Too Many Numbers To Be Added!

$\underbrace{(\sum_{i=0}^{10}i)}_{vertical}\times\underbrace{(\sum_{i=0}^{10}i)}_{horizontal}=55\times55 = 3025$

Note : We use the fact that $1 + 2 + 3 + \cdots + n = \dfrac 12 n(n+1)$ .

Ignore the Zeros. On the first row we have 1+2+3+4+5+6+7+8+9+10=55= X On the second row we have 2X, third row 3X..... Sem it all you have 55X = 3025

We observe that in each row, the sum of the numbers are $\displaystyle \sum_{n=0}^{10} nx$ where $x$ is the corresponding number in the $1^{st}$ column. For example in in row $2$ , the sum of numbers is $\displaystyle \sum_{n=0}^{10} 2n$ .

So, we have to evaluate $\displaystyle \sum_{n=0}^{10} n+2n+3n+ \cdots + 9n + 10n$ .
= $\displaystyle \sum_{n=0}^{10} (\dfrac {10 ×11}{2})n$
= $\displaystyle \sum_{n=0}^{10} 55n$ = $\boxed {3025}$ .

Sum of Row 0 and Row 10 is 550.

$Row 0 + Row 10 = Row 1 + Row 9$ $Row 1 + Row 9 = Row 2 + Row 8$ This pattern continues through the table. (0 and 10, 1 and 9, 2 and 8, 3 and 7, 4 and 6) $550 \times 5\ = 2750$ Sum of Row 5 is 275 $2750 + 275 = 3025$

We have 1+2+3+4+5+...+100 =1(1+2+...+10)+2(1+2+3+...+10)+3(1+2+...+10)+...9(1+2+3+...+10)+10(1+2+3+...+10) =(1+2+3+4+5+6+7+8+9+10)(1+2+3+4+5+6+7+8+9+10) =55*55=3025

Second row = 11x5 Third row = 22x5...... Till last row = 110x5 Take 5 common 5(11+22+33.....110) take 11 common =5x11(1+2+3.....10) =55(55) =3025

You've got numbers from 1 to 10 in the problem and since it's a multiplication square you end up multiplying all the different numbers which is like when you're expanding double brackets the inside and then the outside. Anyway in the end it's just the sum of 1 to 10 squared which is 55 squared which is 3025

Each column is equal to the bottom number multiplied by half the size of the column. 11 numbers in each column makes it: 5.5*(10+20+30+40+50+60+70+80+90+100)

11+110, 22+99....=121 in each pair. 5x121=605 605x5=3025

I know it's a bit blunt but it worked.

You can also do it by assuming that in each row the sum of all the numbers will be $a*10+\frac {a*10}{2}$ where $a$ is a number from 1 to 10. For instance, the sum of the 10 row is $550+50$ , thus you can consider using only the column $a*5$ sum all its numbers and multiplying the result by 10 you obtain $2750$ . At this point you have to add the value of the 5 column to find the answer to this problem $2750+275=\boxed{3025}$ .

S = 5×(1+10) + 5×(2+20) + 5×(3+30) + 5×(4+40) + 5×(5+50) +....+ +... + 5×(10+100)

= (5 + 10 + 15 + 20 + 25 +...+ 50) + (50 +100 + 150 + 200 +...+ 500) =

= 5×(5+50) + 5×(50+500) = 25+250+250+2500

ans: 3025

10! Sqared is how I solved it. Or 10 factorial Sqared.