Find the least number which when divided by 25, 40 and 60, leaves remainder 7 in each case.

Options

A. 607
B. 608
C. 609
D. 606

Answer

To find the least number which, when divided by 25, 40, and 60, leaves a remainder of 7 in each case, we can follow these steps:

### Step 1: Understand the Problem
We need to find a number \( N \) such that:
- \( N \mod 25 = 7 \)
- \( N \mod 40 = 7 \)
- \( N \mod 60 = 7 \)

This means that \( N - 7 \) must be divisible by 25, 40, and 60.

### Step 2: Calculate the LCM of 25, 40, and 60
To find the least common multiple (LCM) of these numbers, we can use their prime factorization:

- \( 25 = 5^2 \)
- \( 40 = 2^3 \times 5^1 \)
- \( 60 = 2^2 \times 3^1 \times 5^1 \)

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

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

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

### Step 3: Find the Required Number
Since \( N - 7 \) is divisible by the LCM (600), we can express \( N \) as:
\[ N = 600k + 7 \]
where \( k \) is a non-negative integer.

To find the least number, we set \( k = 1 \):
\[ N = 600 \times 1 + 7 = 607 \]

### Final Answer
The least number which when divided by 25, 40, and 60 leaves a remainder of 7 is:
\[ \boxed{607} \]