3$. $e\approx 3(1-0.1)$ which can feed into the $(1+x)^n$