IBPS PO Mains 30 Nov 2024 Shift 2 Paper

Concept:Use the given series and conditions to determine $y$, then check odd powers of $y$ modulo $32$.Explanation:From Note III, $2x$ and $2y$ are the roots of $d^2 - 18d + 32 = 0$.Solving: $d = \frac{18 \pm \sqrt{324-128}}{2} = \frac{18 \pm 14}{2}$, so $d = 16$ or $d = 2$.Thus $\{2x, 2y\} = \{16, 2\}$, giving $(x, y) = (1,8)$ or $(8,1)$.Since all series elements are positive, $y - 7.5$ must be positive, so $y \neq 1$.Therefore $x = 1$ and $y = 8$.Now find $B$ using Series 1: $15, B, 18, (4x)^2, 21, y+11$.With $x=1$ and $y=8$, the series becomes $15, B, 18, 16, 21, 19$.Odd positions $15, 18, 21$ increase by $+3$; even positions $B, 16, 19$ also increase by $+3$, so $B=13$.For $Z$, use Series 3: $y-7.5, x, 28, 4, Z^3, 16$, which becomes $0.5, 1, 28, 4, Z^3, 16$.From Note II, $A, B, Z$ are distinct primes less than 20 and $A+B+Z$ is even.Since $B=13$ is odd, $A+Z$ must be odd, so one of them is $2$.If $A=2$, then $Z$ is an odd prime and $Z^3$ is too large for the series; hence $Z=2$.So $Z^3=8$, and $y=8$.Now find the remainder when $y^n = 8^n$ is divided by $32$ for odd $n$.For $n=1$: $8^1 \div 32$ gives remainder $8$.For $n=3$: $8^3 = 512$, and $512 \div 32 = 16$ exactly, giving remainder $0$.Since different odd values of $n$ give different remainders, a single remainder cannot be determined.Answer:Cannot be determined (Option E).