Wake Up, Neo

Let the sum of all entries of a likely matrix of order $k$ be $S(k)$ .

The matrix will look like this:

$\begin{pmatrix} 1 & 2 & \cdots &k \\ 2 & 4 & \cdots & 2k\\ \vdots & \vdots & \ddots & \vdots \\ k & 2k & \cdots & k^2 \end{pmatrix}$

Notice the first column has sum $= (1+2+\cdots+k)$ . The second one has sum $= 2 \times (1+2+\cdots+k)$ , the third one has sum $= 3 \times (1+2+\cdots+k)$ , and the $k$ -th column has sum $= k \times (1+2+\cdots+k)$ .

Adding all up, we get $(1+2+\cdots+k)(1+2+\cdots+k) \Rightarrow (1+2+\cdots+k)^2$ . Note this is the square of the sum of the first $k$ integers, whose closed formula is $\frac{k(k+1)}{2}$ . Jumping some steps, we see the problem asks us not $S(40)$ , but its square root. That means the answer to this problem is $\frac{40 \cdot 41}{2} \Rightarrow = \boxed{820.}$ .

By the definition of the entries of the matrix we have for example $a_{1m}=1,2,3,4,\dots,40,$ so the sum of the first row with all the colums is $\sum_{n=1}^{40}{n}.$ in the second row with all the columns we have $2\sum_{n=1}^{40}{n}.$ And we can do this for every row k with all the columns, the sum is $k\sum_{n=1}^{40}{n}.$ hence we have that the sum of all the entries of the matrix is $S=\sum_{n=1}^{40}{n} + 2\sum_{n=1}^{40}{n}+ \dots + 40\sum_{n=1}^{40}{n}=\sum_{n=1}^{40}{n}\times (1+2+\dots + 40)=\sum_{n=1}^{40}{n}\times \sum_{n=1}^{40}{n}.$ so $S=\frac{40^{2}\times 41^{2}}{2^{2}}$ then $\sqrt{S}=\frac{40 \times 41}{2}=820$

The matrix looks like this:

$1 \times 1$ , $1 \times 2$ , $1 \times 3$ , ........, $1 \times 40$

$2 \times 1$ , $2 \times 2$ , $2 \times 3$ , ........, $2 \times 40$

$3 \times 1$ , $3 \times 2$ , $3 \times 3$ , ........, $3 \times 40$

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$40 \times 1$ , $40 \times 2$ , $40 \times 3$ , ........, $40 \times 40$

So, their addition

$S = (1 \times 1) + (1 \times 2) + (1 \times 3) + ..... + (1 \times 40) + (2 \times 1) + (2 \times 2) + (2 \times 3) + ..... + (2 \times 40) + (3 \times 1) + (3 \times 2) + (3 \times 3) + ..... + (3 \times 40) + ..... + (40 \times 1) + (40 \times 2) + (40 \times 3) + ..... + (40 \times 40)$

$S = 1 \times (1 + 2 + 3 + ..... + 40) + 2 \times (1 + 2 + 3 + ..... + 40) + 3 \times (1 + 2 + 3 + ..... + 40) + ..... + 40 \times (1 + 2 + 3 + ..... + 40)$

$S = (1 + 2 + 3 + ..... + 40) \times (1 + 2 + 3 + ..... + 40)$

$\sqrt{S} = (1 + 2 + 3 + ..... + 40)$

$= \frac{40 \times 41}{2}$

$= \boxed{820}$

That's the answer!

Let's count the sum of every line
Line 1 $\rightarrow$ $L_1=1\times1+1\times2+1\times3+\cdots+1\times40$
$L_1=1(1+2+3+\cdots+40)$
Line 2 $\rightarrow$ $L_2=2\times1+2\times2+2\times3+\cdots+2\times40$
$L_2=2(1+2+3+\cdots+40)$
Line 3 $\rightarrow$ $L_3=3\times1+3\times2+3\times3+\cdots+3\times40$
$L_3=3(1+2+3+\cdots+40)$
$\vdots$ $\rightarrow$ $\vdots$
Line 40 $\rightarrow$ $L_{40}=40\times1+40\times2+40\times3+\cdots+40\times40$
$L_{40}=40(1+2+3+\cdots+40)$

Assuming that $a=1+2+3+\cdots+40$
$L_1=1a$
$L_2=2a$
$L_3=3a$
$\vdots$
$L_{40}=40a$

