Find the remainder when
`123456789101112 . . .` is `97`- digit number divided by 16?

Options

A. 5
B. 6
C. 7
D. 8

Answer

To find the remainder when the 97-digit number formed by concatenating integers from 1 onward is divided by 16, we can follow these steps:

### Step 1: Determine the last four digits of the number
The number is formed by concatenating integers starting from 1. We need to find out how many digits are used up to the point where we have a total of 97 digits.

### Step 2: Count the digits contributed by each range of numbers
- **From 1 to 9**: There are 9 numbers, each contributing 1 digit, so total digits = 9.
- **From 10 to 99**: Each number contributes 2 digits. We need to find out how many numbers we can include from this range to reach a total of 97 digits.

### Step 3: Calculate the remaining digits needed after counting from 1 to 9
After counting the digits from 1 to 9, we have used 9 digits. Therefore, we need \( 97 - 9 = 88 \) more digits.

### Step 4: Determine how many two-digit numbers we can use
Since each two-digit number contributes 2 digits, the number of two-digit numbers we can use is:
\[ \text{Number of two-digit numbers} = \frac{88}{2} = 44 \]
This means we can use the first 44 two-digit numbers, which are from 10 to 53 (inclusive).

### Step 5: Identify the last four digits
The last four digits of our concatenated number will be the last two-digit number (53) and the first two digits of the next number (54). Therefore, the last four digits are:
- From 52: "52"
- From 53: "53"
- From 54: "54" (only the first digit "5" is needed)

So, the last four digits are "5253".

### Step 6: Find the remainder when the last four digits are divided by 16
Now we need to find the remainder of 5253 when divided by 16.

1. Divide 5253 by 16:
\[ 5253 \div 16 = 328.3125 \]
The integer part is 328, so we multiply back:
\[ 328 \times 16 = 5248 \]
2. Subtract this from 5253 to find the remainder:
\[ 5253 - 5248 = 5 \]

### Final Answer
The remainder when the 97-digit number is divided by 16 is **5**.

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