Find the smallest number which when divided by 25, 40 and 60 leaves remainder 8 in each case.

Answer

To find the smallest number which, when divided by 25, 40, and 60, leaves a remainder of 8 in each case, we can follow these steps:

### Step 1: Understand the Problem
We need to find a number \( N \) such that:
- \( N \mod 25 = 8 \)
- \( N \mod 40 = 8 \)
- \( N \mod 60 = 8 \)

This means that when \( N \) is divided by 25, 40, and 60, it leaves a remainder of 8.

### Step 2: Rewrite the Conditions
We can rewrite the conditions as:
- \( N - 8 \) is divisible by 25
- \( N - 8 \) is divisible by 40
- \( N - 8 \) is divisible by 60

This implies that \( N - 8 \) must be a common multiple of 25, 40, and 60.

### Step 3: Find the Least Common Multiple (LCM)
To find \( N - 8 \), we need to calculate the LCM of 25, 40, and 60.

**Finding the prime factorization:**
- \( 25 = 5^2 \)
- \( 40 = 2^3 \times 5^1 \)
- \( 60 = 2^2 \times 3^1 \times 5^1 \)

**Finding the LCM:**
- Take the highest power of each prime factor:
- For \( 2 \): highest power is \( 2^3 \) (from 40)
- For \( 3 \): highest power is \( 3^1 \) (from 60)
- For \( 5 \): highest power is \( 5^2 \) (from 25)

Thus, the LCM is:
\[ LCM = 2^3 \times 3^1 \times 5^2 = 8 \times 3 \times 25 \]

Calculating this:
\[ 8 \times 3 = 24 \]
\[ 24 \times 25 = 600 \]

So, \( LCM(25, 40, 60) = 600 \).

### Step 4: Calculate \( N \)
Since \( N - 8 \) is equal to the LCM we found:
\[ N - 8 = 600 \]
Thus,
\[ N = 600 + 8 = 608 \]

### Conclusion
The smallest number which, when divided by 25, 40, and 60, leaves a remainder of 8 is **608**.

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