2025 CSAT — Q25

First prune the redundant divisor. Every multiple of $9$ is already a multiple of $3$, so “not divisible by $9$” adds nothing once “not divisible by $3$” is imposed. The condition collapses to: not divisible by any of $2, 3, 5, 7$.

Count, by inclusion–exclusion, how many of $1$–$100$ are divisible by at least one of $2,3,5,7$:

Singles: $\lfloor 100/2\rfloor + \lfloor100/3\rfloor + \lfloor100/5\rfloor + \lfloor100/7\rfloor = 50 + 33 + 20 + 14 = 117.$

Pairs (subtract): $\lfloor100/6\rfloor + \lfloor100/{10}\rfloor + \lfloor100/{14}\rfloor + \lfloor100/{15}\rfloor + \lfloor100/{21}\rfloor + \lfloor100/{35}\rfloor = 16 + 10 + 7 + 6 + 4 + 2 = 45.$

Triples (add): $\lfloor100/{30}\rfloor + \lfloor100/{42}\rfloor + \lfloor100/{70}\rfloor + \lfloor100/{105}\rfloor = 3 + 2 + 1 + 0 = 6.$

Quad (subtract): $\lfloor100/{210}\rfloor = 0.$

Not divisible by any: $100 - 78 = 22$.

Answer: (c) 22. (These are $1$ together with the $21$ primes $>7$ up to $100$.)

The exam-fast move is killing the redundant divisor before any counting: $9$ is a subset condition of $3$, so four-way inclusion–exclusion ($2^4-1 = 15$ terms) shrinks to the cleaner three-way-plus-one over $\{2,3,5,7\}$. Mishandling the $9$ — either double-removing its multiples or running a needless $9$-term — is the engineered error. Under the clock, also sanity-check the answer structurally: the survivors are exactly $1$ and the primes in $(7,100]$, of which there are $21$, so $22$ total — a fast independent confirmation of the inclusion–exclusion arithmetic.