Question

If α, β are natural numbers such that 100$^α - 199 β = (100)(100) + (99)(101) + (98)(102) + ⋯ + (1)(199)$, then the slope of the line passing through (α, β) and origin is :

• A. 510
• B. 530
• C. 540
• D. 550

Hint:

The sum of squares $\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$.

Answer 1

The Correct Option is D

Step 1: RHS $= \sum_{r=0}^{99} (100-r)(100+r) = \sum_{r=0}^{99} (100^2 - r^2)$.
Step 2: RHS $= 100 \times 100^2 - \sum_{r=0}^{99} r^2 = 10^6 - \frac{99(100)(199)}{6}$.
Step 3: RHS $= 10^6 - 33(50)(199) = 10^6 - 1650(199) = 1000000 - 328350 = 671650$.
Step 4: $100^\alpha - 199\beta = 671650$. Try $\alpha=3$: $10^6 - 199\beta = 671650 \implies 199\beta = 328350$.
Step 5: $\beta = \frac{328350}{199} = 1650$.
Step 6: Slope $m = \frac{\beta}{\alpha} = \frac{1650}{3} = 550$.