Find the greatest number which divides 2011 and 2623 , leaving remainders 9 and 5 respectively.

Options

A. 153
B. 152
C. 151
D. 150

Answer

To find the greatest number that divides 2011 and 2623, leaving remainders of 9 and 5 respectively, we can follow these steps:

### Step-by-Step Solution:

1. **Subtract the Remainders**:
- From 2011, subtract the remainder 9:
\[ 2011 - 9 = 2002 \]
- From 2623, subtract the remainder 5:
\[ 2623 - 5 = 2618 \]

2. **Identify the Numbers**:
- Now we have two numbers: 2002 and 2618. We need to find the greatest common divisor (GCD) of these two numbers.

3. **Use the Division Algorithm**:
- We will divide the larger number (2618) by the smaller number (2002) and find the remainder:
\[ 2618 \div 2002 = 1 \quad \text{(quotient)} \]
\[ \text{Remainder} = 2618 - (2002 \times 1) = 616 \]

4. **Repeat the Process**:
- Now, we will take the previous divisor (2002) and the remainder (616) and repeat the division:
\[ 2002 \div 616 = 3 \quad \text{(quotient)} \]
\[ \text{Remainder} = 2002 - (616 \times 3) = 154 \]

5. **Continue Until Remainder is Zero**:
- Next, we take 616 and divide it by 154:
\[ 616 \div 154 = 4 \quad \text{(quotient)} \]
\[ \text{Remainder} = 616 - (154 \times 4) = 0 \]

6. **Identify the GCD**:
- Since we have reached a remainder of 0, the last non-zero remainder (154) is the GCD of 2002 and 2618.

7. **Conclusion**:
- Therefore, the greatest number that divides 2011 and 2623, leaving remainders of 9 and 5 respectively, is:
\[ \text{Greatest Number} = 154 \]