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So I've just bought an app from the playstore. It's a mental math app. I need it for my new job that require me to do calculation out of paper.

Have a look at the forth step shown in the image. Why so?

enter image description here

Also, how it chooses 29 ( I know the process until 21 and 29) as the answer by using those numbers, 4, 9, and 4.

Plain_Dude_Sleeping_Alone
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    I presume 'digit sum' refers to the fully reduced answer, so $8+4+1 = 13$, then $1+3=4$. – B. Mehta Dec 14 '17 at 00:03
  • Why so.. what? The digit sum of $8+4 + 1 = $ digit sum of $13 = 1 + 3 = 4$. a nd the digit sum of $(21)^2$ = digit sum of $(2+1)^2 = 3^2 = 9$ wheres the digit sum of $(29)^2$ = digit sum of $(2+9)^2$ = digit sum $(11)^2 = $ digit suo of $(1+1)^2 = 2^2 = 4$. What confuses you? – fleablood Dec 14 '17 at 00:06
  • Phrased another way, the author of the solution pictured is checking to see the remainder of each of $841, 21^2$ and $29^2$ when they are divided by nine. It is generally well known that a number is equivalent mod9 to the sum of its digits mod9. By summing the digits of a number and then repeating by summing the digits of the result over and over, you will be left with a result that is between $0$ and $9$ (and zero only occurs as the result for the number zero itself). Since $841$ is different mod9 than $21^2$ we know that $841\neq 21^2$. – JMoravitz Dec 14 '17 at 00:07
  • Thanks. I didn't notice that before. – Plain_Dude_Sleeping_Alone Dec 14 '17 at 00:07
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    how it chooses 29 Without digit sum, just notice that $841$ is close to $900=30^2,$, then the next lower number to try will check out $29^2=(30-1)^2=30^2-2 \cdot 30 +1=841,$. – dxiv Dec 14 '17 at 00:16
  • I don't see anywhere that this says how we know 841 has a square root at all. If we used this method to frind $\sqrt{801}$ we would get that the first digit will be $2$ and the last digit will be $1$ or $9$ and $DS(801) = 9$ and $DS(21^2) = DS(2+1)^2 = 9$. So we would conclude $\sqrt {801} = 21$. But $801$ is not a perfect square and $\sqrt{801} \ne 21$. – fleablood Dec 14 '17 at 00:29

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Where they write "digit sum", they actually mean the digital root: As long as your result is not single digit, add the digits again.

So you get $841 \rightarrow 8 + 4 + 1 = 13 \rightarrow 1 + 3 = 4$.

Note that the digital root is equal to the reminder from division by $9$, except that if the reminder is $0$, the digital root is $9$ (unless you started with $0$, of course). So calculating the digital root essentially is doing calculation modulo 9.

Of course if two numbers are equal, then their reminder on division by 9 is also the same. The calculation shows that for $841$ and for $29^2$, that reminder is $4$, while $21^9$ is a multiple of $9$. Thus you know that $21^2\ne 841$ because the remainder doesn't agree.

Note that strictly speaking, you don't know yet that $29^2=841$; all you know that if $841$ is the square of a natural number, that number must be $29$. The method never looked at the middle digit, therefore it would have arrived at $29$ also for e.g. $831$, despite $29^2\ne 831$.

celtschk
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What is happening is they are finding is the digital root and not the digit sum. The digital root basically means adding the numbers until it becomes a single number. If $D(x)$ is the digital root of $x$, then $D(987)=9+8+7=24\rightarrow2+4=6$. You know that $841$ has digital root $4$, and that $29$ has digital root $2$. To find $D\left(29^2\right)$, we can simplify it to $D\left(2^2\right)$ the same way when we are finding the last digit of $29^2$ (We know the $2$ will not affect the last digit). So the digital root of $841$ and $29^2$ is $4$, therefore $29^2$ could be equal to $841$.

Kyan Cheung
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Two things about DS (digit sum),

1) it is infinitely recursive.

That is $DS(x) = DS(DS(x)) = DS(DS(DS(x))) = .....$.

So $DS(841) = DS(8 + 4 + 1) = DS(13) = DS(1+3) = 4$.

2) It transfer over multiplication.

So $DS(a*b) = DS(a)*DS(b) = DS(DS(a)*DS(b))$.

So if $sqrt 841 = ab$ of $(ab)^2 = 841$ then $DS(841) = DS(ab^2) = (DS(ab))^2$.

So if $DS(841) \ne (DS(21^2))$ then $841 \ne 21^2$.

So $DS(841) = 4$ and $DS(21^2) = (DS(21))^2 = (DS(2+1))^2 = DS(3)^2 = 3^2 = 9 \ne 4$. SO $29^2 \ne 841$.

But $DS(29^2) = DS(29)^2 = DS(2+9)^2 = DS(11)^2 = DS(1+1)^2 = DS(2)^2 = 2^2 = 4$.

So $29^2$ might be $841$.

Now $841 = 8.41 *100$ so $\sqrt {841} =\sqrt{8.41}*10$ so $20 < \sqrt{841} \le 29$. And IF $841 = 2a^2$ then $a = 1$ or $a = 9$ because otherwise $2a^2 $ does not end with $1$. And $DS(21^2) \ne DS(841)$ but $DS(29^2) = DS(841)$.

SO !!!IF!!! $841$ is a perfect square then $\sqrt{841} = 29$.

!BUT! I do not see anything here that says WHY we should think that $841$ is a perfect square.

fleablood
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