# 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, find numbers

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?

• @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? Oct 5, 2022 at 20:17

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.]
• 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? Oct 5, 2022 at 20:32
• 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 $\{7,11,13\}$
• 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? Oct 5, 2022 at 20:52