



 
Hi Richie, Since the LCM of $6, 15$ and $x$ is $90 = 2 \times 3^2 \times 5$ the prime decomposition of $x$ cannot contain any primes other than $2, 3$ and $5.$ Since x is odd it cannot have $2$ as a prime factor Also since $90 = 2 \times 3^2 \times 5,$ the prime decomposition of $x$ can contain at most two threes and at most one five. The LCM of $6 = 2 \times 3$ and $15 = 3 \times 5$ is $2 \times 3 \times 5 = 30$ and since the LCM of $6, 15$ and $x$ is $90, x$ must contribute another $3.$ Thus $x$ could be $3 \times 3 = 9$ or $3 \times 3 \times 5 = 45.$ Penny  


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