In how many ways a cricketer can make a century with fours and sixes only?

Options

A. 6
B. 9
C. 8
D. 10

Answer

To solve the problem of how many ways a cricketer can make a century (100 runs) using only fours (4 runs) and sixes (6 runs), we can approach it step by step.

### Step-by-Step Solution:

1. **Define the Variables**:
Let \( x \) be the number of fours and \( y \) be the number of sixes. The equation we need to satisfy is:
\[ 4x + 6y = 100 \]

2. **Simplify the Equation**:
We can simplify the equation by dividing everything by 2:
\[ 2x + 3y = 50 \]

3. **Find Possible Values for \( y \)**:
Since \( y \) must be a non-negative integer, we can express \( x \) in terms of \( y \):
\[ 2x = 50 - 3y \implies x = \frac{50 - 3y}{2} \]
For \( x \) to be a non-negative integer, \( 50 - 3y \) must be non-negative and even.

4. **Determine the Range for \( y \)**:
- The maximum value for \( y \) occurs when \( 3y \leq 50 \):
\[ y \leq \frac{50}{3} \approx 16.67 \implies y \leq 16 \]
- The minimum value for \( y \) is 0.

5. **List Possible Values of \( y \)**:
We will check for each integer value of \( y \) from 0 to 16 and see if \( x \) is a non-negative integer:
- For \( y = 0 \): \( x = \frac{50 - 3(0)}{2} = 25 \) (valid)
- For \( y = 1 \): \( x = \frac{50 - 3(1)}{2} = 23.5 \) (not valid)
- For \( y = 2 \): \( x = \frac{50 - 3(2)}{2} = 22 \) (valid)
- For \( y = 3 \): \( x = \frac{50 - 3(3)}{2} = 19.5 \) (not valid)
- For \( y = 4 \): \( x = \frac{50 - 3(4)}{2} = 19 \) (valid)
- For \( y = 5 \): \( x = \frac{50 - 3(5)}{2} = 17.5 \) (not valid)
- For \( y = 6 \): \( x = \frac{50 - 3(6)}{2} = 16 \) (valid)
- For \( y = 7 \): \( x = \frac{50 - 3(7)}{2} = 13.5 \) (not valid)
- For \( y = 8 \): \( x = \frac{50 - 3(8)}{2} = 12 \) (valid)
- For \( y = 9 \): \( x = \frac{50 - 3(9)}{2} = 9.5 \) (not valid)
- For \( y = 10 \): \( x = \frac{50 - 3(10)}{2} = 9 \) (valid)
- For \( y = 11 \): \( x = \frac{50 - 3(11)}{2} = 6.5 \) (not valid)
- For \( y = 12 \): \( x = \frac{50 - 3(12)}{2} = 6 \) (valid)
- For \( y = 13 \): \( x = \frac{50 - 3(13)}{2} = 3.5 \) (not valid)
- For \( y = 14 \): \( x = \frac{50 - 3(14)}{2} = 3 \) (valid)
- For \( y = 15 \): \( x = \frac{50 - 3(15)}{2} = 0.5 \) (not valid)
- For \( y = 16 \): \( x = \frac{50 - 3(16)}{2} = 0 \) (valid)

6. **Count the Valid Combinations**:
The valid combinations of \( (x, y) \) pairs are:
- \( (25, 0) \)
- \( (22, 2) \)
- \( (19, 4) \)
- \( (16, 6) \)
- \( (12, 8) \)
- \( (9, 10) \)
- \( (6, 12) \)
- \( (3, 14) \)
- \( (0, 16) \)

This gives us a total of **9 valid combinations**.

### Final Answer:
The cricketer can make a century in **9 different ways** using only fours and sixes.