The smallest number which when divided by 10,15,20 and 35 leaves 6,11,16 and 31 as remainder is-

Options

A. 416
B. 424
C. 436
D. None of these

Answer

To find the smallest number that leaves specific remainders when divided by certain divisors, we can follow these steps:

### Step 1: Set Up the Problem
We need to find a number \( x \) such that:
- \( x \mod 10 = 6 \)
- \( x \mod 15 = 11 \)
- \( x \mod 20 = 16 \)
- \( x \mod 35 = 31 \)

### Step 2: Rewrite the Congruences
We can rewrite the congruences in terms of \( x \):
- \( x = 10k + 6 \) for some integer \( k \)
- \( x = 15m + 11 \) for some integer \( m \)
- \( x = 20n + 16 \) for some integer \( n \)
- \( x = 35p + 31 \) for some integer \( p \)

### Step 3: Adjust the Remainders
To simplify the problem, we can adjust each equation by subtracting the remainder from both sides:
- \( x - 6 = 10k \)
- \( x - 11 = 15m \)
- \( x - 16 = 20n \)
- \( x - 31 = 35p \)

This means that:
- \( x - 6 \) is a multiple of 10,
- \( x - 11 \) is a multiple of 15,
- \( x - 16 \) is a multiple of 20,
- \( x - 31 \) is a multiple of 35.

### Step 4: Find the Least Common Multiple
Next, we need to find the least common multiple (LCM) of the divisors (10, 15, 20, 35) to find a common solution.

Calculating the LCM:
- The prime factorization of the numbers:
- \( 10 = 2 \times 5 \)
- \( 15 = 3 \times 5 \)
- \( 20 = 2^2 \times 5 \)
- \( 35 = 5 \times 7 \)

The LCM will take the highest power of each prime:
- \( 2^2 \) from 20,
- \( 3^1 \) from 15,
- \( 5^1 \) from any,
- \( 7^1 \) from 35.

Thus,
\[ \text{LCM} = 2^2 \times 3^1 \times 5^1 \times 7^1 = 4 \times 3 \times 5 \times 7 = 420. \]

### Step 5: Solve for \( x \)
Now we can express \( x \) in terms of the LCM:
\[ x = 420k + R, \]
where \( R \) is a constant that satisfies all the original conditions.

We can find \( R \) by substituting back to find a number that satisfies all conditions.

### Step 6: Find the Constant \( R \)
To find \( R \), we can substitute \( k = 0 \) into \( x = 420k + R \) and check for the smallest \( R \) that satisfies:
- \( R \equiv 6 \mod 10 \) (i.e., \( R = 6 \))
- \( R \equiv 11 \mod 15 \) (i.e., \( R = 11 \))
- \( R \equiv 16 \mod 20 \) (i.e., \( R = 16 \))
- \( R \equiv 31 \mod 35 \) (i.e., \( R = 31 \))

### Step 7: Find the Smallest \( x \)
After checking the values, we find that:
- \( R = 6 \) satisfies \( 10 \) and \( 15 \) but not the others.
- Continuing this process, we find that \( R = 416 \) satisfies all conditions.

Thus, the smallest number \( x \) that satisfies all the conditions is:
\[ \boxed{416}. \]