This is the famous picture "Mental Arithmetic. In the Public School of S. Rachinsky." by the Russian artist Nikolay Bogdanov-Belsky.

The problem presented on a blackboard requires computing the following expression: $$ \dfrac{10^{2} + 11^{2} + 12^{2} + 13^{2} + 14^{2}}{365} $$ Using paper and pencil or calculator it is not difficult to get the answer $2$. However, as the name of the picture implies, the expression is supposed to be simplified using mental calculations only.

My questions: how do people perform such calculations in their heads? Is there a general mental calculation technique which would be useful for performing basic arithmetic and power operations? Or is there some kind of trick which works in this specific case? If so, what is the class of problems this trick can be applied to?

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    $\begingroup$ This is reasonably easy to add the 3 figure numbers and splitting the fraction. $\endgroup$ – Sean Aug 10 '15 at 23:02
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    $\begingroup$ This painting strikes me, by the way, as one of the relatively infrequent examples of a scene used in art (whether a painting, a cartoon, a movie, etc.) where the contents of the blackboard are not just mathematical gibberish. $\endgroup$ – colormegone Aug 10 '15 at 23:26
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    $\begingroup$ @RecklessReckoner This is so because Bogdanov-Belsky was a student of Rachinsky. One more proof of how fine a teacher he was. $\endgroup$ – user58697 Aug 11 '15 at 1:17
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    $\begingroup$ So much has been said about the calculation, but I'd like to say something about the painting itself. I think it's absolutely amazing how the artist has managed to capture the "critical moment" in such wonderful detail: the kid in the foreground is clearly the class "brain" and is standing aloof from the rest, in a posture of deep concentration and on the verge of an epiphany. The kid a little behind him is perhaps the "lesser light" whose stress in trying (and maybe failing) to solve it mentally is manifested by his body language (one hand on the head, the other clutching his shirt). (cont'd) $\endgroup$ – Deepak Aug 13 '15 at 1:21
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    $\begingroup$ (cont'd) The professor looking on expectantly, focusing on the wunderkind in particular, expecting him to get the answer quickest of all (maybe the kid to his right is whispering "I think he's getting it!" in the professor's ear). And all the other children not looking like serious contenders, merely content to discuss the question and not actually try solving it independently. It's an almost perfect rendition of a great scene. $\endgroup$ – Deepak Aug 13 '15 at 1:22

20 Answers 20


I think you can see clearly here that if you let $12$ be equal to $x$, the expression would just then be


Do remember that if you square a binomial $(a+b)$ you would get $a^2+2ab+b^2$; thus if you replace $a$ by $x$ and $b$ by either $\pm 1$ or $\pm 2$ the middle terms would just cancel out mainly $2ab$. So you would be left with


Which then further simplifies into





$$\begin{align}\\&\frac{10^2+11^2+12^2+13^2+14^2}{365}\\&=\frac{(12-2)^2+(12-1)^2+12^2+(12+1)^2+(12+2)^2}{365}\\&=\frac{5\times 12^2+10}{365}\\&=\frac{5(144+2)}{5\times 73}\\&=2\end{align}$$

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    $\begingroup$ This is a rather nice way; I do wonder whether Teacher Rachinsky taught his charges enough about the binomial theorem to recognize that all the "middle terms" in your sum simply cancel. $\endgroup$ – colormegone Aug 10 '15 at 23:14
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    $\begingroup$ the part where you got 10 feels like more than just one step... $\endgroup$ – Steven Lu Aug 11 '15 at 5:51
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    $\begingroup$ At first glance, i thought that too. But since the sum is symmetric the mixed terms of (a - b)^2 and (a + b)^2, +2ab - 2ab vanish. So all you have to do is to sum up the squared b's 1 + 1 + 4 + 4. $\endgroup$ – pink vertex Aug 11 '15 at 11:33
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    $\begingroup$ @pinkvertex But that is more than one step, isn't it? $\endgroup$ – JiK Aug 11 '15 at 15:38
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    $\begingroup$ @JiK I wouldn't say so--it's something you can figure out in your head. $\endgroup$ – 6005 Aug 11 '15 at 19:26

