Find the least number which when divided by 35 leaves a remainder 25 , when divided by 45 leaves a remainder 35 , and when divided by 55 leaves a remainder 45 .

Options

A. 3465
B. 4575
C. 3455
D. 3670

Answer

To find the least number which, when divided by 35, leaves a remainder of 25, when divided by 45 leaves a remainder of 35, and when divided by 55 leaves a remainder of 45, we can follow these steps:

### Step 1: Understand the Problem
We need to find a number \( x \) such that:
- \( x \mod 35 = 25 \)
- \( x \mod 45 = 35 \)
- \( x \mod 55 = 45 \)

### Step 2: Rewrite the Conditions
From the conditions, we can rewrite them as:
- \( x = 35k + 25 \) for some integer \( k \)
- \( x = 45m + 35 \) for some integer \( m \)
- \( x = 55n + 45 \) for some integer \( n \)

### Step 3: Find the LCM
Next, we need to find the least common multiple (LCM) of the divisors (35, 45, and 55).

1. **Factorize the numbers:**
- \( 35 = 5 \times 7 \)
- \( 45 = 3^2 \times 5 \)
- \( 55 = 5 \times 11 \)

2. **Identify the highest power of each prime factor:**
- For \( 3 \): \( 3^2 \)
- For \( 5 \): \( 5^1 \)
- For \( 7 \): \( 7^1 \)
- For \( 11 \): \( 11^1 \)

3. **Calculate the LCM:**
\[ \text{LCM} = 3^2 \times 5^1 \times 7^1 \times 11^1 = 9 \times 5 \times 7 \times 11 \]
- First, calculate \( 9 \times 5 = 45 \)
- Then, \( 45 \times 7 = 315 \)
- Finally, \( 315 \times 11 = 3465 \)

So, the LCM of 35, 45, and 55 is 3465.

### Step 4: Find the Least Number
Now, we know that the number \( x \) can be expressed in terms of the LCM:
- The difference between the divisor and the remainder for each case is 10:
- \( 35 - 25 = 10 \)
- \( 45 - 35 = 10 \)
- \( 55 - 45 = 10 \)

Thus, we can subtract this difference from the LCM:
\[ x = 3465 - 10 = 3455 \]

### Step 5: Conclusion
The least number which satisfies all the conditions is **3455**.

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