The question asks for the remainder when the number formed by concatenating integers from 1 to 100 (123456789101112...99100) is divided by 16.
To find the remainder when a large number is divided by 16, we only need to consider the number formed by its last four digits. This is because 16 is a factor of \(10^4\) (10000). Specifically, \(10000 = 16 \times 625\). Therefore, any number can be written as \(N = 10000k + L\), where \(L\) is the number formed by the last four digits. Since \(10000k\) is perfectly divisible by 16, the remainder of \(N\) when divided by 16 is the same as the remainder of \(L\) when divided by 16.
Mathematically, if \(N\) is the large number, we are looking for \(N \pmod{16}\). We know that \(N = 10000k + L\). Since \(10000 \equiv 0 \pmod{16}\), then \(N \equiv 10000k + L \equiv 0 \times k + L \equiv L \pmod{16}\).
The number is formed by writing the integers from 1 to 100 in sequence:
123456789101112... 9899100
The last few numbers concatenated are 98, 99, and 100.
The number ends with ...9899100.
The last four digits of this number are '9100'.
Now, we need to find the remainder when the number formed by the last four digits, which is 9100, is divided by 16.
We perform the division: \(9100 \div 16\).
So, \(9100 = 16 \times 568 + 12\).
The remainder when 9100 is divided by 16 is 12.
According to the divisibility rule for 16, the remainder when the concatenated number (123...99100) is divided by 16 is the same as the remainder when its last four digits (9100) are divided by 16.
Therefore, the remainder is 12.
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2. S is always divisible by 9.
select the correct answer using the code given below: