Find the least number which when divided by 16,36 and 40 leaves 5 as remainder in each case.

Answer

To find the least number which, when divided by 16, 36, and 40, leaves a remainder of 5 in each case, we can follow these steps:

### Step 1: Understand the problem
We need to find a number \( x \) such that:
- \( x \mod 16 = 5 \)
- \( x \mod 36 = 5 \)
- \( x \mod 40 = 5 \)

This means that if we subtract 5 from \( x \), the result should be divisible by 16, 36, and 40.

### Step 2: Set up the equation
Let’s define a new variable:
\[ k = x - 5 \]
Now, we need \( k \) to be divisible by 16, 36, and 40. Therefore, we can express this as:
\[ k \mod 16 = 0 \]
\[ k \mod 36 = 0 \]
\[ k \mod 40 = 0 \]

### Step 3: Find the Least Common Multiple (LCM)
To find \( k \), we need to calculate the LCM of 16, 36, and 40.

**Prime factorization:**
- \( 16 = 2^4 \)
- \( 36 = 2^2 \times 3^2 \)
- \( 40 = 2^3 \times 5^1 \)

**Finding LCM:**
The LCM is found by taking the highest power of each prime factor:
- For \( 2 \): the highest power is \( 2^4 \)
- For \( 3 \): the highest power is \( 3^2 \)
- For \( 5 \): the highest power is \( 5^1 \)

Thus, the LCM is:
\[ \text{LCM} = 2^4 \times 3^2 \times 5^1 \]

### Step 4: Calculate the LCM
Now, let's calculate the LCM:
- \( 2^4 = 16 \)
- \( 3^2 = 9 \)
- \( 5^1 = 5 \)

Now multiply these together:
\[ \text{LCM} = 16 \times 9 \times 5 \]
Calculating step-by-step:
1. \( 16 \times 9 = 144 \)
2. \( 144 \times 5 = 720 \)

So, \( k = 720 \).

### Step 5: Find \( x \)
Now, recall that \( x = k + 5 \):
\[ x = 720 + 5 = 725 \]

### Conclusion
The least number which, when divided by 16, 36, and 40, leaves a remainder of 5 is:
\[ \boxed{725} \]

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