thanks
Divisors
- Thread starter Loki123
- Start date
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
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One is not a prime number. You should know that.
blamocur
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There are reasons not to count 1 as a divisor. With prime numbers there is a unique representation for each natural number [imath]m= p_1^{k_1} p_2^{k_2}...p_n^{k_n}[/imath], e.g. [imath]75=5^2 \cdot 3^1[/imath]. But with 1 any power, including 0, can be included. I.e., the representation is no longer unique.
pka
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Suppose that the number [imath]n[/imath] has the prime factorization:
[imath]n=p_1^f\cdot p_2^j\cdot p_3^k\cdot p_4^h\cdot p_5^g\cdot[/imath].
From that we see that [imath]n[/imath] has [imath](f+1)(j+1)(k+1)(h+1)(g+1)[/imath] divisors.
Example: [imath]396000=2^5\cdot 3^2\cdot 5^3\cdot 11^1[/imath] has [imath](6)(3)(4)(2)=144[/imath] divisors.
SEE HERE
[imath][/imath][imath][/imath][imath][/imath][imath][/imath]