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I wish to teach myself how to do math calculations–like $99\times 58$ or $2048+1296$ or $506+998$–inside my head.

I know there already exist two methods–Abacus and Vedic Maths. I don't know of more, nor I have had experience with any.

Has someone had experience with one or more of such techniques? Which one works best for a person with background of being a high school passout?

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  • $\begingroup$ This question isn't a duplicate of this because the other person's calculations were undoubtedly much simpler (8/4=2; 2x3=6; ...); I have the intuition for numbers, but I need to know of techniques to make calculations simpler. $\endgroup$
    – Gaurang Tandon
    Commented Dec 1, 2017 at 2:25
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    $\begingroup$ Suggest this is a suboptimal use of time. $\endgroup$
    – Daniel R. Collins
    Commented Dec 1, 2017 at 5:05
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    $\begingroup$ If you use an abacus, then you wouldn't be doing all your calculations "inside your head." You might as well use an electronic calculator. $\endgroup$
    – JRN
    Commented Dec 1, 2017 at 6:46
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    $\begingroup$ @JoelReyesNoche have a look at this video I found from a quick google search, in the latter part you'll notice several kids doing calculations without any "physical" abacus. Once practiced sufficiently, they can operate on a "virtual" abacus ^_^ $\endgroup$
    – Gaurang Tandon
    Commented Dec 1, 2017 at 11:45
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    $\begingroup$ @DanielR.Collins I don't see how optimizing the several common calculations you may perform in a day is a suboptimal use of time :/ Please elaborate your comment. $\endgroup$
    – Gaurang Tandon
    Commented Dec 1, 2017 at 11:55

7 Answers 7

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One technique that I like is "chunking", breaking a calculation into easier pieces.

For example,

$$99 \times 58 = (100 - 1) \times 58 = 5800 - 58=5742$$

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  • $\begingroup$ Even though it might sound like preschool math, it might actually help to use your fingers here to keep numbers in mind; e.g. in the given example, while computing $99 \times 50 = 100\times 50 - 1\ times 50 = 5000 - 50 = 4950$, it might be nice to "store" the eight without having to think about it. $\endgroup$
    – Dirk
    Commented Dec 1, 2017 at 10:21
  • $\begingroup$ This is a great technique, but I surely believe this isn't the only out there - since we generally do several more things than multiplications, most common ones being dealing with fractions and square roots. $\endgroup$
    – Gaurang Tandon
    Commented Dec 1, 2017 at 11:52
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For the calculation at the hand, the technique by Boba Fret is probably the way to go. In general, there are different techniques.

A source that I like is the book by Arthur Benjamin and Mike Shermer called "The secrets of mental math".

For completeness: Next to the Abacus and Vedic Maths you mention, one could add the Trachtenberg system; some resources provided at the end.

Since this is a question in MESE, I'd dare to add: When teaching mental math to students, don't miss the opportunity to explain the techniques based on the understanding about place value and (pre)-algebra. Unfortunately I don't see this done in many resources. For example, the resources on Vedic Math usually just state their "sutras" as rules to be used, like the "vertical and crosswise", or "all from nine, last from ten". While this may be great for memorization, it should also be discussed why they work. For example, the distributive rule helps, together with an understanding of place value system, to explain a technique to square numbers ending in 5's, which is simple for 2-digit numbers:

The how: To square, for example $65$, increase the tenth-digit by one and multiply, then append 25: $6 \times 7 = 42$, so $65^2 = 4225$.

The why: Write the number in place values, and use the distributive rule to rearrange:

$$\begin{align*} 65^2 &= (60 + 5)(60+5) \\ &= 60 \times 60 + 5 \times 60 + 5 \times 60 + 5 \times 5\\ &= 60 \times(60 + 5 + 5) + 25\\ &= 6 \times (6+1) \times 100 + 25 \end{align*}$$

Now, the last line can be read to explain the technique to square numbers ending in 5.

This way, techniques for mental calculation can be based on conceptual understanding, as opposed to "mystic rules" from ancient India.

