One of our users asked the following question:-
This is a wonderful sort of problem I found out.
While I got some answers from fellow pals like these (I have attached them at the end), I found that it rather “deceived them”.
My way of solving may be this:-
If 1/a is a natural number greater than 1
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We may take it 2 for example, or simply x. (x is natural)
In that sense a would be equal to ½ or 1/x
So, if we take ½, in place of a,
then nothing but the LCM of a3 a4 & a6
is finding the LCM of ⅛, 1/16 & 1/64
We know the rule of LCM of fractions = (LCM of numerators)/(HCF of denominators)
So, answer will be a3.
Similarly, if you use x simply in place of 2, it would have yielded similar results.
In school exams, you’re appreciated to use x and not taking random numbers.
We used for clarifying that better.
NOTE:- You may directly instead apply the lemma that the LCM of fractions if with same numerator, will give it for lowest power.
But, this is not for school exams.
The given info is just a sort of trick trying to deceive you and mislead….
This sum will go on with just normal method of doing L.C.M
Lcm will be a⁶