# mental maths 9C4, 6^5 etc. [closed]

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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## closed as too localized by Austin Mohr, Ilya, azarel, Pedro Tamaroff♦, Grigory MAug 27 '12 at 18:48

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It sounds as though you want to be able to compute quickly in general, not just for these examples. In this case, there are entire books dedicated to the subject, making your question too broad for this community. – Austin Mohr Aug 27 '12 at 14:40
@AustinMohr People ask how they can get started in certain areas all the time, or say they've just learned calculus out of some calculus book and ask where they should go next, questions that are much more broad than this, yet do not seem to be too broad for this community. – Graphth Aug 27 '12 at 14:55
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. – Jyrki Lahtonen Aug 27 '12 at 15:31

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. 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. – process91 Aug 27 '12 at 15:26

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 $(1+x)^n$

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