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A number has 101 composite factors. How many prime factors at max A number could have ?

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    What will strengthen your concepts is reading what's been written already and trying to understand it. In particular, you have no shot at this question if you don't understand the relation between the prime factorization of a number and the number of factors it has.2012-06-25

1 Answers 1

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Suppose $m = p_1^{a_1} ... p_n^{a_n}$ has evactly $101$ composite factors.

Then $101 + (1+n) = (a_1 + 1)(a_2+1) ... (a_n+1)$.

But the RHS is at least $2^n$ and it is easily checked that the inequality:

$102 + n \geq 2^n$

fails for $n \geq 7$. So there can be at most $6$ primes in the factorisation of $m$.

We now try to decompose the numbers $101 + (1+n)$ into a product of exactly $n$ integers for $n=1,2,3,4,5,6$, in order to see whether the $a_i$ can actually exist in each case.

We see that:

$108 = 2^2 \times 3^3$

$107$ is prime

$106 = 2\times 53$

meaning that the cases for $n=4,5,6$ cannot work.

However the number $105 = 3\times 5\times 7$ does have such a representation as a product of three numbers. Hence the biggest number of primes you may have in $m$ is $3$ in order to have exactly 101 composite factors.

Such a number is given by $m = p_1^2 p_2^4 p_3^6$ for any three different primes you wish.

As an aside, all such numbers $m$ must be of one of the following forms:

$p_1^2 p_2^4 p_3^6$

$p_1^7 p_2^{12}$

$p_1^3 p_2^{25}$

$p_1 p_2^{51}$

$p_1^{102}$

Where $p_1,p_2,p_3$ are distinct primes.

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    Thanks, I certainly agree with what you say and will try to enforce it in future.2012-06-25