Divisors

[Loki123]

It is. I hope you understand why and not just making a lucky guess.

thanks

 

[Loki123]

It is. I hope you understand why and not just making a lucky guess.

hey sorry for getting this active again but i did this problem again today and i have a question
i found 5 prime divisors 2, 3, 5, 167, 401
2^n = 2^5 = 32 - this is the answer

however
what about 1? shouldn't we include it in prime divisors?

 

[BigBeachBanana]

hey sorry for getting this active again but i did this problem again today and i have a question
i found 5 prime divisors 2, 3, 5, 167, 401
2^n = 2^5 = 32 - this is the answer

however
what about 1? shouldn't we include it in prime divisors?

One is not a prime number. You should know that.

 

[blamocur]

hey sorry for getting this active again but i did this problem again today and i have a question
i found 5 prime divisors 2, 3, 5, 167, 401
2^n = 2^5 = 32 - this is the answer

however
what about 1? shouldn't we include it in prime divisors?

There are reasons not to count 1 as a divisor. With prime numbers there is a unique representation for each natural number $m= p_1^{k_1} p_2^{k_2}...p_n^{k_n}$, e.g. $75=5^2 \cdot 3^1$. But with 1 any power, including 0, can be included. I.e., the representation is no longer unique.

 

[pka]

hey sorry for getting this active again but i did this problem again today and i have a question
i found 5 prime divisors 2, 3, 5, 167, 401
2^n = 2^5 = 32 - this is the answer

however
what about 1? shouldn't we include it in prime divisors?

Suppose that the number $n$ has the prime factorization:
$n=p_1^f\cdot p_2^j\cdot p_3^k\cdot p_4^h\cdot p_5^g\cdot$.
From that we see that $n$ has $(f+1)(j+1)(k+1)(h+1)(g+1)$ divisors.
Example: $396000=2^5\cdot 3^2\cdot 5^3\cdot 11^1$ has $(6)(3)(4)(2)=144$ divisors.
SEE HERE
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