Edit as requested, some resources:

Resources for the Trachtenberg System:

Resources for Vedic Math:

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    $\begingroup$ +1 for the point that you should not only teach rules but also why they work. It happens way too often that I see people posting rules that only work in very special cases (e.g. three digit numbers above 900) and assuming that they are generally true, sometimes even defending an obviously wrong result they obtained with them. $\endgroup$
    – Dirk
    Commented Dec 1, 2017 at 10:19
  • $\begingroup$ Thank You very much. This is exactly my experience too and the reason adding it in my answer. $\endgroup$
    – SCS
    Commented Dec 1, 2017 at 10:34
  • $\begingroup$ You made a good point. Apart from the book you mentioned, do you have links to good sources to study the other three systems (abacus, vedic maths, tracternberg)? $\endgroup$
    – Gaurang Tandon
    Commented Dec 1, 2017 at 11:54
  • $\begingroup$ Thanks. To the abacus, I don't have any specific sources. To the others, I'll edit my answer accordingly. $\endgroup$
    – SCS
    Commented Dec 1, 2017 at 16:13
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    $\begingroup$ By the way, an imho a bit easier justification of the $65^2$ result using the third binomial formula is $$60\cdot 70=(65-5)(65+5)=65^2-5^2=65^2-25$$ so $$65^2=60\cdot 70+25$$ $\endgroup$
    – Photon
    Commented Dec 2, 2017 at 9:09
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When I do math in my head, I usually combine a few approaches:

  • Always keep track of units.
  • Try to re-use math I have memorized (such as the multiplication tables, powers of 2, powers of 3, common square roots, common trig ratios, et cetera)
  • Use "carrying" to split each number into a round number and an "error" term. Often, the first operations will be on the number(s) that are closest to round numbers.
  • An "error term" can either be a number (such as 2 or -4 or 0.216), or a percentage of the round number. For example, 0.3048 is 1.6% more than 0.3. In the example below, I convert "0.3048" to "0.3 (+ 1.6%)". A more standard way of writing this would be "0.3 * (1 + 1.6%)".
  • Use memorized values to handle round number inputs to square roots or trig functions, and interpolation to adjust for the "error" terms.
  • Chunk four-digit numbers into pairs of digits.
  • Use my fingers to store values as Roman numerals. (This allows storing a pair of digits on my hands: Right fingers = Is; right thumb = V; left fingers = Xs; left thumb = L.)

For example:

  • 99 * 58 = (100 - 1) * 58 = 5800 - 58 = 5700 + 100 - 58 = 5742
  • 2048 + 1296 = 2048 + 1300 - 4 = 2044 + 1300 = 3344
  • 506 + 998 = 506 + 1000 - 2 = 504 + 1000 = 1504

Since the original poster asked (in a comment) about square roots:

$$ \sqrt{\frac{2 \cdot 0.3048 \text{ meters}}{(9.80665 \frac{\text{meters}}{\text{second}^2})}} $$ $$= \sqrt{\frac{2 \cdot 0.3048 \text{ second}^2}{9.80665}} $$ $$= 1 \text{ second} \sqrt{\frac{2 \cdot 0.3048}{9.80665}} $$ $$\approx 1 \text{ second} \sqrt{\frac{2 \cdot 0.3 (+ 1.6\%)}{10 (- 2\%)}} $$ $$\approx 1 \text{ second} \sqrt{\frac{0.6}{10} (+1.6\% +2\%)}$$ $$= 1 \text{ second} \sqrt{0.06 (+1.6\% +2\%)}$$ $$= 0.1 \text{ second} \sqrt{6 (+3.6\%)}$$ $$= 0.1 \text{ second} \sqrt{6.216}$$ $$= 0.01 \text{ second} \sqrt{621.6}$$ $$\approx 0.01 \text{ second}\cdot (\sqrt{625}-\frac{3.4}{625-576})$$ $$\approx 0.01 \text{ second}\cdot (25-\frac{3.4}{50})$$ $$= 0.01 \text{ second}\cdot (25-\frac{6.8}{100})$$ $$\approx 0.01 \text{ second}\cdot (25-0.07)$$ $$= 0.01 \text{ second}\cdot 24.93$$ $$= 0.2493 \text{ seconds}. $$ (Actual value is 0.249322 seconds, at 45° latitude on Earth.)

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  • $\begingroup$ I am unsure how exactly you're going about this - "Use my fingers to store values as Roman numerals. (This allows storing a pair of digits on my hands.)" Can you please give an example? $\endgroup$
    – Gaurang Tandon
    Commented Dec 5, 2017 at 1:43
  • $\begingroup$ +1 for your detailed math approximation method but I believe, in your approximation, you could've said the given expression is reasonably close to $\sqrt{0.03\cdot2/10}=\sqrt{0.06}=0.24494$ which is accurate upto two decimal places. $\endgroup$
    – Gaurang Tandon
    Commented Dec 5, 2017 at 1:55
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    $\begingroup$ @GaurangTandon -- Actually, my first mental approximation was $\sqrt{\frac{2 \cdot 0.3}{10}} = \sqrt{0.06} \approx \sqrt{0.0625} = \sqrt{\frac {1}{16}} = \frac {1}{4}$ second. This value is good for many purposes, including sanity-checking the more complicated mental approximation. $\endgroup$
    – Jasper
    Commented Dec 5, 2017 at 6:45
  • $\begingroup$ Oh, that's also pretty close, thanks! $\endgroup$
    – Gaurang Tandon
    Commented Dec 5, 2017 at 10:37
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Adding to the proposal of BobaFret: I would first calculate the big part of the number, then the small part. For example: $$2048+1296=?$$ First do the big part: $$2000+1200=3200$$ Then the small part: $$48+96=44+100=144$$ So in total it is $$2048+1296=3200+144=3344$$ where in the small part calculation I transferred a 4 from 48 to 96 to get to a nice number. This transferring is quite helpful as well, actually:

