Numerical algebra
Find the total possible number of integral values of
so that the the expression
has integral value, if k =
where
are distinct primes and
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The answer is 128.
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1 solution
When we simplify the expression, we get - 6 x + 6 + x k as the answer.Now , 6x + 6 will have an integral value for any value of x , so we now need to concentrate on only the third term, that is - x k .Now this term will be an integer only when the numerical part of the value of x is a factor of k.The total factors of k= (a+1)(b+1)(c+1) = 64, which is given ! But the answer does not end here.The total integral possible values of x is just double of the above value, which is 64*2=128 .Why? Let us take an example to make this clear.2 gives an integer value when divided 4 integers, -1,1,-2, and 2 though it has only 2 factors = 1 and 2.The case in the above problem is similar.