Asked by david
Find the smallest positive integer not relatively prime to 2015 that has the same number of positive divisors as 2015.
Answers
Answered by
bob
the answer is 30
Answered by
joe
no because both 2o15 and 30 are divisible by 3
Answered by
BOb Joe
joe, you are wrong... read the question... it says that are not relatively prime. it is 30
Answered by
david bob joe BOb Joe
the answer is 30
Answered by
a person
2015 is not divisible by 3
Answered by
a person
however I agree that 300 is correct
Answered by
another person
how do you know its 30
Answered by
ur mom
like this cus ur mom
Answered by
yo
its 30
Answered by
billy bob joe
the answer is 30
Since $2015=5 \cdot 13 \cdot 31$, we know $2015$ has $(1+1)(1+1)(1+1)=8$ positive divisors. To create a number $n$ with 8 divisors, we have a few choices for its prime factorization. If $n$ is divisible by only one prime, it must look like $p^7$. With two primes, $n$ will have the form $p^3q$, and with three primes, $n$ must be of the form $pqr$, where $p$,$q$, and $r$ are distinct primes.
Now, we try to make each of these factorizations as small as possible while forcing some overlap with the factorization of $2015$. This will ensure that our new number is not relatively prime with 2015, or $\mathrm{gcd}(n,2015)>1$. Since $2015$ involves the primes $5$, $13$, and $31$, our new number should involve the prime $5$. To maintain minimality, we will use $2$ and $3$ if additional primes are needed. Remember, though, that $5$ must appear somewhere in the factorization!
Making these substitutions, we have $p^7=5^7=78125$, $p^3q=2^3 \cdot 5=40$, and $pqr=2 \cdot 3 \cdot 5 = 30$. So, the minimal choice is $\boxed{30}$.
Since $2015=5 \cdot 13 \cdot 31$, we know $2015$ has $(1+1)(1+1)(1+1)=8$ positive divisors. To create a number $n$ with 8 divisors, we have a few choices for its prime factorization. If $n$ is divisible by only one prime, it must look like $p^7$. With two primes, $n$ will have the form $p^3q$, and with three primes, $n$ must be of the form $pqr$, where $p$,$q$, and $r$ are distinct primes.
Now, we try to make each of these factorizations as small as possible while forcing some overlap with the factorization of $2015$. This will ensure that our new number is not relatively prime with 2015, or $\mathrm{gcd}(n,2015)>1$. Since $2015$ involves the primes $5$, $13$, and $31$, our new number should involve the prime $5$. To maintain minimality, we will use $2$ and $3$ if additional primes are needed. Remember, though, that $5$ must appear somewhere in the factorization!
Making these substitutions, we have $p^7=5^7=78125$, $p^3q=2^3 \cdot 5=40$, and $pqr=2 \cdot 3 \cdot 5 = 30$. So, the minimal choice is $\boxed{30}$.
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