Why digit sum of 841 is 4 instead of 13? 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?

Also, how it chooses  29 ( I know the process until  21 and 29) as the answer by using those numbers, 4, 9, and 4.
 A: 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 $29^2\ne 831$.
A: 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 $841$.
A: 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.
