Question:
The number of multiples of 6 lying between 25 and 363 is :
The number of multiples of 6 lying between 25 and 363 is :
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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.
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.
Updated On: Jul 7, 2026
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The topic of this question is Arithmetic Progressions (AP).
We are required to find the total count of numbers that are divisible by 6 and lie strictly between 25 and 363.
The terms that are multiples of 6 will form an arithmetic progression where the common difference between consecutive terms is 6.
We need to determine the first term, the last term, and then calculate the total number of terms in this sequence.
Step 2: Key Formula or Approach:
The general term of an Arithmetic Progression is given by the formula:
\[ a_n = a + (n - 1)d \]
Where:
- \( a \) represents the first term of the arithmetic progression.
- \( d \) represents the common difference between the consecutive terms.
- \( a_n \) represents the \( n \)-th term (or the last term) of the arithmetic progression.
- \( n \) represents the total number of terms in the sequence.
Step 3: Detailed Explanation:
1. First, we find the first multiple of 6 that is strictly greater than 25.
Dividing 25 by 6 gives 4 with a remainder of 1.
The next higher multiple of 6 is \( 6 \times 5 = 30 \).
Therefore, the first term of our arithmetic progression is \( a = 30 \).
2. Next, we find the last multiple of 6 that is strictly less than 363.
Dividing 363 by 6 gives 60 with a remainder of 3.
Subtracting the remainder from 363 gives \( 363 - 3 = 360 \).
This is \( 6 \times 60 = 360 \), which is less than 363.
Therefore, the last term of our arithmetic progression is \( a_n = 360 \).
3. The common difference \( d \) is 6, since we are dealing with multiples of 6.
4. Now, we substitute these values into the general term formula:
\[ 360 = 30 + (n - 1) \times 6 \]
5. Subtract 30 from both sides of the equation to isolate the term with \( n \):
\[ 360 - 30 = (n - 1) \times 6 \]
\[ 330 = (n - 1) \times 6 \]
6. Divide both sides of the equation by 6:
\[ \frac{330}{6} = n - 1 \]
\[ 55 = n - 1 \]
7. Solve for \( n \) by adding 1 to both sides:
\[ n = 55 + 1 \]
\[ n = 56 \]
8. Thus, there are 56 multiples of 6 that lie between 25 and 363.
Step 4: Final Answer:
The number of multiples of 6 lying between 25 and 363 is 56, which matches option (A).
The topic of this question is Arithmetic Progressions (AP).
We are required to find the total count of numbers that are divisible by 6 and lie strictly between 25 and 363.
The terms that are multiples of 6 will form an arithmetic progression where the common difference between consecutive terms is 6.
We need to determine the first term, the last term, and then calculate the total number of terms in this sequence.
Step 2: Key Formula or Approach:
The general term of an Arithmetic Progression is given by the formula:
\[ a_n = a + (n - 1)d \]
Where:
- \( a \) represents the first term of the arithmetic progression.
- \( d \) represents the common difference between the consecutive terms.
- \( a_n \) represents the \( n \)-th term (or the last term) of the arithmetic progression.
- \( n \) represents the total number of terms in the sequence.
Step 3: Detailed Explanation:
1. First, we find the first multiple of 6 that is strictly greater than 25.
Dividing 25 by 6 gives 4 with a remainder of 1.
The next higher multiple of 6 is \( 6 \times 5 = 30 \).
Therefore, the first term of our arithmetic progression is \( a = 30 \).
2. Next, we find the last multiple of 6 that is strictly less than 363.
Dividing 363 by 6 gives 60 with a remainder of 3.
Subtracting the remainder from 363 gives \( 363 - 3 = 360 \).
This is \( 6 \times 60 = 360 \), which is less than 363.
Therefore, the last term of our arithmetic progression is \( a_n = 360 \).
3. The common difference \( d \) is 6, since we are dealing with multiples of 6.
4. Now, we substitute these values into the general term formula:
\[ 360 = 30 + (n - 1) \times 6 \]
5. Subtract 30 from both sides of the equation to isolate the term with \( n \):
\[ 360 - 30 = (n - 1) \times 6 \]
\[ 330 = (n - 1) \times 6 \]
6. Divide both sides of the equation by 6:
\[ \frac{330}{6} = n - 1 \]
\[ 55 = n - 1 \]
7. Solve for \( n \) by adding 1 to both sides:
\[ n = 55 + 1 \]
\[ n = 56 \]
8. Thus, there are 56 multiples of 6 that lie between 25 and 363.
Step 4: Final Answer:
The number of multiples of 6 lying between 25 and 363 is 56, which matches option (A).
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