Calvin is Bored -_-

Let $a,b,c$ and $d$ be the ages of each of the $4$ students. With that in mind, we get the following system by forming all the triples of students

$\left\{ \begin{array}{l l} a+b+c=44 \quad\\ a+b+d=51 \quad\\ a+c+d=53 \quad\\ b+c+d=56 \quad\\ \end{array} \right.$

If we add all the equations from the system together, we will get

$3a+3b+3c+3d=204 \Rightarrow 3(a+b+c+d)=204 \Rightarrow$

$\Rightarrow a+b+c+d=68$ , which is the sum we are looking for.

Let their ages be a,b,c,d. Then we have a+b+c=44 b+c+d=51 c+d+a=53 d+a+b=56 Adding all these we have 3(a+b+c+d)=204 So a+b+c+d=68 So ans is 68

A+B+C=44 , A+B+D=51 , A+C+D=53 , B+C+D=56 than we sum them all 3A+3B+3C+3D=204 , A+B+C+D=68

From our information, we find the following system of equations, where $x_i$ is the score of the $i$ th student:

$x_1+x_2+x_3=44$

$x_1+x_2+x_4=51$

$x_1+x_3+x_4=53$

$x_2+x_3+x_4=56$

(Note that it does not matter which students are which.)

Adding all of these equations together, we find $3x_1+3x_2+3x_3+3x_4=204 \rightarrow \boxed{x_1+x_2+x_3+x_4=68}$

Let w, x, y, and z be the ages of the 4 students. We can create the sum 3w+3x+3y+3z=44+51+53+56 when we add the 4 equations together. Dividing both sides by 3 gives us the sum w+x+y+z, which is 68

Let the ages of 4 students be a,b,c,d. Sum of 3 at a time (a+b+c) + (b+c+d)+(c+d+a)+(d+a+b) is 44+51+53+56=204 now solving the above we get 3 times a+b+c+d equals 204.. sum a+b+c+d equals 68.

Now by the data given above, we kvow that d is bigger than c -- 7

And we also know that c is bigger than b -- 2

We also know that * b* is bigger than a --3

So, by some trial and error , a = 12 b = 15 c = 17 d = 24

Thus , 12+15+17+24 = 68

Let x, y , z be the digits of N , then N = xyz = 100x + 10y +z = a^{2} and by increasing the digits 100(x+1) + 10(y+2) + z + 3 = b^{2} say .
Now (b+a)(b-a) = b^{2} - a^{2} = 123 = 3\times41 . taking (b+a) = 41 and (b-a)= 3 we get 2a = 38 or a = 19 so a^{2}= 361 = N

let the 4 ages be $a$ , $b$ , $c$ and $d$ .

$a + b + c = 44$ $[1]$

$a + b + d = 51$ $[2]$

$a + c + d = 53$ $[3]$

$b + c + d = 56$ $[4]$

from $[1] + [2] + [3] + [4]$ , we get the equation $3a + 3b + 3c + 3d = 204$ . By dividing the whole equation by 3, we get $a + b + c + d = 68$ , which is what we are looking for as $a + b + c + d$ is the sum of the 4 ages

Let the students ages be a,b,c,d. From the details and assumptions,let

        a+b+c=44
        b+c+d=51
        c+d+a=53
        a+b+d=56

Summing all four will give 3(a+b+c+d)=204 which implies

        Sum of their ages=a+b+c+d=68

Hence,answer is 68.

$w + x + y = 44$

$x + y + z = 51$

$y + z + w = 53$

$z + w + x = 56$

$3w + 3x + 3y + 3z = 204$

$w + x + y + z = 68$

Let the ages of four students be a , b , c , d respectively, then according to question,we have 1. a + b + c = 44 2. b + c + d = 51 3. a + c * * + d = 53 4. a + b + d = 56 Adding equations 1 , 2 , 3 ,& 4 ,we have-

3 * ( a + b + c + d ) = 44 + 51 + 53 + 56 ( *a * + b + c + d ) = 204 / 3 = 68

Let, a b c and d be the ages, given that: a+b+c=44 , a+b+d=51 , a+c+d=53, b+c+d=56 , adding up all these equations gives us: 3a+3b+3c+3d=204 => 3(a+b+c+d)=204 => a+b+c+d=68

The sum of all partial sums is equal to thrice the total age.

Let us assume the ages of the children be - a,b,c,d

we see that the sum of the ages of the first 3 children namely a,b,c=44

So a+b+c=44 so b+c+d=51 so c+d+a=53 so d+a+b=56

on adding the equations- 3(a+b+c+d)=204 so a+b+c+d= 204/3 =68

(44+51+53+56)/3=68

Let $a+b+c=44,b+c+d=51,c+d+a=53,d+a+b=56$ so $3a+3b+3c+3d=204$ , $a+b+c+d=68$ _ $ans$