let the number of times $-1$ occurs in the list be $a$ let the number of times $0$ occurs in the list be $b$ let the number of times $1$ occurs in the list be $c$ let the number of times $2$ occurs in the list be $d$ Thus, $a+b+c+d=2023$, $-a+c+2d=19$, and $a+c+4d=99$ Adding the second and third equations gives $c+3d=59$ Subtracting the second from the third gives $a+d=40 \implies a=40-d$ The expression to find the sum of the cubes of the terms is $-a+c+8d\implies -a+5d+59\implies 6d+19$ by substitution Therefore, to find the max and min of the sum, we simply need to find the max and min of $d$ the smallest possible value of $d$ is $0$, so lets assume that $d=0$. Then $-a+c=19$ and $a+c=99$ It is easy to see that all of the variables will have positive integer values so the min value is $6*0+19=\fbox{19}$ Notice that $d$ is maximized when $b$ and $c$ are minimized. Upon closer inspection, the smallest values of $b$ and $c$ that result in the variables having integer values are $b=0$ and $c=1$. Thus, $a+d=2022$ and $-a+2d=18$. Solving this system for $d$ yields $d=680$, so the max of the sum of the cubes of the 2023 numbers is $6*680+19=\fbox{4099}$.

You have a list of $2023$ numbers, where each one can be $-1$, $0$, $1$ or $2$. The sum of all numbers is $19$ and the sum of their squares is $99$. What are the minimum and maximum values of the sum of the cubes of those $2023$ numbers? a + b + c + d = 2023 –a + 0 + c + 2d = 19 a × (–1)^2 + b × 0^2 + c × 1^2 + d × 2^2 = a + c + 4d = 99 2a + 2d = 80 ===> a + d = 40 ===> b + c = 1983 c + 3d = 59 0 ≤ a, b, c, d ≤ 1983 ===> 0 ≤ d ≤ 19 and 21 ≤ a ≤ 40 While 2 ≤ c ≤ 59 and 1924 ≤ b ≤ 1981 Minimum sum of cubes : 1) minimise d , and , 2) maximise a , and , 3) maximise b Since 1) and 3) are in conflict with each other because they would move in the same direction together, we use rule number 2) to resolve this, which resulted in minimised d, maximised a and best possible maximised b ===> d = 0, a = 40 and b = 1924 Part 1 answer = 40 × –1 + 1924 × 0 + 59 × 1 + 0 × 8 = –40 + 59 = 19 Maximum sum of cubes : 1) maximise d , and , 2) minimise a , and , 3) minimise b Since 1) and 3) are in conflict with each other because they would move in the same direction together, we use rule number 2) to resolve this, which resulted in maximised d, minimised a and best possible minimised b ===> d = 19, a = 21 and b = 1981 Part 2 answer = 21 × –1 + 1981 × 0 + 2 × 1 + 19 × 8 = –21 + 2 + 152 = 133