(I didn't check the last case but it seems to be obvious so sorry if there is a mistake here) Let this positive integer be $n$. Assume there exists $n$ satisfying these condition. Claim 1: $n$ must be even. Proof. Assume $n$ is odd Then, Last two digits of each divisor of $n$ must be odd Since there are $50$ numbers of two digits odd. Which means last two digits of the divisor of $n$ is permutation of $\{01,03,\dots,99\}$ (otherwise, there must exists two distinct divisors of $n$ which has identical last two digits desired a contradiction.) Since there exists a divisor of $n$ which ends with $75$. Thus, $5^2 \mid n$ if $5^3 \mid n$ then, $100 \mid 125-25$ a contradiction. Then $v_5(n)=2$ which is impossible since $3 \nmid 50$. $\square$ Case 1. $v_2(n)=1$ Observed that there are $25$ of the divisors of $n$ which is even Moreover, These $25$ divisors couldn't ends with $\{00,04,08,\dots,96\}$ Thus, For each element in $\{02,06,10,\dots,98\}$ there must exists exactly one divisor of $n$ which ends with this element. Therefore, $5^2 \mid n$ but $5^3$ can't. desired a contradiction since $3 \nmid 50$ Case 2. $v_2(n) = 4$ Let $n = 2^4 \times p_1 \times p_2^4$ ($2^4 \times p^9$ is similar) Observed that there are $40$ divisors of $n$ which is even , other $10$ is odd Notice that $5 \mid n$ doesn't works. if $5 \nmid n$ then each element in $\{00,10,\dots,90\}$ is not reachable from last two digits of each divisor Thus, Last two digits of these $40$ divisors is permutation of $\{02,04,06,08,12,\dots,96,98\}$ Moreover, there are $20$ divisors which is divisible by $2$ but not $4$ or it can be written as $2\times p_1^{a_1} \times p_2 ^{a_2}$ but there at most $2 \times 5 =10$ divisors which divisible by $2$ but not $4$ contradiction. Case 3. $v_2(n) >4$ This case is quite obvious since only possible value of $v_2(n)$ is $9,24,49$ which desired contradiction every cases. hence, there exists no $n$ satisfying these condition.