An integer is chosen at random from the first two hundred digits. What is the probability that the integer chosen is divisible by 6 or 8 ?

Answer

To solve the problem of finding the probability that an integer chosen at random from the first 200 integers is divisible by 6 or 8, we can follow these steps:

### Step 1: Define the total number of integers
The total number of integers we are considering is from 1 to 200, which gives us:
\[ N = 200 \]

### Step 2: Find the count of integers divisible by 6
To find how many integers from 1 to 200 are divisible by 6, we can use the formula for the number of terms in an arithmetic sequence.

The first integer divisible by 6 is 6, and the last integer less than or equal to 200 that is divisible by 6 is 198 (since \( 6 \times 33 = 198 \)).

The sequence of integers divisible by 6 is:
\[ 6, 12, 18, \ldots, 198 \]

This is an arithmetic sequence where:
- First term \( a = 6 \)
- Common difference \( d = 6 \)
- Last term \( l = 198 \)

To find the number of terms \( n \):
\[ l = a + (n-1) \cdot d \]
\[ 198 = 6 + (n-1) \cdot 6 \]
\[ 198 - 6 = (n-1) \cdot 6 \]
\[ 192 = (n-1) \cdot 6 \]
\[ n-1 = \frac{192}{6} = 32 \]
\[ n = 32 + 1 = 33 \]

So, there are 33 integers from 1 to 200 that are divisible by 6.

### Step 3: Find the count of integers divisible by 8
Next, we find how many integers from 1 to 200 are divisible by 8.

The first integer divisible by 8 is 8, and the last integer less than or equal to 200 that is divisible by 8 is 200.

The sequence of integers divisible by 8 is:
\[ 8, 16, 24, \ldots, 200 \]

This is also an arithmetic sequence where:
- First term \( a = 8 \)
- Common difference \( d = 8 \)
- Last term \( l = 200 \)

To find the number of terms \( n \):
\[ l = a + (n-1) \cdot d \]
\[ 200 = 8 + (n-1) \cdot 8 \]
\[ 200 - 8 = (n-1) \cdot 8 \]
\[ 192 = (n-1) \cdot 8 \]
\[ n-1 = \frac{192}{8} = 24 \]
\[ n = 24 + 1 = 25 \]

So, there are 25 integers from 1 to 200 that are divisible by 8.

### Step 4: Find the count of integers divisible by both 6 and 8 (i.e., divisible by 24)
Now, we need to find how many integers are divisible by both 6 and 8. The least common multiple (LCM) of 6 and 8 is 24.

The first integer divisible by 24 is 24, and the last integer less than or equal to 200 that is divisible by 24 is 192.

The sequence of integers divisible by 24 is:
\[ 24, 48, 72, \ldots, 192 \]

This is an arithmetic sequence where:
- First term \( a = 24 \)
- Common difference \( d = 24 \)
- Last term \( l = 192 \)

To find the number of terms \( n \):
\[ l = a + (n-1) \cdot d \]
\[ 192 = 24 + (n-1) \cdot 24 \]
\[ 192 - 24 = (n-1) \cdot 24 \]
\[ 168 = (n-1) \cdot 24 \]
\[ n-1 = \frac{168}{24} = 7 \]
\[ n = 7 + 1 = 8 \]

So, there are 8 integers from 1 to 200 that are divisible by both 6 and 8.

### Step 5: Use the principle of inclusion-exclusion to find the count of integers divisible by 6 or 8
Using the principle of inclusion-exclusion:
\[ |A \cup B| = |A| + |B| - |A \cap B| \]
Where:
- \( |A| = 33 \) (divisible by 6)
- \( |B| = 25 \) (divisible by 8)
- \( |A \cap B| = 8 \) (divisible by both)

Calculating:
\[ |A \cup B| = 33 + 25 - 8 = 50 \]

### Step 6: Calculate the probability
The probability \( P \) that a randomly chosen integer from 1 to 200 is divisible by 6 or 8 is given by:
\[ P = \frac{|A \cup B|}{N} = \frac{50}{200} = \frac{1}{4} \]

### Final Answer
The probability that the integer chosen is divisible by 6 or 8 is:
\[ \frac{1}{4} \]