Question

The number of multiples of 6 lying between 25 and 363 is :

• A. 56
• B. 56.5
• C. 57
• D. 58

Hint:

To quickly find the number of multiples of a number \( k \) between \( A \) and \( B \) (where both are exclusive):
Calculate the difference between the greatest integer functions of the boundaries.
Use the formula: \( \lfloor \frac{B - 1}{k} \rfloor - \lfloor \frac{A}{k} \rfloor \).
For this question, we calculate: \( \lfloor \frac{362}{6} \rfloor - \lfloor \frac{25}{6} \rfloor = 60 - 4 = 56 \).
This fast calculation is extremely helpful in saving time during competitive examinations.

Answer 1

The Correct Option is A

Step 1: Set up the counting problem.
We need every multiple of 6 that sits strictly between 25 and 363. Instead of building an AP and using the $n$th term formula, we can count these multiples directly using the idea that the number of multiples of 6 up to any number $N$ is given by $\left\lfloor \frac{N}{6} \right\rfloor$, the greatest whole number of times 6 divides into $N$.

Step 2: Count multiples of 6 up to 363.
Divide 363 by 6.
\[ 363 \div 6 = 60 \text{ remainder } 3 \]
So $6 \times 60 = 360$, which is the largest multiple of 6 not exceeding 363. This means there are 60 multiples of 6 from 6 up to 360.

Step 3: Count multiples of 6 up to 25.
Divide 25 by 6.
\[ 25 \div 6 = 4 \text{ remainder } 1 \]
So $6 \times 4 = 24$, the largest multiple of 6 not exceeding 25. This means there are 4 multiples of 6 from 6 up to 24, and all of these are less than or equal to 25, so none of them lie in our required range.

Step 4: Subtract to get the count strictly between 25 and 363.
The multiples of 6 lying between 25 and 363 are exactly the multiples up to 360 minus the multiples up to 24 (since 360 is less than 363 and 24 is less than 25, none of the boundary values 25 or 363 is itself a multiple of 6, so we don't need to adjust for that).
\[ \text{Required count} = 60 - 4 = 56 \]

Step 5: Check the answer makes sense.
The first multiple after 25 is 30, and the last multiple before 363 is 360. Using the count formed this way, $30, 36, 42, \ldots, 360$ should have 56 terms, which matches our subtraction.

Final Answer:
The number of multiples of 6 between 25 and 363 is 56, so option (A) is correct.
\[ \boxed{56} \]