The remainder when 888222888222888222… upto 9235 digits is divided by `5^3` is :

Options

A. 103
B. 38
C. can't be determined
D. none of these

Answer

To find the remainder when the number formed by the digits "888222" repeated up to 9235 digits is divided by \(5^3\), we can follow these steps:

### Step 1: Identify the repeating pattern
The number "888222" has a repeating pattern of 6 digits: 888222.

### Step 2: Determine the total number of complete cycles
To find how many complete cycles of "888222" fit into 9235 digits, we divide 9235 by 6:

\[ 9235 \div 6 = 1539 \quad \text{(complete cycles)} \quad \text{with a remainder of } 1. \]

This means there are 1539 complete cycles of "888222" and 1 extra digit.

### Step 3: Write out the total number of digits
The total number of digits contributed by the complete cycles is:

\[ 1539 \times 6 = 9234 \quad \text{(digits from complete cycles)}. \]

Adding the extra digit gives us:

\[ 9234 + 1 = 9235 \quad \text{(total digits)}. \]

### Step 4: Find the contribution of the complete cycles
The contribution of each complete cycle "888222" to the number can be calculated as follows. Each cycle contributes:

\[ 888222 \text{ (in decimal)}. \]

### Step 5: Calculate the value of the complete cycles
The complete cycles contribute:

\[ 1539 \times 888222. \]

### Step 6: Add the contribution of the extra digit
The extra digit is "8" (the first digit of the repeating pattern). Thus, we need to add this to the total contribution from the complete cycles.

### Step 7: Calculate the total number
The total number can be expressed as:

\[ N = 1539 \times 888222 + 8. \]

### Step 8: Find the remainder when divided by \(5^3\)
Now, we need to find \(N \mod 125\) (since \(5^3 = 125\)).

1. First, calculate \(888222 \mod 125\):
- \(888222 \div 125 = 7105.776\) (take the integer part, which is 7105)
- \(7105 \times 125 = 888125\)
- \(888222 - 888125 = 97\)
- So, \(888222 \mod 125 = 97\).

2. Now calculate \(1539 \times 97 \mod 125\):
- \(1539 \div 125 = 12.312\) (take the integer part, which is 12)
- \(12 \times 125 = 1500\)
- \(1539 - 1500 = 39\)
- So, \(1539 \mod 125 = 39\).
- Now, \(39 \times 97 = 3783\).

3. Finally, calculate \(3783 \mod 125\):
- \(3783 \div 125 = 30.264\) (take the integer part, which is 30)
- \(30 \times 125 = 3750\)
- \(3783 - 3750 = 33\).
- So, \(3783 \mod 125 = 33\).

4. Now add the contribution of the extra digit:
- \(N \mod 125 = (33 + 8) \mod 125 = 41\).

### Final Answer
The remainder when the number formed by the digits "888222" repeated up to 9235 digits is divided by \(5^3\) is **41**.

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