To find the smallest number that meets the given conditions, we can follow these steps:
### Step 1: Understand the Problem
We need to find a number \( N \) such that:
- \( N \mod 5 = 2 \)
- \( N \mod 10 = 2 \)
- \( N \mod 12 = 2 \)
- \( N \mod 15 = 2 \)
- \( N \mod 7 = 0 \)
### Step 2: Rewrite the Conditions
From the first four conditions, we can express \( N \) as:
\[ N = k \cdot \text{lcm}(5, 10, 12, 15) + 2 \]
where \( k \) is some integer.
### Step 3: Calculate the LCM
To find the least common multiple (LCM) of the numbers 5, 10, 12, and 15:
- The prime factorization is:
- \( 5 = 5^1 \)
- \( 10 = 2^1 \cdot 5^1 \)
- \( 12 = 2^2 \cdot 3^1 \)
- \( 15 = 3^1 \cdot 5^1 \)
Taking the highest power of each prime:
- \( 2^2 \) from 12
- \( 3^1 \) from 12 and 15
- \( 5^1 \) from 5, 10, and 15
Thus, the LCM is:
\[ \text{lcm}(5, 10, 12, 15) = 2^2 \cdot 3^1 \cdot 5^1 = 4 \cdot 3 \cdot 5 = 60 \]
### Step 4: Express \( N \)
Now we can express \( N \) as:
\[ N = 60k + 2 \]
### Step 5: Apply the Condition for 7
Next, we need \( N \) to be divisible by 7:
\[ 60k + 2 \equiv 0 \mod 7 \]
This simplifies to:
\[ 60k \equiv -2 \mod 7 \]
Calculating \( 60 \mod 7 \):
\[ 60 \div 7 = 8 \quad \text{(remainder 4)} \]
So,
\[ 60 \equiv 4 \mod 7 \]
Thus, we have:
\[ 4k \equiv -2 \mod 7 \]
This can be rewritten as:
\[ 4k \equiv 5 \mod 7 \]
### Step 6: Solve for \( k \)
To solve for \( k \), we can find the multiplicative inverse of 4 modulo 7. The inverse of 4 is 2 because:
\[ 4 \cdot 2 = 8 \equiv 1 \mod 7 \]
Now, multiply both sides of \( 4k \equiv 5 \) by 2:
\[ k \equiv 10 \mod 7 \]
This simplifies to:
\[ k \equiv 3 \mod 7 \]
### Step 7: Find the Smallest \( k \)
The smallest non-negative integer \( k \) that satisfies \( k \equiv 3 \mod 7 \) is \( k = 3 \).
### Step 8: Calculate \( N \)
Substituting \( k = 3 \) back into the equation for \( N \):
\[ N = 60 \cdot 3 + 2 = 180 + 2 = 182 \]
### Conclusion
The smallest number that meets all the conditions is:
\[ \boxed{182} \]
