[UPLIFTJEE]

If the set $\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}+5 \mathrm{~b}=42, \mathrm{a}, \mathrm{b} \in \mathrm{N}\}$ has m elements and $\sum_{\mathrm{n}=1}^{\mathrm{m}}\left(1-\mathrm{i}^{\mathrm{n}!}\right)=\mathrm{x}+\mathrm{i}$ y, where $\mathrm{i}=\sqrt{-1}$, then the value of $\mathrm{m}+\mathrm{x}+\mathrm{y}$ is
[JEE (Main) 8 April 2024 S I]
(A) 8
(B) 5
(C) 4
(D) 12

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[UPLIFTJEE]

Solution

$\mathrm{a}+5 \mathrm{~b}=42, \mathrm{a}, \mathrm{b} \in \mathrm{N} \Rightarrow \mathrm{a}=42-5 \mathrm{~b} \Rightarrow \mathrm{~b}=1, \mathrm{a}=37 \Rightarrow \mathrm{~b}=2, \mathrm{a}=32$

\[ \begin{aligned} & \mathrm{b}=3, \mathrm{a}=27 \\ & \vdots \\ & \mathrm{~b}=8, \mathrm{a}=2 \Rightarrow \mathrm{R} \text { has } 8 \text { elements, hence } \mathrm{m}=8 \\ & \sum_{\mathrm{n}=1}^{\mathrm{m}}\left(1-\mathrm{i}^{\mathrm{n}!}\right)=\mathrm{x}+\mathrm{i} y \text {, for } \mathrm{n} \geq 4, \mathrm{i}^{\mathrm{n}!}=1 \Rightarrow(1-\mathrm{i})+\left(1-\mathrm{i}^{2!}\right)+\left(1-\mathrm{i}^{3!}\right)+0+0=\mathrm{x}+\mathrm{iy} \Rightarrow 1-\mathrm{i}+1+1+1+1=5-\mathrm{i}=\mathrm{x}+\mathrm{iy} \end{aligned} \]

we get $\mathrm{x}=5, \mathrm{y}=-1 \Rightarrow \mathrm{~m}+\mathrm{x}+\mathrm{y}=8+5-1=12$

[UPLIFTJEE]

Answer

D

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