$$506+998=504+1000=1504$$

The reason for doing the big part first is that it is usually the easier part (3200 is easier to memorize than 144) and also it gives you a rough estimate of the result which makes it easier to memorize as well (3200 feels like it is of the right order of magnitude for the result of 2048+1296 while 144 is just some random number appearing in the calculation).

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    $\begingroup$ Surprisingly, I've always done the reverse way since my childhood. So, I instead would have started off by adding 48 and 96 together. This helped me keep sequential track of my carry overs as well. Side note: in the given example in question, I would instead have simply done $2048+1296=2044+1300=3344$... $\endgroup$
    – Gaurang Tandon
    Commented Dec 5, 2017 at 1:58
  • $\begingroup$ Seems like a matter of taste then! $\endgroup$
    – Photon
    Commented Dec 5, 2017 at 8:37
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Here's how I did mental multiplication. When I was in my grade 9 math class, my teacher was asking single digit multiplications and I keep getting the answer really quickly and I think I kept being the first to answer the question because I had already memorized the single digit multiplication table. I'm not sure whether I made the conscious effort to memorize it. Later in 2003 while I was at Magic Camp located at Camp White Pine, I was getting asked multiplication questions. I think some of them were 3 digits by 3 digits and I mentally calculated the answer by the following method. I applied the property that $\forall a \in \mathbb{R}\forall b \in \mathbb{R}\forall c \in \mathbb{R}\forall d \in \mathbb{R}\forall e \in \mathbb{R}\forall f\in \mathbb{R} (100a + 10b + c) \times (100d + 10e + f) = 10,000ad + 1,000(ae + bd) + 100(af + be + cd) + 10(bf + ce) + cf$ I wrote that as a 5 term sum. I treated the digits like those variables and added the last term first then the second last term and retained what I got so far and kept going. I think I got the right answer to so many multiplication questions.

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As was already mentioned by @SCS, Art Benjamin is an exemplar:

      Turn arithmetic into mathemagic .

And here you cannot but marvel at: $388 \times 388 = 150544$, which he answers instantaneously, even under the pressure of Colbert on national TV: Colbert .

Colbert checks him, and verifies the answer. But I think Art answered $388 \times 3\textbf{8}8$ while the posed question was $388 \times 3\textbf{9}8$ instead. Easy to confuse in that environment.

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Are you aware of this one: $(a+b)\cdot (a-b)=a^2-b^2$, which means nothing else than that the product of two numbers equals the difference of the squares of the average and the distance to that average.

Come again???

Let's take 37 and 43:
the average is 40, the distance to that average is $40-37=43-40=3$, hence:
$37 \cdot 43 = (40 - 3) \cdot (40 + 3) = 40^2 - 3^2 = 1600 - 9 = 1591$.

(This obviously only works when it's relatively easy to calculate the square of the average.)

... and these two: $(a±1)^2 = a^2 ± 2a + 1$ and $(a±2)^2 = a^2 ± 4 \cdot (a±1)$, leading to:
$61^2 = 60^2 + 2 \cdot 60 + 1 = 3721$
$48^2 = 50^2 - 4 \cdot (50 - 1) = 2304$

Yes, I know what you're about to say:
$a^2 - b^2 = (a+b) \cdot (a-b)$, hence:
$a^2 = (a+b) \cdot (a-b) + b^2$, so:
$48^2 = (48 + 2) \cdot (48 - 2) + 2^2 = 50 \cdot 46 + 4 = 2300 + 4 = 2304$

Oh, sorry, you meant that people should learn the first powers of $2$ by heart, so you get:

$48^2 = (16 \cdot 3)^2 = (2^4 \cdot 3)^2 = 2^8 \cdot (8 + 1) = 2^8 \cdot (2^3 + 1) = 2^{11} + 2^8 = 2048 + 256 = 2304$

Oh, you still know another way to get it done?
Don't hesitate editing this answer :-)

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