2025 CSAT — Q6

$p - q = q - r$ says $p, q, r$ are in arithmetic progression with middle term $q$, so $p + q + r = 3q$. We hunt prime APs with all terms $< 20$.

Reading A — three distinct primes (the natural reading of “three prime numbers $p, q, r$”). Because a $3$-term AP of primes (other than $3,5,7$) needs an even common difference (a $d$ that is odd would force an even term, and $2$ cannot sit in the middle of three primes $<20$), the prime APs are:

with middle terms $q = 5, 7, 11, 13, 11$. The sums $3q$ are $15, 21, 33, 39, 33$ — the distinct values are $\{15, 21, 33, 39\}$, i.e. $4$.

Reading B — primes “not necessarily distinct” (allowing $d = 0$, i.e. $p=q=r$). Then any prime $q$ gives $(q,q,q)$ with sum $3q \in \{6,9,15,21,33,39,51,57\}$; the non-constant APs add no new sums. That gives $8$ distinct values (option (d)).

Neither natural reading yields $6$. My independent blind-solve gives $4$ (distinct primes, the reading I judge intended) or $8$ (repeats allowed).

The mathematics is clean — $p+q+r = 3q$, so we count distinct middle terms, not triples — but the reading is the trap, and this is a live example of why we never reverse-engineer an answer from a published key. The number of prime-AP triples ($5$) differs from the number of distinct sums ($4$, since $33$ recurs), and the “repeats allowed” reading gives $8$. A disciplined solver states the reading explicitly, counts distinct $3q$ values, and — finding the official $6$ irreproducible — flags it rather than bending the math. That honesty is the scoring discipline.