Question

let y = 3x+ 10 - x^2, where x and y are positive whole no.
find sum of all possible values of y.

Answer 1

\[y =\left( x +2\right)\left( 5-x \right)\]

Answer 2

y should be greater than or equal to zero. so x<5
given \[x \ge 0\]
so substitute values between 0-5

Answer 3

Solve for x from the equation
x^2 - 3x +(y-10) =0

=> x = [ 3+- sqrt(49 - 4y) ]/2

First thing to take care of is 49-4y should be a perfect square for some whole number y

-> 49 - 4y = m^2 (for some integer m)
-> y = (49-m^2)/4
->y =(7+m)(7-m)/4

clearly, m has to be odd number
i.e. m^2 can only take values 1^2 ,3^2 ,5^2 ..
keep in mind 49-4y >= 0 i.,e.y<= 12.25

max value of y can be 12. min can be anything for real number, but since y is a whole number, it can be minimum 0.
for m^2 =1, y=12
m^2= 9 , y=10
m^2 = 25, y=6
m^2 = 49 , y=0

now check for each y, whether x is a whole number or not,

x = [ 3+- sqrt(49 - 4y) ]/2

ofcorse it will be since sqrt(49 -4y) is always odd and 3 is also add, there addition or subtraction will yield an integer which is a multiple of 2.

hence required ans = 12+10+6+0 = 38

Answer 4

28*

Answer 5

thank u @shubhamsrg