Find the least number which when divided by 25, 40 and 60 leaves 9 as the remainder ineach case.

Answer

To find the least number which, when divided by 25, 40, and 60, leaves a remainder of 9 in each case, we can follow these steps:

### Step 1: Understand the problem
We need to find a number \( N \) such that:
- \( N \mod 25 = 9 \)
- \( N \mod 40 = 9 \)
- \( N \mod 60 = 9 \)

This means that \( N - 9 \) should be divisible by 25, 40, and 60.

### Step 2: Set up the equation
Let \( M = N - 9 \). Then, we need to find \( M \) such that:
- \( M \mod 25 = 0 \)
- \( M \mod 40 = 0 \)
- \( M \mod 60 = 0 \)

This means \( M \) must be a common multiple of 25, 40, and 60.

### Step 3: Find the LCM of 25, 40, and 60
To find the least common multiple (LCM), we can factor each number into its prime factors:
- \( 25 = 5^2 \)
- \( 40 = 2^3 \times 5^1 \)
- \( 60 = 2^2 \times 3^1 \times 5^1 \)

Now, we 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)

Now, we can calculate the LCM:
\[ \text{LCM} = 2^3 \times 3^1 \times 5^2 \]

### Step 4: Calculate the LCM
Calculating the LCM:
\[ 2^3 = 8, \quad 3^1 = 3, \quad 5^2 = 25 \]
Now multiply these together:
\[ \text{LCM} = 8 \times 3 \times 25 \]
First, calculate \( 8 \times 3 = 24 \).
Then, calculate \( 24 \times 25 = 600 \).

So, \( M = 600 \).

### Step 5: Find \( N \)
Now, recall that \( N = M + 9 \):
\[ N = 600 + 9 = 609 \]

### Final Answer
The least number which when divided by 25, 40, and 60 leaves a remainder of 9 is \( \boxed{609} \).

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