Number of primes in prime factorization

Question

If $C$ is a positive integer where $C < 2^n$, then $C$ has less than $n$ primes in its prime factorization.

I know that the Prime Number Theorem states that the number of primes less or equal $n$ is given by $\pi(x) \geq \dfrac{x}{\log x}, \ \forall x\geq 11$. However, I'm not sure if I can use this expression to prove this.

Any guidance or solution is appreciated. Thanks.

Answer 1

Of course you don't need such a high powered theorem like the prime number theorem! Argue by contradiction. Assume $C$ had $n$ primes in its factorization; then $p_1p_2\cdots p_n \mid C$ for $p_1, ..., p_n$ some primes. So, $C \geq p_1p_2\cdots p_n \geq 2^n$ since $p_i\geq 2$ for each prime $p_i.$ This contradicts $C < 2^n,$ though.

Answer 2

The smallest natural number with $n$ prime factors (possibly duplicated) is of course $2^n$. Thus any other natural number with $n$ prime factors must be larger than that.

Answer 3

Sure you can use the prime number theorem, but that's taking the scenic route. If you still want to get the result the long way, but not that long, try asking this question: What's the smallest positive odd number with exactly $n$ prime factors, not necessarily distinct?

That would be $3^n$. Notice:

• Positive odd numbers with one prime factor are the odd primes: 3, 5, 7, 11, 13, 17, 19, 23, 29, ...
• Positive odd numbers with two prime factors are: 9, 15, 21, 25, 33, 35, ... (A046315)
• Positive odd numbers with three prime factors are: 27, 45, 63, 75, 99, ... (A046316)
• Positive odd numbers with four prime factors are: 81, 135, 189, 225, ... (A046317)
• etc.

And in each case $3^n > 2^n$. Therefore $2^n$ is the smallest positive number with $n$ prime factors.