How do I input this into my calculator?

Let $f(x)$ denote a monic $100^{\text{th}}$ degree polynomial with roots $1,2,3, \ldots , 100$

We use the following identities

$\displaystyle \sum_{k=1}^n k = \frac {n(n+1)}{2}$ , $\displaystyle \sum_{k=1}^n k^2 = \frac {n(n+1)(2n+1)}{6}$ , $\displaystyle \sum_{k=1}^n k^3 = \left ( \sum_{k=1}^n k \right )^2$

Let $P_k$ denote the sum of $k^{\text{th}}$ powers of roots of $f(x)$

Then we have

$P_1 = 1 + 2 + \ldots + 100 = \frac {100(100+1)}{2} = 5050$

$P_2 = 1^2 + 2^3 \ldots + 100^2 = \frac {100(100+1)(2(100)+1)}{6} = 338350$

$P_3 = 1^3 + 2^3 + \ldots + 100^3 = P_1 ^2 = 5050^2$

Let $S_n$ denote the $n^{ \text{th}}$ symmetric sum of numbers $1,2,3, \ldots , 100$

Then $S_1 = P_1 = 5050, S_2 = \frac {1}{2} (S_1^2 - P_2 ) = 12582075$

And lastly $S = S_3$ , apply this identity :

$P_3 - 3S_3 = S_1^3 - 3S_1 S_2 \Rightarrow S_3 = \frac {1}{3} (P_3 - S_1^3 + 3S_1 S_2) \equiv \boxed{50}$

