How do people perform mental arithmetic for complicated expressions?

Question

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

The problem on the blackboard is: $$ \dfrac{10^{2} + 11^{2} + 12^{2} + 13^{2} + 14^{2}}{365} $$

The answer is easy using paper and pencil: $2$. However, as the name of the picture implies, the expression ought be simplified only mentally.

My questions:

1. Are there general mental calculation techniques useful for performing basic arithmetic and exponents?

2. Or is there some trick which works in this case?

3. If so, what is the class of problems this trick can be applied to?

Answer 1

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

$$\frac{(x-2)^2+(x-1)^2+x^2+(x+1)^2+(x+2)^2}{365}$$

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

$$\frac{(x^2+4)+(x^2+1)+x^2+(x^2+1)+(x^2+4)}{365}$$

Which then further simplifies into

$$\frac{5x^2+10}{365}$$

$$\frac{720+10}{365}$$

$$=2$$

Answer 2

There may be an easier way:

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

$500+20(0+1+2+3+4)+1+4+9+16$

$=500+20(10)+1+4+9+16$

$=700+1+4+9+16$

$=730$

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.

Answer 3

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

Answer 4

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.

Answer 5

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 $$169+196=365$$

Answer 6

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.

Answer 7

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.

Answer 8

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 $2$.

Answer 9

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

Answer 10

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.

Answer 11

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.

Answer 12

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

Answer 13

$$\color{Green}{n^2=1+3+5+\cdots+(2n-1)}$$ 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 $$10^2+11^2+12^2+13^2+14^2=500+84+69+50+27=730$$ $$\color{Green}{10^2+11^2+12^2+13^2+14^2=2\times365.}$$

Answer 14

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).

Answer 15

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.

Answer 16

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.

Answer 17

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 $\,2$.

Answer 18

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 $2$.

Answer 19

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} $ .

Answer 20

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.

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• 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.