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I solved this problem by listing all two-digit integers and going through each one. Is there an easy way to solve the problem?

How many positive two-digit integers have exactly 8 positive factors?

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If $n$ has prime power factorization $n=p^aq^br^c$ then the number of factors for $n$ is $(a+1)(b+1)(c+1)$ ( A number with 4 or more distinct prime factors will exceed 100). For numbers of the type $p^aq^b$ the count is $(a+1)(b+1)$.

Now see how to get $8$ as product of the above type $(a+1)(b+1)(c+1)$, or $(a+1)(b+1)$ then choose primes $p,q,r$ to ensure the number is $<100$.

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  • $\begingroup$ You should mention that it is $\tau(n)$ that is working behind. $\endgroup$ Commented Jun 12, 2016 at 3:06
  • $\begingroup$ Also with only one prime factor, $p^7$ is larger than $100$. $\endgroup$
    – 2'5 9'2
    Commented Jun 12, 2016 at 3:07
  • $\begingroup$ @KushalBhuyan: It would be fun to generalize and prove for oneself such a formula from the two specific cases given. I also left out why a simple prime power <100 cannot have 8 factors. $\endgroup$ Commented Jun 12, 2016 at 3:09

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