What is the least number which leaves remainder 3 and 7 respectively when divided by 7 and 11 ?

Options

A. 80
B. 73
C. 93
D. 150

Answer

To solve the problem of finding the least number that leaves a remainder of 3 when divided by 7 and a remainder of 7 when divided by 11, we can follow these steps:

### Step 1: Understand the conditions
We need to find a number \( x \) such that:
- \( x \equiv 3 \mod 7 \)
- \( x \equiv 7 \mod 11 \)

### Step 2: Express the conditions in terms of equations
From the first condition:
\[ x = 7k + 3 \]
for some integer \( k \).

From the second condition:
\[ x = 11m + 7 \]
for some integer \( m \).

### Step 3: Set the two equations equal to each other
Since both expressions represent \( x \), we can set them equal:
\[ 7k + 3 = 11m + 7 \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ 7k - 11m = 4 \]

### Step 5: Find integer solutions
We need to find integer values of \( k \) and \( m \) that satisfy this equation. We can try different values for \( k \) and solve for \( m \).

### Step 6: Try values for \( k \)
Let's try \( k = 2 \):
\[ 7(2) - 11m = 4 \]
\[ 14 - 11m = 4 \]
\[ 11m = 10 \]
\[ m = \frac{10}{11} \] (not an integer)

Now let's try \( k = 3 \):
\[ 7(3) - 11m = 4 \]
\[ 21 - 11m = 4 \]
\[ 11m = 17 \]
\[ m = \frac{17}{11} \] (not an integer)

Now let's try \( k = 4 \):
\[ 7(4) - 11m = 4 \]
\[ 28 - 11m = 4 \]
\[ 11m = 24 \]
\[ m = \frac{24}{11} \] (not an integer)

Now let's try \( k = 5 \):
\[ 7(5) - 11m = 4 \]
\[ 35 - 11m = 4 \]
\[ 11m = 31 \]
\[ m = \frac{31}{11} \] (not an integer)

Now let's try \( k = 6 \):
\[ 7(6) - 11m = 4 \]
\[ 42 - 11m = 4 \]
\[ 11m = 38 \]
\[ m = \frac{38}{11} \] (not an integer)

Now let's try \( k = 7 \):
\[ 7(7) - 11m = 4 \]
\[ 49 - 11m = 4 \]
\[ 11m = 45 \]
\[ m = \frac{45}{11} \] (not an integer)

Now let's try \( k = 8 \):
\[ 7(8) - 11m = 4 \]
\[ 56 - 11m = 4 \]
\[ 11m = 52 \]
\[ m = \frac{52}{11} \] (not an integer)

Now let's try \( k = 9 \):
\[ 7(9) - 11m = 4 \]
\[ 63 - 11m = 4 \]
\[ 11m = 59 \]
\[ m = \frac{59}{11} \] (not an integer)

Now let's try \( k = 10 \):
\[ 7(10) - 11m = 4 \]
\[ 70 - 11m = 4 \]
\[ 11m = 66 \]
\[ m = 6 \] (this is an integer)

### Step 7: Substitute back to find \( x \)
Now substitute \( k = 10 \) back into the equation for \( x \):
\[ x = 7(10) + 3 = 70 + 3 = 73 \]

### Step 8: Verify the solution
Check if \( x = 73 \) satisfies both original conditions:
1. \( 73 \mod 7 = 3 \) (correct)
2. \( 73 \mod 11 = 7 \) (correct)

Thus, the least number which leaves a remainder of 3 when divided by 7 and a remainder of 7 when divided by 11 is:
\[ \boxed{73} \]