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Bunuel
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Bunuel
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Hello Bunuel,

Does the fact of divisibility in (1) not mean the m and n must be multiples of 4 and 2 for statement 1. Given that X is divisible by 144 there would be no remainders when x/144 ie 2^m.3^n /by 2^4.3^2 - hence all 2s and 3s in the numerator must be cancelled?

Similarly, when we say x is not divisible by 648 in (2) - does that not mean x/648 will leave a remainder?

Please advise If this way of looking at the Q works.
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Hello Bunuel,

Does the fact of divisibility in (1) not mean the m and n must be multiples of 4 and 2 for statement 1. Given that X is divisible by 144 there would be no remainders when x/144 ie 2^m.3^n /by 2^4.3^2 - hence all 2s and 3s in the numerator must be cancelled?

Similarly, when we say x is not divisible by 648 in (2) - does that not mean x/648 will leave a remainder?

Please advise If this way of looking at the Q works.

1. No, m does not necessarily need to be a multiple of 4 and n does not necessarily not need to be a multiple of 3. A simply example, can show that: m = 5 and n = 3, then \(\frac{2^m3^n}{2^43^2}=\frac{2^53^3}{2^43^2}=2*3=6\).

2. Yes, x is not divisible by 648 means that \(\frac{x}{648} = \frac{2^m3^n}{2^33^4}\) will leave a non-zero remainder. This will occur when either \(m < 3\) or \(n < 4\), or both. You can again play with some numbers to clearly see that.
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