Answer:
Explanation:
Let the 100-digit number be represented as
$a_1a_2a_3\ldots a_{99}a_{100}$
We need to prove that there exist 88 digits that we can delete from this number such that the remaining number is divisible by 101.
We can represent the given number as a sum of powers of 10 as follows:
$a_1 \times 10^{99} + a_2 \times 10^{98} + a_3 \times 10^{97} + \cdots + a_{99} \times 10^1 + a_{100} \times 10^0$
We can use the fact that $10 \equiv -1 \pmod{101}$ to simplify the above expression as follows:
$a_1 \times (-1)^{99} + a_2 \times (-1)^{98} + a_3 \times (-1)^{97} + \cdots + a_{99} \times (-1)^1 + a_{100} \times (-1)^0$
which is equivalent to:
$-a_1 + a_2 - a_3 + \cdots + (-1)^{99} a_{99} + (-1)^{100} a_{100}$
We can group the terms as follows:
$(-a_1 + a_2 - a_3 + \cdots + a_{98} - a_{99}) + (a_{99} - a_{100})$
Notice that the first group of terms is the sum of 49 pairs of terms, each pair having a difference of $-1$ or $1$. Therefore, the sum of the first group of terms is either $-49$ or $49$.
Since none of the digits are 0, we know that $a_{99} \\eq 0$ and $a_{100} \\eq 0$, which implies that $a_{99} - a_{100}$ is not divisible by 101. Therefore, either $a_{99} - a_{100} = 1$ or $a_{99} - a_{100} = -1$.
Now consider the two cases:
Case 1: The sum of the first group of terms is $49$
In this case, we can delete the digits corresponding to the terms in the first group that have a negative sign. This will remove 49 digits, leaving us with a number that is equal to $(a_{99} - a_{100}) \times 10^1 = \pm 10$, which is divisible by 101.
Case 2: The sum of the first group of terms is $-49$
In this case, we can delete the digits corresponding to the terms in the first group that have a positive sign. This will remove 49 digits, leaving us with a number that is equal to $(-a_{99} + a_{100}) \times 10^1 = \pm 10$, which is divisible by 101.
Therefore, in either case, we can always delete 88 digits from the original number to obtain a number that is divisible by 101.