Mixing
Probability of $f(n) - varphi(n)$ to be an even number.
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Let $f(n)$ denotes the product of all the positive divisors of n, and $varphi(n)$ denotes the sum of all the positive divisors of $n$. If number $n$ is to be randomly selected from the first $100$ positive integers, what is the probability that $f(n) - varphi(n)$ is an even number?
So far, I know that $f(n)$ and $varphi(n)$ need to be either both odd or both even. If $n$ is odd, then $f(n)$ must also be odd; if $n$ is even, then $f(n)$ must also be even. However, I have difficulty calculating when $varphi(n)$ is odd or even. It appears to be completely random.
probability elementary-number-theory arithmetic-functions
asked Jan 20 at 5:04
Chris.Chris.
132
$endgroup$
add a comment |
$begingroup$
Let $f(n)$ denotes the product of all the positive divisors of n, and $varphi(n)$ denotes the sum of all the positive divisors of $n$. If number $n$ is to be randomly selected from the first $100$ positive integers, what is the probability that $f(n) - varphi(n)$ is an even number?
So far, I know that $f(n)$ and $varphi(n)$ need to be either both odd or both even. If $n$ is odd, then $f(n)$ must also be odd; if $n$ is even, then $f(n)$ must also be even. However, I have difficulty calculating when $varphi(n)$ is odd or even. It appears to be completely random.
probability elementary-number-theory arithmetic-functions
asked Jan 20 at 5:04
Chris.Chris.
132
$endgroup$
$begingroup$
Hello, and welcome to MSE. Although your title says you have no clue, please tell us what have your tried so far? Here is a hint that might help you: The probability would be the # of values of $n$ where $fleft(nright) - phileft(nright)$ is even divided by the total # of values being checked, this being $100$ in your case.
$endgroup$
– John Omielan
Jan 20 at 5:11
add a comment |
$begingroup$
Let $f(n)$ denotes the product of all the positive divisors of n, and $varphi(n)$ denotes the sum of all the positive divisors of $n$. If number $n$ is to be randomly selected from the first $100$ positive integers, what is the probability that $f(n) - varphi(n)$ is an even number?
So far, I know that $f(n)$ and $varphi(n)$ need to be either both odd or both even. If $n$ is odd, then $f(n)$ must also be odd; if $n$ is even, then $f(n)$ must also be even. However, I have difficulty calculating when $varphi(n)$ is odd or even. It appears to be completely random.
probability elementary-number-theory arithmetic-functions
asked Jan 20 at 5:04
Chris.Chris.
132
$endgroup$
Let $f(n)$ denotes the product of all the positive divisors of n, and $varphi(n)$ denotes the sum of all the positive divisors of $n$. If number $n$ is to be randomly selected from the first $100$ positive integers, what is the probability that $f(n) - varphi(n)$ is an even number?
So far, I know that $f(n)$ and $varphi(n)$ need to be either both odd or both even. If $n$ is odd, then $f(n)$ must also be odd; if $n$ is even, then $f(n)$ must also be even. However, I have difficulty calculating when $varphi(n)$ is odd or even. It appears to be completely random.
probability elementary-number-theory arithmetic-functions
probability elementary-number-theory arithmetic-functions
asked Jan 20 at 5:04
Chris.Chris.
132
asked Jan 20 at 5:04
Chris.Chris.
132
edited Jan 20 at 6:22
Chris.
asked Jan 20 at 5:04
Chris.Chris.
132
asked Jan 20 at 5:04
Chris.Chris.
132
asked Jan 20 at 5:04
Chris.Chris.
132
132
$begingroup$
Hello, and welcome to MSE. Although your title says you have no clue, please tell us what have your tried so far? Here is a hint that might help you: The probability would be the # of values of $n$ where $fleft(nright) - phileft(nright)$ is even divided by the total # of values being checked, this being $100$ in your case.
$endgroup$
– John Omielan
Jan 20 at 5:11
add a comment |
$begingroup$
Hello, and welcome to MSE. Although your title says you have no clue, please tell us what have your tried so far? Here is a hint that might help you: The probability would be the # of values of $n$ where $fleft(nright) - phileft(nright)$ is even divided by the total # of values being checked, this being $100$ in your case.
$endgroup$
– John Omielan
Jan 20 at 5:11
$begingroup$
Hello, and welcome to MSE. Although your title says you have no clue, please tell us what have your tried so far? Here is a hint that might help you: The probability would be the # of values of $n$ where $fleft(nright) - phileft(nright)$ is even divided by the total # of values being checked, this being $100$ in your case.
$endgroup$
– John Omielan
Jan 20 at 5:11
$begingroup$
Hello, and welcome to MSE. Although your title says you have no clue, please tell us what have your tried so far? Here is a hint that might help you: The probability would be the # of values of $n$ where $fleft(nright) - phileft(nright)$ is even divided by the total # of values being checked, this being $100$ in your case.
