How many numbers lie between 100 and 10000 which when successive divided by 7, 11 and 13 leaves the respective remainders of 5,6 and 7 ?

Answer

To solve the problem, we need to find numbers between 100 and 10,000 that, when divided by 7, 11, and 13, leave remainders of 5, 6, and 7 respectively.

### Step 1: Set Up the Congruences
We can express the conditions given in the problem using modular arithmetic:
- Let \( x \) be the number we are looking for.
- The conditions can be written as:
\[ x \equiv 5 \mod{7} \]
\[ x \equiv 6 \mod{11} \]
\[ x \equiv 7 \mod{13} \]

### Step 2: Solve the System of Congruences
We can solve these congruences using the method of successive substitutions or the Chinese Remainder Theorem.

1. **Start with the first two congruences:**
- From \( x \equiv 5 \mod{7} \), we can express \( x \) as:
\[ x = 7k + 5 \quad \text{for some integer } k \]
- Substitute this into the second congruence:
\[ 7k + 5 \equiv 6 \mod{11} \]
\[ 7k \equiv 1 \mod{11} \]
- To solve for \( k \), we need the multiplicative inverse of 7 modulo 11. The inverse is 8 since \( 7 \cdot 8 \equiv 1 \mod{11} \).
- Thus,
\[ k \equiv 8 \mod{11} \implies k = 11m + 8 \quad \text{for some integer } m \]
- Substitute back to find \( x \):
\[ x = 7(11m + 8) + 5 = 77m + 56 + 5 = 77m + 61 \]

2. **Now include the third congruence:**
- Substitute \( x = 77m + 61 \) into \( x \equiv 7 \mod{13} \):
\[ 77m + 61 \equiv 7 \mod{13} \]
- Calculate \( 77 \mod{13} \) and \( 61 \mod{13} \):
\[ 77 \equiv 12 \mod{13} \quad \text{and} \quad 61 \equiv 9 \mod{13} \]
- Thus, the equation becomes:
\[ 12m + 9 \equiv 7 \mod{13} \]
\[ 12m \equiv -2 \equiv 11 \mod{13} \]
- The multiplicative inverse of 12 modulo 13 is 12 (since \( 12 \cdot 12 \equiv 1 \mod{13} \)):
\[ m \equiv 12 \cdot 11 \mod{13} \equiv 132 \mod{13} \equiv 2 \mod{13} \]
- Therefore,
\[ m = 13n + 2 \quad \text{for some integer } n \]
- Substitute back to find \( x \):
\[ x = 77(13n + 2) + 61 = 1001n + 154 + 61 = 1001n + 215 \]

### Step 3: Find Valid Values of \( x \)
Now we need to find values of \( n \) such that:
\[ 100 < 1001n + 215 < 10000 \]

1. **Lower Bound:**
\[ 1001n + 215 > 100 \implies 1001n > -115 \implies n \geq 0 \]

2. **Upper Bound:**
\[ 1001n + 215 < 10000 \implies 1001n < 9785 \implies n < 9.76 \implies n \leq 9 \]

### Step 4: Count the Values of \( n \)
The possible integer values for \( n \) are \( 0, 1, 2, \ldots, 9 \). This gives us a total of:
\[ n = 10 \text{ values} \]

### Final Answer
Thus, there are **10 numbers** between 100 and 10,000 that satisfy the given conditions.

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