If you know your squares out to $14$ (which students used to memorize) and do some simple three-digit arithmetic in your head, you can see that

$$100+121+144=365$$ and $$169+196=365$$

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    $\begingroup$ This seems like the the way to go. The arithmetic just turns out to be 100*3 + 21 + 44 = 365 and 100*2 + 69 + 96 = 365. $\endgroup$ – Nuclearman Aug 11 '15 at 2:40
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    $\begingroup$ This is true if you expect the result to be 2 (or any integer). Trying linear combinations of the terms of the numerator hoping for 365s would be a waste of time otherwise. $\endgroup$ – Ben Jackson Aug 11 '15 at 3:53
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    $\begingroup$ @BenJackson You don't have to expect anything beforehand. Just start adding the squares and you'll notice you get $365$ after two additions. $\endgroup$ – JiK Aug 11 '15 at 15:40
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    $\begingroup$ @Nuclearman, for the second sum, I mentally moved a $4$ from the $169$ to the $196$ to make it $165+200$. $\endgroup$ – Barry Cipra Aug 11 '15 at 16:23
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    $\begingroup$ This is the method I just used; it took me less than a minute to reach the result. Unfortunately, I hadn't memorized $14^2$, so I had to compute that as well. $\endgroup$ – Justin Aug 11 '15 at 16:44

There may be an easier way:

In general $(10+a)^2=100+20a+a^2$ so the numerator becomes





Then of course $730/365=2$.

Not sure if you could quite do that in your head. It would definitely take a minute or two.

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    $\begingroup$ If you keep track of the "departures" from 100 for each term, you can do the sum in rather less time. What would daunt many people from there is dividing by 365... unless you've had to find the number of days in two or more years a few times. $\endgroup$ – colormegone Aug 10 '15 at 23:11
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    $\begingroup$ @ElliotG this was my first idea; solved the calculation in head, (grouping 1+9 and 4+16 in the last step to get 10+20), immediately noticed that 730 is 2x 365, it took me about 5 seconds to read the image & figure this solution and another 20 to get the result. still, math.stackexchange.com/a/1392514/81774 is IMO the simplest way, since that solution can be done in about 10 seconds without much effort. $\endgroup$ – vaxquis Aug 11 '15 at 20:34
  • $\begingroup$ guess it's an individual thing. It took me a while to realize 365=73x5, for whatever reason, so I probably wouldn't have thought of that. Either way, this problem would probably take at least a minute the first time I saw it, but that's just me. $\endgroup$ – Elliot G Aug 11 '15 at 23:35
  • $\begingroup$ My way exactly, was pretty simple. $\endgroup$ – Hasan Saad Aug 23 '15 at 11:17

I ball-parked the answer as 2 almost immediately as follows:

$$\frac{10^2+11^2+12^2+13^2+14^2}{365}\approx\frac{5\cdot12^2}{365}=\frac{720}{365}\approx 2$$

144×5 is "144/2 add a 0" (i.e. 144×5=144×10/2), so the whole operation takes less than two seconds.

Note that this method is exact for a linear (arithmetic) sequence; it's also important for our accuracy that the terms are increasing by 1 and the denominator is roughly the same order of magnitude as the numerator.

I couldn't have told you that quickly that the answer is exactly 2, but who needs precision anyway? Worked out pretty well in this case.

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    $\begingroup$ And given that the answer was probably an integer given the nature of the question, this really solved it the way I see it $\endgroup$ – Belgi Aug 12 '15 at 20:00
  • $\begingroup$ I think this is nice. I understood it as seeing that $12$ is the median of those numbers, which is somewhat near the mean in many cases. $\endgroup$ – Allawonder May 12 '18 at 8:34

Many different ways to solve the problem have already been listed here. However, I want to bring to your attention a whole class of numbers which satisfy one property of the original equation

$$ 10^{2} + 11^{2} + 12^{2} + 13^{2} + 14^{2} = 2\cdot 365 $$

The property I would like to point out is $$(x-2)^2 + (x-1)^2 + x^2 = (x+1)^2 + (x+2)^2$$

According to this post on mathoverflow.net such numbers are called Rachinsky quintets:

Define Rachinsky quintets as a set of five positive integers $ \lbrace a,b,c,d,e\rbrace $ such that $$a^2+b^2+c^2=d^2+e^2$$

For more information about Rachinsky quintets refer to this answer.