$I\quad am\quad going\quad to\quad generalize\quad the\quad result.\\ Consider\quad :\\ 1.1.1\quad +\quad 1.1.2\quad +\quad 1.1.3\quad ...\quad 1.1.n\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 1.1(1+2+3..n)\quad =\quad 1.1.\sum { n } \\ 1.2.1\quad +\quad 1.2.2\quad +\quad 1.2.3\quad ...\quad 1.2.n\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 1.2(1+2+3...n)\quad =\quad 1.2.\sum { n } \\ .......\\ 1.n.1\quad +\quad 1.n.2\quad +\quad 1.n.3\quad ...\quad 1.n.n\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 1.52(1+2+3...n)\quad =\quad 1.n.\sum { n } \\ 2.1.1\quad +\quad 2.1.2\quad +\quad 2.1.3\quad ...\quad 2.1.n\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 2.1(1+2+3...n)\quad =\quad 2.1.\sum { n } \\ .......\\ .......\\ n.n.1\quad +\quad n.n.2\quad +\quad n.n.3\quad ...\quad n.n.n\quad \quad \quad =n.n.(1+2+3....n)\quad =\quad n.n.\sum { n } \\ The\quad total\quad sum\quad above\quad =\quad N\quad =\quad { \left[ \sum { n } \right] }^{ 3 }\\ Now\quad consider\quad the\quad numbers\quad of\quad the\quad form\quad \alpha \alpha \beta \quad where\quad \alpha =\{ 1,2...n\} \quad and\quad \beta =\{ 1,2...n\} \\ There\quad are\quad 3\quad numbers\quad of\quad the\quad above\quad form\quad giving\quad the\quad same\quad product.\\ That\quad is\quad 1.1.2\quad is\quad the\quad same\quad as\quad 1.2.1\quad and\quad 2.1.1\quad .Thus,\quad from\quad N,\quad we\quad must\quad remove\quad 3\times M\\ where\quad M\quad is\quad numbers\quad of\quad the\quad form\quad \alpha \alpha \beta .\\ M\quad =\quad 1.1.1\quad +\quad 1.1.2\quad .....\quad 1.1.n\quad \quad \quad =\quad { 1 }^{ 2 }\quad \sum { n } \quad \\ \quad \quad \quad \quad \quad 2.2.1\quad +\quad 2.2.2\quad .....\quad 2.2.n\quad \quad \quad =\quad { 2 }^{ 2 }\quad \sum { n } \\ \quad \quad \quad \quad \quad ...........\\ \quad \quad \quad \quad \quad n.n.1\quad +\quad n.n.2\quad .....\quad n.n.n\quad \quad \quad =\quad { n }^{ 2 }\quad \sum { n } \\ Therefore\quad M\quad =\quad \sum { { n }^{ 2 } } .\sum { n } \\ When\quad we\quad take\quad this\quad sum\quad 3\quad times,\quad all\quad the\quad cubes\quad have\quad been\quad \\ subtracted\quad two\quad more\quad times,\quad and\quad we\quad do\quad not\quad want\quad any\quad cubes\\ in\quad the\quad final\quad result.\\ To\quad compensate\quad for\quad the\quad defecit,\quad we\quad add\quad 2.\sum { { n }^{ 3 } } \\ The\quad remaining\quad numbers\quad are\quad of\quad the\quad form\quad \alpha \beta \gamma .\\ Similar\quad to\quad the\quad previous\quad case,\quad there\quad are\quad 6\quad numbers\quad which\quad give\quad \\ the\quad same\quad sum.\quad i.e.\quad \alpha \beta \gamma \quad =\quad \alpha \gamma \beta \quad =\quad \beta \alpha \gamma \quad =\quad \beta \gamma \alpha \quad =\quad \gamma \alpha \beta \quad =\quad \gamma \beta \alpha \quad \\ So\quad we\quad divide\quad by\quad 6.\\ Thus,\quad final\quad result\quad =\quad \frac { 1 }{ 6 } \left\{ N\quad -\quad 3M\quad +\quad 2.\sum { { n }^{ 3 } } \right\} \quad \\ =\quad \frac { 1 }{ 6 } \left\{ { \left[ { \sum { n } } \right] }^{ 3 }\quad -\quad 3.\sum { { n }^{ 2 } } .\sum { n } \quad +\quad 2.\sum { { n }^{ 3 } } \right\} \\ =\frac { 1 }{ 6 } \left\{ { \left[ \frac { n(n+1) }{ 2 } \right] }^{ 3 }\quad -\quad 3.\frac { n(n+1)(2n+1) }{ 6 } .\frac { n(n+1) }{ 2 } \quad +\quad 2.{ \left[ \frac { n(n+1) }{ 2 } \right] }^{ 2 } \right\} \\ After\quad factorizing,\quad we\quad get\quad \frac { { n }^{ 2 }{ (n+1) }^{ 2 }(n-1)(n-2) }{ 48 } \\ put\quad n=100\quad to\quad get\quad the\quad result.\\ \frac { { 100 }^{ 2 }{ (101) }^{ 2 }(99)(98) }{ 48 } mod(100)=20,618,771,250\quad mod(100)\quad =\quad \boxed { 50 } \\ \\ \quad$

We can cancel out almost everything modulo 100.

Step 1: Remove all products which include 100.

Step 2: Cancel out each term where no two terms add to 100. To do this we take the value farthest from 50 and flip it to get a complement. e.x. (27 x 36 x 86) cancels with (14 x 27 x 36). If we are given the former, we flip 86 to find the complement since it is farthest from 50. If we are given the latter, we flip 14 since it is farthest from 50. These cancel out since they sum to (86+14)(27)(36)

After this step all remaining terms contain exactly one pair of numbers which add to 100.

Step 3: Cancel out each term where the number other than the pair is not 50. To do this we just flip what this other number is to get a complement term.

After this step we are left with only terms of the form (n x 50 x (100-n)). These are 0 mod 100 when n is even and 50 mod 100 when n is odd. There are 25 cases when n is odd and so our final sum should be 50 mod 100.

It's an interesting problem, being a programmer I wrote this code in Python:

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sum=0
for num1 in range(1,99):
    for num2 in range(num1+1,100):
        for num3 in range(num2+1,101):
            #print(num1,num2,num3)
            sum+=(num1*num2*num3)

print(sum,sum%100)

Thus sum is 20618771250 and mod 100 gives 50.