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Let a,b,c be digits such that the six digit number $abcabc$ has 4 prime factors and only one prime factor out of the four has a power of 3 (say $\mathrm{k}^3$ ). If there are $\mathrm{n}$ such numbers find $n$.


$abcabc$ can be re-written as $7 \times 11 \times 13 \times(100a+10b+c)$. Hence $3$ of those primes are $7,11,13$ and so the prime with power of $3$ has to be $100a+10b+c$. Let it be equal to $k^{3}$, then $100 \leq k^{3} \leq 999$. Hence $k=5,6,7,8,9$ and from there we get $5$ cases for $a,b,c$. But the answer key says there are $23$ such numbers. What did I get wrong /missed?

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  • $\begingroup$ @JohnOmielan Hello! But how can they 2 more factors of $7,11,13$ if it's said that only one of the four primes factors have a power of 3? $\endgroup$ Oct 5, 2022 at 20:17

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Well, but what about $3$-digit [or fewer] numbers $x$ of the form $x=7^e11^f13^g3^h$; $e,f,g,h$ nonnegative integers; furthermore $h \ge 1$? [Keep in mind no one said $a>0$ or $b>0$ here.]

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  • $\begingroup$ Which 3 digit number are you considering (what is the role of $x$)? Also why is there a $3^{h}$? Where are you getting $3$ as a prime factor? $\endgroup$ Oct 5, 2022 at 20:32
  • $\begingroup$ As you wrote the problem: Let $y$ be the integer. Then $y$ can be written $y=1001x$. The $4$ factors of the integer $y=1001x$; $1 \le x \le 999$, are $7,11,13$, and $3$. So $x$ is precisely any integer in $[1,999]$ that is divisible by $3$ and whose other prime factors, if any, are in the set $\{7,11,13\}$ $\endgroup$
    – Mike
    Oct 5, 2022 at 20:35
  • $\begingroup$ I guess you're letting $abc=x$ but then $100 \leq x \leq 999$ and not $1$ if I'm not wrong. And I still don't get why $3$ has to be a prime factor of $y$? It is said on the problem that only one prime factor has a power of 3 and that prime factor is not necessarily 3 by itself right? $\endgroup$ Oct 5, 2022 at 20:52

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