$endgroup$
– John Omielan
Jan 20 at 5:11
add a comment |
1 Answer
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oldest
votes
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$f(n)$ is odd whenever $n$ is odd and even whenever $n$ is even because when $n$ is even there is a factor $2$.
$varphi(n)$ is multiplicative-we have $varphi(nm)=varphi(n)varphi(m)$ whenever $n$ and $m$ are coprime. If $n=p^k$ for $k$ an odd prime, $varphi(p^k)$ is even when $k$ is odd, but $varphi(2^k)$ is always odd. This means that $varphi(n)$ is even when $n$ has at least one factor of an odd prime raised to an odd power and odd otherwise. For example, all multiples of $3$ that are not multiples of $9$ have an even sum of factors because we have all the factors without a factor $3$ and all the factors with a factor $3$, making $4$ times the sum of all the factors without a factor $3$.
For odd $n$ the only ones with $varphi(n)$ odd are $9,25,49$, so $47$ odd numbers have $f(n)-varphi(n) odd.
For even $n$ the only ones with $varphi(n)$ odd are $18,36,54,72,90,50,80$ so $7$ even numbers have $f(n)-varphi(n)$ odd.
The chance a number $n$ in the range $1$ to $100$ inclusive has $f(n)-varphi(n)$ even is $frac {51}{100}$
answered Jan 20 at 5:32
Ross MillikanRoss Millikan
298k23198371
$endgroup$
$begingroup$
This is not right... Why is o(n) multiplicative? If you substitute 2 and 3 in, it is not right. o(2*3) ≠ o(2)*o(3)
$endgroup$
– Chris.
Jan 20 at 11:03
$begingroup$
@Chris: yes it is. $varphi(2)=1+2=3, varphi(3)=1+3=4, varphi=1+2+3+6=12=3cdot 4$
$endgroup$
– Ross Millikan
Jan 20 at 14:57
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$f(n)$ is odd whenever $n$ is odd and even whenever $n$ is even because when $n$ is even there is a factor $2$.
$varphi(n)$ is multiplicative-we have $varphi(nm)=varphi(n)varphi(m)$ whenever $n$ and $m$ are coprime. If $n=p^k$ for $k$ an odd prime, $varphi(p^k)$ is even when $k$ is odd, but $varphi(2^k)$ is always odd. This means that $varphi(n)$ is even when $n$ has at least one factor of an odd prime raised to an odd power and odd otherwise. For example, all multiples of $3$ that are not multiples of $9$ have an even sum of factors because we have all the factors without a factor $3$ and all the factors with a factor $3$, making $4$ times the sum of all the factors without a factor $3$.
For odd $n$ the only ones with $varphi(n)$ odd are $9,25,49$, so $47$ odd numbers have $f(n)-varphi(n) odd.
For even $n$ the only ones with $varphi(n)$ odd are $18,36,54,72,90,50,80$ so $7$ even numbers have $f(n)-varphi(n)$ odd.
The chance a number $n$ in the range $1$ to $100$ inclusive has $f(n)-varphi(n)$ even is $frac {51}{100}$
answered Jan 20 at 5:32
Ross MillikanRoss Millikan
298k23198371
$endgroup$
$begingroup$
This is not right... Why is o(n) multiplicative? If you substitute 2 and 3 in, it is not right. o(2*3) ≠ o(2)*o(3)
$endgroup$
– Chris.
Jan 20 at 11:03
$begingroup$
@Chris: yes it is. $varphi(2)=1+2=3, varphi(3)=1+3=4, varphi=1+2+3+6=12=3cdot 4$
$endgroup$
– Ross Millikan
Jan 20 at 14:57
add a comment |
$begingroup$
$f(n)$ is odd whenever $n$ is odd and even whenever $n$ is even because when $n$ is even there is a factor $2$.
$varphi(n)$ is multiplicative-we have $varphi(nm)=varphi(n)varphi(m)$ whenever $n$ and $m$ are coprime. If $n=p^k$ for $k$ an odd prime, $varphi(p^k)$ is even when $k$ is odd, but $varphi(2^k)$ is always odd. This means that $varphi(n)$ is even when $n$ has at least one factor of an odd prime raised to an odd power and odd otherwise. For example, all multiples of $3$ that are not multiples of $9$ have an even sum of factors because we have all the factors without a factor $3$ and all the factors with a factor $3$, making $4$ times the sum of all the factors without a factor $3$.
For odd $n$ the only ones with $varphi(n)$ odd are $9,25,49$, so $47$ odd numbers have $f(n)-varphi(n) odd.
