Out of the numbers divisible by 3 between 14 and 95 if the numbers with 3 at unit's place are removed, then how many numbers will remain?

### Concept & Strategy This problem tests your ability to filter sequences based on specific digit constraints. First, determine the total number of multiples of a given divisor within the specified range. Then, manually identify the specific terms that violate the secondary condition (in this case, having a unit digit of $3$) and subtract their count from the total. ### Step-by-Step Solution * **Given:** The range is between $14$ and $95$. We need numbers divisible by $3$, but must exclude any number ending in $3$. * **Step 1: Total numbers divisible by 3 in the range.** The first multiple of $3$ after $14$ is $15$. The last multiple of $3$ before $95$ is $93$. Using the A.P. formula: $n = \frac{Last - First}{3} + 1$ $$n = \frac{93 - 15}{3} + 1$$ $$n = \frac{78}{3} + 1 = 26 + 1 = 27$$ So, there are $27$ multiples of $3$ in this range. * **Step 2: Identify and remove numbers with 3 at the unit's place.** List the numbers in the range ending in $3$: $23, 33, 43, 53, 63, 73, 83, 93$. Now, check which of these are actually divisible by $3$ (sum of digits must be a multiple of $3$): * $33$ ($3+3=6$, Yes) * $63$ ($6+3=9$, Yes) * $93$ ($9+3=12$, Yes) There are exactly $3$ numbers that fit both criteria. * **Step 3: Calculate the final remaining numbers.** Remaining = (Total multiples of $3$) - (Multiples of $3$ ending in $3$) $$Remaining = 27 - 3 = 24$$ ### Exam Strategy & Shortcut An arithmetic progression with a common difference of $3$ repeats its unit digits in a cycle of $10$ (e.g., $3 \times 1 = 3$, $3 \times 11 = 33$, $3 \times 21 = 63$). Within every $30$ numbers, there will be exactly one multiple of $3$ ending in $3$. From $14$ to $95$ spans roughly $80$ numbers, meaning there should be $2$ or $3$ such occurrences. Spot-checking $33, 63,$ and $93$ is immediate and removes the need to write out non-multiples like $23$ or $43$. ### Common Pitfall Students often subtract *all* numbers ending in $3$ (which is $8$ numbers) from the total $27$, resulting in $19$. You must remember that you are only removing numbers that were in your original list of multiples of $3$. ### Final Answer **Therefore, the correct answer is 24.**