If two numbers 121 and 199 are divided by two digit number then leaves SAME remainder in each case. How many two digit number of such type are possible?

Options

A. 4
B. 6
C. 2
D. 5

Answer

To solve the problem, we need to find how many two-digit numbers leave the same remainder when dividing both 121 and 199.

### Step 1: Find the difference between the two numbers.
First, we calculate the difference between the two numbers:
\[ 199 - 121 = 78 \]
**Hint:** The difference between two numbers is key to finding a common divisor.

### Step 2: Identify the divisors of the difference.
Next, we need to find the divisors of 78. The divisors are the numbers that can divide 78 without leaving a remainder.

To find the divisors, we can factor 78:
\[ 78 = 2 \times 39 = 2 \times 3 \times 13 \]
Now, we can list the divisors of 78:
1. 1
2. 2
3. 3
4. 6
5. 13
6. 26
7. 39
8. 78

**Hint:** Finding the prime factorization helps in determining all the divisors.

### Step 3: Filter the two-digit divisors.
Now, we need to filter out the two-digit divisors from the list we found:
- The only two-digit divisor from our list is 78.

**Hint:** Remember, we are only interested in two-digit numbers, so any single-digit or three-digit numbers should be excluded.

### Step 4: Conclusion.
Since 78 is the only two-digit number that divides the difference (78) and leaves the same remainder when dividing both 121 and 199, the total count of such two-digit numbers is:
\[ \text{Total two-digit numbers} = 1 \]

**Final Answer:** There is only **1 two-digit number** that satisfies the condition.