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If $\log_3M=a_1+b_1$ and $\log_5 M=a_2+b_2$, where $a_1,a_2$ are natural numbers and $b_1,b_2 \in [0,1)$. If $a_1a_2=6$, then find the number of integral values of M. What so I do in the problem. I don't think anything can be computed. It looks like an estimation problem. How to approach? Thanks.

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    $\begingroup$ Note that $3^{a_1}\leq M < 3^{a_1+1}$ and $5^{a_2}\leq M \leq 5^{a_2+1}$ $\endgroup$ – Bob Krueger Apr 12 '15 at 15:48
  • $\begingroup$ Oh, thanks. This means I will have only one case-$a_1=3,a_2=2$. Isn't it. $\endgroup$ – user167045 Apr 12 '15 at 15:54
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Hint:

Since $a_1a_2=6$ and $a_2<a_1$, either $a_1=3$ and $a_2=2$, or $a_1=6$ and $a_2=1$.

1) If $a_1=3$ and $a_2=2$, we have $27\le M<81$ and $25\le M<125$.

2) If $a_1=6$ and $a_2=1$, then $3^6\le M<3^7$ and $5\le M<25$.

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