The question is to find out the number of values of $k$ such that $$LCM(6^6,8^8,k)=(12)^{12}$$

I tried using the formula $$abc=\frac{LCM(6^6,8^8,k) HCF(6^6,k) HCF(8^8,k)HCF(k,6^6)}{HCF(6^6,8^8,k)}$$ but couldn't figure out anything from this.Any help shall be highly appreciated. Thanks.


Let $k=2^{a} 3^b$

Since LCM of $2^63^6 ~(6^6)$ and $2^{24} ~(8^8)$ and $2^{a} 3^b (k)$ is $2^{24}3^{12} ~(12^{12})$

LCM of $6,24$ and $a$ should be $24$ and LCM of $6$ and $b$ should be $12$.

Can you proceed now?

  • $\begingroup$ Thanks for your answer:). it was a easy problem:( $\endgroup$ – Navin Apr 16 '17 at 19:18

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