The least number which when divided by 12,16,24 and 48 leaves a remainder of 3,7,15 and 39 respectively , is _______

Options

A. 16
B. 48
C. 57
D. 39

Answer

To find the least number which, when divided by 12, 16, 24, and 48, leaves remainders of 3, 7, 15, and 39 respectively, we can follow these steps:

### Step 1: Set Up the Equations
We can express the conditions given in the problem as equations:
- Let the least number be \( x \).
- From the problem, we have:
- \( x \equiv 3 \mod 12 \)
- \( x \equiv 7 \mod 16 \)
- \( x \equiv 15 \mod 24 \)
- \( x \equiv 39 \mod 48 \)

### Step 2: Rewrite the Equations
We can rewrite these equations in terms of \( x \):
- \( x = 12k + 3 \) for some integer \( k \)
- \( x = 16m + 7 \) for some integer \( m \)
- \( x = 24n + 15 \) for some integer \( n \)
- \( x = 48p + 39 \) for some integer \( p \)

### Step 3: Solve the First Two Equations
Start with the first two equations:
1. \( x = 12k + 3 \)
2. \( x = 16m + 7 \)

Setting them equal to each other:
\[ 12k + 3 = 16m + 7 \]
Rearranging gives:
\[ 12k - 16m = 4 \]
Dividing the entire equation by 4:
\[ 3k - 4m = 1 \]

### Step 4: Find Integer Solutions
We can find integer solutions for \( k \) and \( m \). By trial and error or using the Extended Euclidean Algorithm, we can find:
- One solution is \( k = 3 \) and \( m = 2 \).

### Step 5: General Solution for \( k \) and \( m \)
The general solution for \( k \) and \( m \) can be expressed as:
\[ k = 3 + 4t \]
\[ m = 2 + 3t \]
for some integer \( t \).

### Step 6: Substitute Back to Find \( x \)
Substituting \( k \) back into the equation for \( x \):
\[ x = 12(3 + 4t) + 3 = 36 + 48t + 3 = 39 + 48t \]

### Step 7: Use the Third Equation
Now we substitute \( x \) into the third equation:
\[ 39 + 48t \equiv 15 \mod 24 \]
This simplifies to:
\[ 15 + 0t \equiv 15 \mod 24 \]
This holds true for any \( t \).

### Step 8: Use the Fourth Equation
Now we substitute \( x \) into the fourth equation:
\[ 39 + 48t \equiv 39 \mod 48 \]
This is also true for any \( t \).

### Step 9: Find the Least Positive \( x \)
The smallest value of \( x \) occurs when \( t = 0 \):
\[ x = 39 + 48(0) = 39 \]

### Final Answer
Thus, the least number which when divided by 12, 16, 24, and 48 leaves a remainder of 3, 7, 15, and 39 respectively is **39**.

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