2023 CSAT — Q14

The key fact: signs and even coefficients don’t affect parity. A minus sign is "$+$" mod $2$, and any even multiplier kills its term mod $2$.

The set $\{p,q,r,s,t\}$ has three even and two odd numbers (the labels of which are odd are not fixed).

Statement 1: $p+q+r-s-t \equiv p+q+r+s+t \pmod 2$. The total parity is (three evens) $+$ (two odds) $=$ even $+$ even $=$ even, regardless of which two are odd. So this is definitely even. True.

Statement 2: $2p + q + 2r - 2s + t \equiv q + t \pmod 2$ (the $2p, 2r, 2s$ terms vanish). The parity of $q + t$ depends on which of the five are the odd ones:

• if $q,t$ are the two odds: $q+t$ even;
• if $q,t$ are both even: $q+t$ even;
• if exactly one of $q,t$ is odd: $q+t$ odd.

All three arrangements are allowed, so $2p+q+2r-2s+t$ is not definitely anything. False.

Answer: (a) 1 only.

Reduce everything mod $2$ on sight: drop even-coefficient terms, turn every $-$ into $+$. Statement 1 collapses to “parity of all five $=$ 3 evens $+$ 2 odds $=$ even” — a fixed fact. Statement 2 collapses to $q+t$, whose parity is not fixed because the problem never says which letters are odd. The discipline “definitely means for every admissible arrangement” is what separates a confident (a) from the traps.