JEE Main 2026 — 24 January, Evening Shift — Question 23

Given $z = (1+i)(1+2i)(1+3i)\ldots(1+ni)$.

$$|z|^2 = \prod_{r=1}^n |1+ri|^2 = \prod_{r=1}^n (1+r^2).$$

Given $|z|^2 = 44200$. Factorize 44200: $44200 = 442 \times 100 = 2 \times 221 \times 100 = 2 \times 13 \times 17 \times 100 = 2 \times 13 \times 17 \times 2^2 \times 5^2 = 2^3 \times 5^2 \times 13 \times 17$. Now, $1+r^2$ for $r=1,2,3,\ldots$: $1+1^2=2$, $1+2^2=5$, $1+3^2=10=2\times5$, $1+4^2=17$, $1+5^2=26=2\times13$. Product up to $n=5$: $2 \times 5 \times 10 \times 17 \times 26 = 2 \times 5 \times (2\times5) \times 17 \times (2\times13) = 2^3 \times 5^2 \times 13 \times 17 = 44200$. Thus, $n=5$.