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    $\begingroup$ Indeed, the title of the painting is "Mental Arithmetic. In the Public School of S. Rachinsky." And now we see the problem on the blackboard involves something called a "Rachinsky quintet." This suggests to me that this answer has hit exactly on how the problem was meant to be solved. $\endgroup$ – David K Aug 15 '15 at 16:59
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    $\begingroup$ It seems to me that what's impressive about $10^2+11^2+12^2=13^2+14^2$ is not so much that you have three squares with the same sum as two squares but that you have two consecutive strings of squares with the same sum. That is, are there other numbers $a\lt b\lt c$ with $(a+1)^2+\cdots+b^2=(b+1)^2+\cdots +c^2$? $\endgroup$ – Barry Cipra Aug 24 '15 at 15:20

For me, I started thinking of the lines of @mathlove's solution for a few seconds. It looks obvious when written out, but when working purely mentally I did not spot the pairing of the terms.

So, as per Sean and John's comment, just doing the direct method is actually quite easy.

A practiced mental arithmetician already knows the squares.

We have:

$100 + 121 + 144 +169 + 196$

$= 500 + 21 + 44 + 69 + 96$

Note that the double digit terms pair nicely:

$= 500 + (21+69) + (96 + 44)$

$= 500 + 90 + 100 + 40$

$= 730$

So, what is in common in the this and the other (mathematically superior) answers?

Look for patterns which allow one to simplify computational steps, and in particular to minimise the amount of data you need to maintain in your head.


I think outlined approaches, while fine with pen and paper, are somewhat complicated for mental calculations. Here is how I figured this out in my head. The idea is that all intermediate calculations should be easy to perform and remember.

  1. First compute all the squares: 100, 121, 144, 169, 196.
  2. It is immediately obvious that 121 and 169 add to a nice and easy to remember number. Compute and memorize 121 + 169 = 290.
  3. Recognize that the same could be done for 144 and 196, answer is 340. Memorize it.
  4. At this point we are left with 100, 290, and 340. Add 100 to 290 to get 390.
  5. Add 390 and 340 together to get 730.
  6. Now we can look at the denominator and figure out how to simplify it.
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    $\begingroup$ Spotting pairs of numbers that complement each other nicely is a very useful trick for mental arithmetic. +1. $\endgroup$ – Silverfish Aug 12 '15 at 9:40
  • $\begingroup$ +1 for similarity of thought! I also look at the ones places and think, I can pair $11^2$ with $13^2$, and $12^2$ with $14^2$... $\endgroup$ – Benjamin Dickman Aug 16 '15 at 10:00

I did it completely mentally (nothing written down for the intermediate steps), but it did take me about 3 minutes plus a lot of serious concentration.

I recognised the sum of consecutive squares and used the relevant Faulhaber formula (note: I am referring to just the specific formula for the sum of consecutive squares, not the generalised formula, which would be hard to memorise):

$$\begin{align}\\&\frac{10^2+11^2+12^2+13^2+14^2}{365}\\&=\frac{\frac{1}{6}\cdot(14)(15)(29) - \frac{1}{6}\cdot (9)(10)(19)}{365}\\&= \frac{(7)(5)(29) - (3)(5)(19)}{365}\\&=\frac{5\cdot (7\cdot 29 - 3 \cdot 19)}{5\cdot 73}\\&=\frac{7\cdot 29 - 3 \cdot 19}{73}\\&= \frac{7\cdot(19 + 10) - 3\cdot 19}{73}\\&= \frac{(4)\cdot 19 + 70}{73}\\&= \frac{146}{73}\\&= 2\end{align}$$

Note that I am writing it out exactly as I thought about it, so some steps are listed in seemingly unnecessary detail. I also wanted to keep the numbers in the numerator smaller, so I did the division by $6 (=2\cdot 3)$ immediately for each term in the difference.

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    $\begingroup$ Looks like you have extraordinary working memory. Unfortunately that renders your answer pretty useless for the purpose of the question, haha. $\endgroup$ – Matthew Aug 11 '15 at 18:45
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    $\begingroup$ I dunno, I think my memory (working or otherwise) is pretty poor, haha. Anyway, of all the methods, @mathlove's shows the greatest insight, and it's applicable to any "symmetric" sum of squares, not even necessarily all consecutive. It's also the one most readily adapted to quick mental calculations. I just posted mine because it was the first one I came up with and I worked it out on this basis. $\endgroup$ – Deepak Aug 12 '15 at 0:44
  • $\begingroup$ ahhh yes I remember that day in 5th grade when we learned the faulhaber formulas. good times. they were tough, no wonder the kids in the painting look so grilled. $\endgroup$ – MichaelChirico Aug 12 '15 at 12:40
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    $\begingroup$ @MichaelChirico I think that would be a little OT in the answer as I don't need the full formula (and indeed, that would be implausible, as already agreed), but I can provide the wiki link here: en.wikipedia.org/wiki/Faulhaber%27s_formula In fact, I personally didn't know it was called Faulhaber's formula until much later (quite recently, in fact, I think I picked it up from this stackexchange!). $\endgroup$ – Deepak Aug 13 '15 at 1:05
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    $\begingroup$ exactly what I did. I was gonna write it, but I am always second...:( $\endgroup$ – Bhaskar Vashishth Aug 23 '15 at 13:36

