The number of terms between 11 and 200 which are divisible by 7 but not by 3 are

### Concept & Logic This problem applies the principle of set theory and divisibility. To find numbers divisible by $X$ but NOT by $Y$, you must first find the total count of numbers divisible by $X$, and then subtract the count of numbers that are divisible by both $X$ and $Y$. Numbers divisible by both $7$ and $3$ are exactly the numbers divisible by their Least Common Multiple (LCM). $$LCM(7, 3) = 21$$ ### Step-by-Step Solution * **Given:** Range is between $11$ and $200$. Target is divisible by $7$ but not $3$. * **Step 1: Find total numbers divisible by 7.** The first multiple of $7$ after $11$ is $14$. The last multiple of $7$ before $200$ is $196$ (since $200 \div 7 = 28$ remainder $4$, $200 - 4 = 196$). Count = $\frac{196 - 14}{7} + 1 = \frac{182}{7} + 1 = 26 + 1 = 27$. * **Step 2: Find numbers divisible by both 7 and 3 (i.e., divisible by 21).** The first multiple of $21$ after $11$ is $21$. The last multiple of $21$ before $200$ is $189$ (since $200 \div 21 = 9$ remainder $11$, $200 - 11 = 189$). Count = $\frac{189 - 21}{21} + 1 = \frac{168}{21} + 1 = 8 + 1 = 9$. * **Step 3: Subtract the common multiples.** Terms divisible by $7$ but not $3$ = (Total divisible by $7$) - (Total divisible by $21$) $$Count = 27 - 9 = 18$$ ### Exam Strategy & Shortcut Use the quotient method for ranges: Multiples of $7$ up to $200$ = $\lfloor \frac{200}{7} \rfloor = 28$. Multiples of $7$ up to $11$ = $\lfloor \frac{11}{7} \rfloor = 1$. Total multiples of $7$ = $28 - 1 = 27$. Multiples of $21$ up to $200$ = $\lfloor \frac{200}{21} \rfloor = 9$. Multiples of $21$ up to $11$ = $\lfloor \frac{11}{21} \rfloor = 0$. Total multiples of $21$ = $9 - 0 = 9$. Required answer = $27 - 9 = 18$. This prevents having to deduce the exact starting and ending multiples. ### Common Pitfall A common error is trying to exclude numbers that are just multiples of $3$ without recognizing that we only care about the overlap (multiples of $21$). Subtracting all multiples of $3$ from the range instead of from the multiples of $7$ will yield completely incorrect negative values. ### Final Answer **Therefore, the correct answer is 18.**