In the 1973 MAA Annual High School Mathematics Examination (Part 3, Problem #26), students were challenged with:

"The number of sets of two or more consecutive positive integers whose sum is 100 is: (A) 1 (B) 2 (C) 3 (D) 4 (E) 5."

Let a sequence of $k$ consecutive positive integers start at $a \ge 1$: $$a, \quad a + 1, \quad a + 2, \quad \dots, \quad a + k - 1$$ The sum of these $k$ numbers is: $$\text{Sum} = k \cdot a + \frac{(k - 1)k}{2} = \frac{k(2a + k - 1)}{2} = 100$$ Multiplying both sides by 2 gives: $$k(2a + k - 1) = 200$$ This is a product of two positive integers whose product is $200$. Notice two key mathematical properties:

1. Since $a \ge 1$, we have $2a + k - 1 \ge k + 1 > k$.
2. The difference between the two factors is $(2a + k - 1) - k = 2a - 1$, which is always odd!

Therefore, one factor must be odd and the other must be even. Factoring $200 = 2^3 \times 5^2$: The odd factor can be $1$, $5$, or $25$:

• Odd factor = 1: $k = 1 \implies 2a = 200 \implies a = 100$. But the problem specifies two or more integers ($k \ge 2$), so this is excluded!
• Odd factor = 5: Setting $k = 5$ gives $2a + 4 = 40 \implies 2a = 36 \implies a = 18$.
  The 5 numbers are: $18 + 19 + 20 + 21 + 22 = 100$.
• Odd factor = 25: Setting $2a + k - 1 = 25$ and $k = 8$ gives $2a + 7 = 25 \implies 2a = 18 \implies a = 9$.
  The 8 numbers are: $9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 100$.

Thus, there are exactly 2 sets!

In this lesson, we write nested loops and divisor algorithms in Lua to search for consecutive sequences that sum to 100, and generalize to any target number!