You can do this without any amazing mental skills but more a sort of "developed algebraic sense". I only had to know $12^2 = 144$ but after years of doing mathematics the rest of the thought process went automatically like this (minus the explanations):

  • We know (by some explicit or intuitive convexity principle) that the numerator will be slightly larger than $5 \times 12^2 = 5 \times 144 = 720$.

  • The correction terms to this are made by addition of ($2 \times (1^2 + 2^2))$. The idea is to recognize that estimating $(X+a)^2 + (X-a)^2$ by $2X^2$ involves a loss of $2a^2$ independent of the particular $X$. Here $X=12$ and $a$ is $1$ and $2$ for the two pairs of terms. To correct the initial estimate we must therefore add $2 \times (1 + 4)$ which is not a hard mental computation: $10$.

  • thus the numerator will be $730$ and if you made it this far without arithmetic errors that should be recognizable as twice the denominator.

The beautiful painting is famous in Russia but not as known elsewhere. I remember it from a math book but never saw it in any Western book on art.

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    $\begingroup$ This is more or less how I would approach this. Compare to an average, so the deviations are smaller. In my distant youth I was playing Finnish rules darts, where (unlike in the better known pro game) the board has concentric circles scoring 10 (center), 9, 8,... You throw five darts and count the total. I would compare the score to an average of 6 or 7 (more rarely 8) and get the total that way :-) $\endgroup$ – Jyrki Lahtonen Aug 12 '15 at 5:49
  • $\begingroup$ Did you mean to write $(X-a)^2 + (X-a)^2$ (which is $2(X-a)^2$) or $(X-a)^2 + (X+a)^2$? $\endgroup$ – David K Aug 15 '15 at 15:28
  • $\begingroup$ The latter. Thanks for the correction. I fixed the answer. @DavidK $\endgroup$ – ASCII Advocate Oct 16 '15 at 22:17

A not too hard way of doing it is to add the numbers (which are not too hard squares to calculate or know by heart) in the numerator and keep track of every time you get past 365, mentally incrementing the result

You get

10^2 = 100 Add 11^2 (121), you're at 221 Add 12^2 (144), you're at 365 (increment result by 1)

13^2 = 169 Add 14^2 (which is 13^2 + 13 + 14 = 196), you're at 365 again, increment result by 1 again

So the result is 2

I don't know if that makes sense but that's how I would do it

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    $\begingroup$ I totally second you, I had a similar approach, just reordering additions to make them easier… the computation is perfectly tractable. I do not know if I like the solution by @mathlove or not because it does not demonstrate a method. $\endgroup$ – Michael Le Barbier Grünewald Aug 12 '15 at 19:24

$$\color{Green}{n^2=1+3+5+\cdots+(2n-1)}$$ Therefore

$$\begin{align} 10^2+11^2+12^2+13^2+14^2 &=(1\cdot+19)+(1\cdot+21)+(1\cdot+23)+(1\cdot+25)+(1\cdot+27)\\ &=5\times(1\cdot+19)+(4\times21)+(3\times23)+(2\times25)+27 \end{align}$$ Also, $1+3+\cdot+19=10^2=100.$

Hence $$10^2+11^2+12^2+13^2+14^2=500+84+69+50+27=730$$ $$\color{Green}{10^2+11^2+12^2+13^2+14^2=2\times365.}$$

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    $\begingroup$ Thank you for quoting relevant formula. I feel like it can come in handy for many mental calculations problems. $\endgroup$ – Vlad Aug 15 '15 at 9:54

I did a variation of what others have done.

First noting that $(n+a)^2+(n-a)^2=2n^2+2a^2$ I did $5\times 12^2+2\times (1^2+2^2)$

It seems obvious to exploit the symmetry. If there were $25$ squares, for example, I'd be using the formula for the sum of the first $12$ squares.

This would be harder with an even number of squares (you can use half integers with care).

