A four digit number is formed by repeating a \[2\] digit number such as \[2525\], \[3232\] etc. Any number of this form is always exactly divisible by
\[\left( 1 \right)\] \[7\]
\[\left( 2 \right)\] \[11\]
\[\left( 3 \right)\] \[13\]
\[\left( 4 \right)\] smallest \[3\] digit prime number

Hint: We have to find the number which always divides the number of the form of a four digit number which has a repeating \[2\] digit number like \[2525\], \[3232\] etc. We will solve this question using Euclid's division lemma. We will simply check the divisibility by all the given numbers. We will split both the numbers given as an example by using Euclid’s division lemma with each of the options given. The number which gives us the remainder zero is the required number for the given problem.

Complete step-by-step solution:
Given :
The two numbers given as examples are \[2525\] and \[3232\].
We will solve both the numbers making \[4\] cases as :
We know that a number can be represented in terms of Euclid’s division lemma as :
\[a = bq + r\]
Where \[a\] is the divisor, \[b\] is the dividend, \[q\] is the quotient and \[r\] is the remainder.
\[Case{\text{ }}1{\text{ }}:\]
Using Euclid’s division lemma, we get
\[a = 2525\], \[b = 7\]
We express a and b in terms of Euclid’s division lemma as :
\[2525 = 7 \times 360 + 5\]
As the remainder is not equal to zero, we conclude that \[2525\] is not exactly divisible by \[7\].
Similarly for \[3232\]
\[a = 3232\], \[b = 7\]
We express a and b in terms of Euclid’s division lemma as :
\[3232 = 7 \times 461 + 5\]
As the remainder is not equal to zero, we conclude that \[3232\] is not exactly divisible by \[7\].
Hence, we conclude that \[7\] is not the required answer.
\[Case{\text{ }}2{\text{ }}:\]
Using Euclid’s division lemma, we get
\[a = 2525\], \[b = 11\]
We express a and b in terms of Euclid’s division lemma as :
\[2525 = 11 \times 229 + 6\]
As the remainder is not equal to zero, we conclude that \[2525\] is not exactly divisible by \[11\].
Similarly for \[3232\]
\[a = 3232\], \[b = 11\]
We express a and b in terms of Euclid’s division lemma as :
\[3232 = 11 \times 293 + 9\]
As the remainder is not equal to zero, we conclude that \[3232\] is not exactly divisible by \[11\].
Hence, we conclude that \[11\] is not the required answer.
\[Case{\text{ }}3{\text{ }}:\]
Using Euclid’s division lemma, we get
\[a = 2525\], \[b = 13\]
We express a and b in terms of Euclid’s division lemma as :
\[2525 = 13 \times 194 + 3\]
As the remainder is not equal to zero, we conclude that \[2525\] is not exactly divisible by \[13\].
Similarly for \[3232\]
\[a = 3232\], \[b = 13\]
We express a and b in terms of Euclid’s division lemma as :
\[3232 = 13 \times 248 + 8\]
As the remainder is not equal to zero, we conclude that \[3232\] is not exactly divisible by \[13\].
Hence, we conclude that \[13\] is not the required answer.
\[Case{\text{ }}4{\text{ }}:\]
Now, we have to find the smallest \[3\] digit prime number.
We know that the smallest \[3\] digit prime number is \[101\].
Using Euclid’s division lemma, we get
\[a = 2525\], \[b = 101\]
We express a and b in terms of Euclid’s division lemma as :
\[2525 = 101 \times 25 + 0\]
As the remainder is equal to zero, we conclude that \[2525\] is exactly divisible by \[101\].
Similarly for \[3232\]
\[a = 3232\], \[b = 101\]
We express a and b in terms of Euclid’s division lemma as :
\[3232 = 101 \times 32 + 0\]
As the remainder is equal to zero, we conclude that \[3232\] is exactly divisible by \[101\].
Hence, we conclude that \[101\] is the required answer.
Thus, the correct option is \[\left( 4 \right)\].

Note: We have evaluated the smallest \[3\] digit prime number as \[101\]. As we know that 100 is the smallest \[3\] digit number but it is not a prime number as it has more than two factors from \[1\] and \[100\]. And the next smallest number is \[101\].
We could have solved this question by showing the actual division. We showed the division using Euclid’s division lemma for the two numbers. Similarly we would have obtained the result for any other number of the same type.