To get the sum of all its entries $(S)$ we need to sum up $L_1$ until $L_{40}$
$S=L_1+L_2+L_3+\cdots+L_{40}$
$S=1a+2a+3a+\cdots+40a$
$S=(1+2+3+\cdots+40)a$
$S=(a)a$
$S=a^2$
$\sqrt{S}=a$
$\sqrt{S}=1+2+3+\cdots+40$
Note : $1+2+3+\cdots+n=\frac {n(n+1)} {2}$
$\sqrt{S}=\frac {40(41)} {2}$
$\sqrt{S}=\boxed{821}$

We note that in order to have every combination of $m \times n$ , we must only write $S = (m_1 + m_2 +...+ m_m)(n_1+n_2+...+n_n)$ . This then simplifies to $(1+2+...+40)^2$ . By the formula for an arithmetic series, we know that the sum in the parentheses is $\sqrt S = (1+40)*\frac{40}{2}=41 \times 20 =820$ .

Firstly, we have to notice that we have to find the square-root of sum of each entry in the square matrix of order 40 , which can't be found in a quick manner by normal formulae.

Of course, one could painstakingly write down all the entries, sum them and then take the square-root. But let me shorten the process by a day.

Notice :

First Row = $1\times (1+2+3+4+...40)$

Second Row = $2\times(1+2+3+4+...40)$

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Fortieth Row = $40\times(1+2+3+4+...40)$

Adding all the rows yields :

S = $1\times (1+2+3+4+...40) + 2\times (1+2+3+4+...40) .... + 40\times (1+2+3+4+...40)$

Factoring out the (1+2+3+4+...40) from each term yields :

S = $(1+2+3+4+...40)\times (1+2+3+4+...40)$ = $(1+2+3+4+....40)^{2}$

Hence * $\sqrt{S}$ * = (1+2+3+4+....40)

Using the formula for sum of N natural numbers we get :

* $\sqrt{S}$ * = * $\frac{(40)(41)}{(2)}$ * = * $\boxed{820}$ *

Since $a_{mn}=m\times n$ , the first row shall consists of elements $1,2,3....40$ . Similarly the second row shall consists of elements of the form $2\times 1,2\times 2,....2\times 40$ . Thus the $n^{th}$ row shall contain elements of the form $n\times 1,n\times 2,....n\times 40$ . Hence $S=(1+2+3+....+40)+2\times (1+2+3+.....+40)+....+n(1+2+3+....+40)+....+40(1+2+3+.....+40)$ $S=(1+2+3+....+40)(1+2+3+.....+40)=20^2\times 41^2$ $\therefore \sqrt S=41\times 20=820.$

The solution is as follows first of all consider an entire row .Since the entire row has it's row number i.e, the number being multiplied in common. we just take it out sum up the remaining numbers which is obviously the sum of first 40 natural numbers as provided SUM OF THAT ROW BEING CONSIDERED=rn*(40x41)/2 (rn represents that row number) repeat same procedure for all rows you then sum up all the rn's you get square of 40x41/2=S^2 which implies S=40x41/2=820

Am,n=m n...i.e..,A1,1=1 1=1,A1,2=1 2......A1,40=1 40=40 for 1st row sum=1+2+...+40=820 similarly for A2,1=2 1=2,A2,2=2 2...A2,40=2 40....sum=2(1+2+..+40)=2(820) soon for A40,1=40 1.....A40,40=40 40......sum=40(1+2+...+40)=40(820) there fore,total sum, s=1(820)+2(820)+...+40(820) s=820(1+2+...+40) s=820 820 s^1/2=820

s={n(n+1)/2}*{n(n+1)/2} where n=40 since it is a square matrix the rows wil be like 1.1,1.2........1.40 same wit columns

$\left[ \begin{array}{c} 1 & 2 & 3 & 4 &\cdots & 40 \\ 2 & 4 & 6 & 8 & \cdots & 80 \\ \vdots & & & \ddots & & \vdots \\ 40 & 80 & 120 & 160 & \cdots & 1600 \end{array}\right] \begin{array}{l} R1 \\ R2 = 2R1 \\ \vdots \\ R40 = 40R1 \end{array}$ For finding the sum of all the entries in R1 we can define little s as the sum of the simple arithmetic sequence $s=1+2+3+....40 = \sum_{n=1}^{40} n$ $\sum_{n=1}^{40} n = \frac{40}{2}(1+40) = 820$ You also know that $R2 = 2R1$ therefor the sum of Row 2 is just 2s, the sum of Row 3 is 3s, etc. Using the formula for the sum of an arithmetic sequence again, you find that the sum of all rows, S, is $S=s+2s+3s+....40s =s(1+2+3+....+40)= s\sum_{n=1}^{40} n$ $S=s\frac{40}{2}(1+40) = s(820) = 820^2$ Then $\sqrt{S} = \sqrt{820^2} = 820$