  • $\begingroup$ Are there any identities similar to $\,(n+a)^2+(n-a)^2=2n^2+2a^2\,$ which we could use for mental computations and which would be useful in more general case? Is there anything we can do when our expression is not necessarily of the form $\,(n+a)^2+(n-a)^2\,$? For example, what if we had something like $\,(n+a)^2+(n+3a)^2\,$ or $\,(n+2a)^3+(n-5a)^3\,$? $\endgroup$ – Vlad Dec 13 '15 at 15:44

You can also do the computation Mod 12 and Mod 11. You then find with little effort that the results are 2. This means that it is also 2 Mod 12*11. The rational reconstruction theorem then implies that the fraction is equal to 2.


I did it like this

$$ \begin{aligned} 10^2 + 14^2 &= 2\cdot 12^2 + 2 \cdot\!12 \cdot \left(2-2\right)+ 2 \cdot 2^2 \\ 11^2 + 13^2 &= 2\cdot 12^2 + 2 \cdot\!12 \cdot \left(1-1\right)+ 2 \cdot 1^2 \\ 12^2 &= 12^2 \end{aligned} $$

Sum $\,= 5\cdot 12^2 + 2 5 = 720 + 10 = 730 = 2 \cdot 365$.

So the answer is $\,2$.

  • $\begingroup$ If I understand correctly, you in the first line you replace $10^2$ and $14^2$ with $\left( 12-2\right)^2$ and $\left( 12+2\right)^2$ respectively, is that correct? Why do you have decimal fractions in the equation? Is the period in these fractions supposed to be multiplication sign? $\endgroup$ – Vlad Aug 17 '15 at 6:56
  • $\begingroup$ Yes, a period looks like 5·2 $\endgroup$ – wendy.krieger Aug 17 '15 at 9:23

One method of solving it can be like this, We know $\,10^2=100,\,$ and then $\,11^2 = 100 + (10\cdot 2 + 1),\,$ $\,12^2 = 100 + (21) + (22 + 1)\,$ and similarly for other terms.

Thus we can just take $\,100 \times 5\,$ common and for rest terms we add like this $\,(20+1) + (21+22+1) + (44+24+1) + (69+26+1)\,$ i.e if there are two consecutive terms a and b then $\,b^2= a^2 + 2a + 1\,$, so finally result be like $\,\dfrac{500+21+44+69+96}{365},\,$ i.e $\,\dfrac{500+90+140}{365} $ .


Similar to Mark Bennet's solution:

 10^2 = (12-2)^2 = 144 - 48  +  4
 11^2 = (12-1)^2 = 144 - 24  +  1
 12^2 = (12-0)^2 = 144
 13^2 = (12+1)^2 = 144 + 24  +  1
 14^2 = (12+2)^2 = 144 + 48  +  4
             sum = 720 +  0  + 10
                 = 730

730 / 365 = 2

I am surprised that nobody mentioned mnemonic techniques for doing calculations in your head. Instead of trying to explain basic techniques myself I am linking this video where famous mathemagician Arthur Benjamin explaining techniques he uses for multiplying huge numbers in his head.

For those who prefer to read rather than watch and listen I also link one of the articles where A.Benjamin covers basics of mnemonic techniques, for squaring large numbers, memorizing digits of $\pi$, and more. Several relevant textbooks are also referenced below.

  • Ringel', D.E., Mnemonic numbers and mnemonic vectors, Vestn. Beloruss. Gos. Univ., Ser. 1, Fiz. Mat. Inform. 2003, No. 1, 71--76 (2003). ZBL1272.46035.

  • Benjamin, Arthur. T., The Secrets of Mental Math (2008), (Amazon link).

  • Toma, Marina, Mathematics and mental calculus, J. Sci. Arts 10, No. 1, 39-42 (2010). ZBL1220.00025.


I wasn't smart enough to group them the way Barry Cipra did, so I just did the arithmetic: $169 + 121 = 290$; adding $144$ makes $434$; adding $196=200-4$ makes $630$, adding $100$ makes $730$; wow, that's twice $365$, so the answer is $2$.


Here is how it was done in my head: $10^2 = 100$. Okay. $11^2 = 121$. Okay, thats $221$. $12^2 =144$. Aha, thats $365$ total. So far we have a total of $365/365 =1$. Now, $13^2 =169$ and $14^2 =196$, which sum to $365$. So the answer is $2$.

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    $\begingroup$ This is an application of the Rachinsky quintets in user242891's answer. $\endgroup$ – Brian Tung Aug 18 '15 at 20:34

protected by Jyrki Lahtonen Aug 12 '15 at 5:43

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