Pairwise Sums Of 4 Variables

Let $w$ , $x$ , $y$ and $z$ be the 4 variables. Then

$w+x=10$

$w+y=13$

$w+z=16$

$x+y=19$

$x+z=22$

$y+z=25$

The algebra is pretty basic. Let us first solve for $w$ .

$w=10-x$

Plugging in gives

$10-x+y=13$

Now let us rearrange $x+y=19$ :

$y=19-x$

Substituting gives

$10-x+19-x=13$

Solving, we find $x=8$ . We can use this to find all other values, which are 2, 11, and 14. Multiplying $2*8*11*14$ is $\boxed{2464}$

Just another method :)

Let $a, b, c, d$ be the four positive integers. Then, by summing up all of the possible unique pairs, we can say that,

• $3a + 3b + 3c + 3d = 105$ ,

which can be simplified to,

• $a + b + c + d = 35$ .

Thus, we can say that, (the variables may be replaced with the other ones)

• $(a + b) = 35 - (c + d)$
• $(a + c) = 35 - (b + d)$
• $(a + d) = 35 - (b + c)$

Then, we can find the three pairs of numbers that satisfy the conditions above.

• 10 & 25
• 13 & 22
• 16 & 19

We can assign the pairs to any equation above (in this solution, I made it corresponding with their orders):

• $a + b = 10$ & $c + d = 25$
• $a + c = 13$ & $b + d = 22$
• $a + d = 16$ & $b + c = 25$

Then, bwalah! In my solution, the results are:

• $a = 2$
• $b = 8$
• $c = 11$
• $d = 14$

Then, by multiplying these positive integers,

• $abcd = 2464$ .

Let a, b, c, d be four variables such that:

a < b < c < d

None of pairs sum to the same amount implies that none of the variables are equal.

Now, either

a + b < a + c < a + d < b + c < b + d < c + d

or

a + b < a + c < b + c < a + d < b + d < c + d

It turns out former is the case, you won't find an integer solution for second one so that explains it.

14 ,11 , 8 , 2 are the four integers

14 * 11 * 8 * 2 = 2464

Lets assume a,b,c & d are the integers.

Now, a+b= 10, a+c+13, a+d=16, b+c=19, b+d=22, c+d= 25

Now, a-b= (a+d)- (b+d)= 10-16= -6

(a+b) + (a-b)= 2a= 10+(-6)= 4 => a=2 Hence, b= 8, c= 11 & d= 14 from the above equations.

Now their product= 1x b x c x d=2464

a+b=10 a+c=13 a+d=16 b+c=19 a+d=25 by soving above eqns we get a=2, b=8, c=11, d=14 so a b c*d= 2464

Adding all sum will give 3*(a+b+c+d) =105 which gives us clue that adding two sums must be 35. (i.e 10 ,25 & 16,19 & 22 , 13). From this you can solve easily for the unknown

A+B = 10, A+C=13, A+D=16, B+C= 19, B+D=22, C+D = 25 After solving above equations 4 integers(+) come out: 2,8,11,14 4 integers product= 2464 (2 8 11*14).

Let the numbers are a,b,c,d

a+b=10

a+c=13

a+d=16

b+c=19

b+d=22

c+d=25

solving these we get

a=2

b=8

c=11

d=14

and their product is 2 8 11*14=2464

let the 4 integers be a,b,c and d. thus, a+b=10, a+c=13, a+d=16, b+c=19, b+d=22, c+d=25, solving which we get a=2, b=8, c=11 and d=14 whose product is 2464.

Let..a+b=10.......a+c=13.....a+d=16....b+c=19.....b+d=22.....c+d=25
By adding all the equation above we will get...3a+3b+3c+3d=105
Factorize....3(a+b+c+d)=105
Then....a+b+c+d=35
a+b+a+c+a+d=39..........》》2a+a+b+c+d=39》》2a+35=39
Then....a=2..b=8...c=11...d=14......myltiply all of them..get...2464

a+b =10, a+c=13, a+d=16, b+c=19, b+d=22, c+d=25 solving a=2, b=8, c=11, d=14 so abcd=2464

four variables (a, b, c, d)

a+b=10, a+c=13, a+d=16, b+c=19, b+d=22, c+d=25

c=13-a, d=16-a

thus (13-a)+(16-a)=25 = -2a=-4 a=2

2+b=10 b=8

2+c=13 c=11

11+d=25 d=14

2 8 11*14=2464

Let $a,b,c$ and $d$ be the four variables. Then $a+b=10$ $a+c=13$ $a+d=16$ $b+c=19$ $b+d=22$ $c+d=25$

$b,c,d$ are in arithmetic progression $b+3=c,b+6=d$

Also $a+b+a+c+a+d+b+c+b+d+c+d=10+13+16+19+22+25=105$ $3(a+b+c+d=105)$ $a+b+c+d=35$

$\Rightarrow$ $a+b+b+3+b+6=a+3b+9=35$ $a+b=10$

$10-b+3b=10+2b=26$ $\Rightarrow$ $b=8, a=2, c=11,d=14$

$a*b*c*d=2*8*11*14=2464$

let the 4 integers be a,b,c and d. Such that a<b<c<d then a+b = 10 let this be eq.1 a+c = 13 let this be eq.2 a+d = 16 let this be eq.3 b+c = 19 let this be eq.4 b+d = 22 let this be eq.5 c+d = 25 let this be eq.6 eq.2 - eq.1 gives a+c-a-b = 13-10 c-b = 3 c = b+3 let this be eq.7

Substitute eq.7 in eq.4 b+b+3 = 19 2b = 16 therefore , b = 8

since, a+b = 10 ; a = 2 (since b = 8) and a+c = 13 so c = 11 (since a = 2) and a+d = 16 so d = 14 (since a = 2)

so the product of the integers = 2 8 11*14 = 2464

Hello...

let 4 positive integers = a,b,c,d....

so the pairs are (a,b),(a,c),(a,d),(b,c),(b,d),(c,d)=10,13,16,19,22,25

a + b =10

a + c = 13

a + d = 16

b + c = 19

b + d = 22

c + d =25

let's we take (a+b)-(a+c) = 10 -13

b-c = -3 ----------> compare with b + c =19,

-3 + c + c =19

2c = 22 c=11(as you're getting the value of one of the integers,you can substitute to find the other 3 integers),

b = 19 -11 = 8

a = 10 -8 = 2

d = 25 - 11 =14,

therefore,4 integers are = 2, 8 , 11 ,14,their product = 2 x 8 x 11 x 14 = 2464

thanks...