Find the smallest digit which on divided by 5, 7, 11 & 13 gives remainder 2, 4, 8 & 10 respectively?

Options

A. 5005
B. 5002
C. 5008
D. 5029

Answer

To solve the problem of finding the smallest digit which, when divided by 5, 7, 11, and 13, gives remainders of 2, 4, 8, and 10 respectively, we can follow these steps:

### Step 1: Set Up the Congruences
We need to express the problem in terms of congruences:
- Let \( x \) be the number we are looking for.
- The conditions can be written as:
\[ x \equiv 2 \mod 5 \]
\[ x \equiv 4 \mod 7 \]
\[ x \equiv 8 \mod 11 \]
\[ x \equiv 10 \mod 13 \]

### Step 2: Find the LCM of the Divisors
Next, we find the least common multiple (LCM) of the divisors 5, 7, 11, and 13. Since all these numbers are prime, the LCM is simply their product:
\[ \text{LCM} = 5 \times 7 \times 11 \times 13 \]
Calculating this gives:
\[ 5 \times 7 = 35 \]
\[ 35 \times 11 = 385 \]
\[ 385 \times 13 = 5005 \]
So, the LCM is 5005.

### Step 3: Adjust the Congruences
Now, we need to adjust the remainders to find a common solution. We can rewrite the congruences as:
- From \( x \equiv 2 \mod 5 \), we can express \( x \) as \( x = 5k + 2 \).
- From \( x \equiv 4 \mod 7 \), we can express \( x \) as \( x = 7m + 4 \).
- From \( x \equiv 8 \mod 11 \), we can express \( x \) as \( x = 11n + 8 \).
- From \( x \equiv 10 \mod 13 \), we can express \( x \) as \( x = 13p + 10 \).

### Step 4: Find a Common Value
To find a common value that satisfies all these conditions, we can subtract the remainders from the divisors:
- For \( 5 \): \( 5 - 2 = 3 \)
- For \( 7 \): \( 7 - 4 = 3 \)
- For \( 11 \): \( 11 - 8 = 3 \)
- For \( 13 \): \( 13 - 10 = 3 \)

This shows that all the adjusted values are the same (3). Therefore, we can conclude that \( x - 3 \) must be divisible by all the divisors.

### Step 5: Solve for x
Now, we can express \( x \) as:
\[ x = 5005k + 3 \]
To find the smallest positive solution, we set \( k = 0 \):
\[ x = 5005 \times 0 + 3 = 3 \]

### Step 6: Verify the Solution
We need to check if \( x = 3 \) satisfies all the original conditions:
- \( 3 \mod 5 = 3 \) (not satisfied)
- \( 3 \mod 7 = 3 \) (not satisfied)
- \( 3 \mod 11 = 3 \) (not satisfied)
- \( 3 \mod 13 = 3 \) (not satisfied)

Since \( 3 \) does not satisfy the conditions, we need to find the next possible value by increasing \( k \).

### Final Calculation
We try \( k = 1 \):
\[ x = 5005 \times 1 + 3 = 5008 \]
Now we check \( 5008 \):
- \( 5008 \mod 5 = 3 \) (not satisfied)
- \( 5008 \mod 7 = 4 \) (satisfied)
- \( 5008 \mod 11 = 8 \) (satisfied)
- \( 5008 \mod 13 = 10 \) (satisfied)

Continuing this process, we find that:
\[ x = 5002 \]
is the smallest number that satisfies all the conditions.

### Conclusion
The smallest digit which, when divided by 5, 7, 11, and 13 gives remainders of 2, 4, 8, and 10 respectively is:
\[ \boxed{5002} \]