Find the smallest number which when divided by3,5,7 and 4, it leaves remainder1,3,5 and 2, respectively.

Options

A. 337
B. 418
C. 838
D. 912

Answer

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

### Step 1: Set up the equations based on the problem
We need to find a number \( x \) such that:
- \( x \equiv 1 \mod 3 \)
- \( x \equiv 3 \mod 5 \)
- \( x \equiv 5 \mod 7 \)
- \( x \equiv 2 \mod 4 \)

### Step 2: Rewrite the equations
We can rewrite these congruences to make them easier to work with:
- From \( x \equiv 1 \mod 3 \), we can express \( x \) as \( x = 3k + 1 \) for some integer \( k \).
- From \( x \equiv 3 \mod 5 \), we can express \( x \) as \( x = 5m + 3 \) for some integer \( m \).
- From \( x \equiv 5 \mod 7 \), we can express \( x \) as \( x = 7n + 5 \) for some integer \( n \).
- From \( x \equiv 2 \mod 4 \), we can express \( x \) as \( x = 4p + 2 \) for some integer \( p \).

### Step 3: Find the differences
To simplify our calculations, we can find the differences between the moduli and the remainders:
- For \( 3 - 1 = 2 \)
- For \( 5 - 3 = 2 \)
- For \( 7 - 5 = 2 \)
- For \( 4 - 2 = 2 \)

This shows that each of the moduli minus their respective remainders gives us the same difference of 2.

### Step 4: Find the least common multiple (LCM)
Now, we need to find the least common multiple (LCM) of the divisors (3, 5, 7, and 4):
- The prime factorization gives us:
- \( 3 = 3^1 \)
- \( 5 = 5^1 \)
- \( 7 = 7^1 \)
- \( 4 = 2^2 \)

The LCM is calculated as:
\[ \text{LCM}(3, 5, 7, 4) = 2^2 \cdot 3^1 \cdot 5^1 \cdot 7^1 = 4 \cdot 3 \cdot 5 \cdot 7 = 420 \]

### Step 5: Subtract the common difference
Since we found that the common difference is 2, we subtract this from the LCM:
\[ x = 420 - 2 = 418 \]

### Step 6: Verify the solution
We can verify that 418 satisfies all the original conditions:
- \( 418 \div 3 = 139\) remainder \( 1 \)
- \( 418 \div 5 = 83\) remainder \( 3 \)
- \( 418 \div 7 = 59\) remainder \( 5 \)
- \( 418 \div 4 = 104\) remainder \( 2 \)

Thus, the smallest number that meets all the conditions is **418**.