For even $n$ the only ones with $varphi(n)$ odd are $18,36,54,72,90,50,80$ so $7$ even numbers have $f(n)-varphi(n)$ odd.
The chance a number $n$ in the range $1$ to $100$ inclusive has $f(n)-varphi(n)$ even is $frac {51}{100}$
answered Jan 20 at 5:32
Ross MillikanRoss Millikan
298k23198371
$endgroup$
$begingroup$
This is not right... Why is o(n) multiplicative? If you substitute 2 and 3 in, it is not right. o(2*3) ≠ o(2)*o(3)
$endgroup$
– Chris.
Jan 20 at 11:03
$begingroup$
@Chris: yes it is. $varphi(2)=1+2=3, varphi(3)=1+3=4, varphi=1+2+3+6=12=3cdot 4$
$endgroup$
– Ross Millikan
Jan 20 at 14:57
add a comment |
$begingroup$
$f(n)$ is odd whenever $n$ is odd and even whenever $n$ is even because when $n$ is even there is a factor $2$.
$varphi(n)$ is multiplicative-we have $varphi(nm)=varphi(n)varphi(m)$ whenever $n$ and $m$ are coprime. If $n=p^k$ for $k$ an odd prime, $varphi(p^k)$ is even when $k$ is odd, but $varphi(2^k)$ is always odd. This means that $varphi(n)$ is even when $n$ has at least one factor of an odd prime raised to an odd power and odd otherwise. For example, all multiples of $3$ that are not multiples of $9$ have an even sum of factors because we have all the factors without a factor $3$ and all the factors with a factor $3$, making $4$ times the sum of all the factors without a factor $3$.
For odd $n$ the only ones with $varphi(n)$ odd are $9,25,49$, so $47$ odd numbers have $f(n)-varphi(n) odd.
For even $n$ the only ones with $varphi(n)$ odd are $18,36,54,72,90,50,80$ so $7$ even numbers have $f(n)-varphi(n)$ odd.
The chance a number $n$ in the range $1$ to $100$ inclusive has $f(n)-varphi(n)$ even is $frac {51}{100}$
answered Jan 20 at 5:32
Ross MillikanRoss Millikan
298k23198371
$endgroup$
$f(n)$ is odd whenever $n$ is odd and even whenever $n$ is even because when $n$ is even there is a factor $2$.
$varphi(n)$ is multiplicative-we have $varphi(nm)=varphi(n)varphi(m)$ whenever $n$ and $m$ are coprime. If $n=p^k$ for $k$ an odd prime, $varphi(p^k)$ is even when $k$ is odd, but $varphi(2^k)$ is always odd. This means that $varphi(n)$ is even when $n$ has at least one factor of an odd prime raised to an odd power and odd otherwise. For example, all multiples of $3$ that are not multiples of $9$ have an even sum of factors because we have all the factors without a factor $3$ and all the factors with a factor $3$, making $4$ times the sum of all the factors without a factor $3$.
For odd $n$ the only ones with $varphi(n)$ odd are $9,25,49$, so $47$ odd numbers have $f(n)-varphi(n) odd.
For even $n$ the only ones with $varphi(n)$ odd are $18,36,54,72,90,50,80$ so $7$ even numbers have $f(n)-varphi(n)$ odd.
The chance a number $n$ in the range $1$ to $100$ inclusive has $f(n)-varphi(n)$ even is $frac {51}{100}$
answered Jan 20 at 5:32
Ross MillikanRoss Millikan
298k23198371
edited Jan 20 at 14:59
answered Jan 20 at 5:32
Ross MillikanRoss Millikan
298k23198371
answered Jan 20 at 5:32
Ross MillikanRoss Millikan
298k23198371
answered Jan 20 at 5:32
Ross MillikanRoss Millikan
298k23198371
298k23198371
$begingroup$
This is not right... Why is o(n) multiplicative? If you substitute 2 and 3 in, it is not right. o(2*3) ≠ o(2)*o(3)
$endgroup$
– Chris.
Jan 20 at 11:03
$begingroup$
@Chris: yes it is. $varphi(2)=1+2=3, varphi(3)=1+3=4, varphi=1+2+3+6=12=3cdot 4$
$endgroup$
– Ross Millikan
Jan 20 at 14:57
add a comment |
$begingroup$
This is not right... Why is o(n) multiplicative? If you substitute 2 and 3 in, it is not right. o(2*3) ≠ o(2)*o(3)
$endgroup$
– Chris.
Jan 20 at 11:03
$begingroup$
@Chris: yes it is. $varphi(2)=1+2=3, varphi(3)=1+3=4, varphi=1+2+3+6=12=3cdot 4$
$endgroup$
– Ross Millikan
Jan 20 at 14:57
$begingroup$
This is not right... Why is o(n) multiplicative? If you substitute 2 and 3 in, it is not right. o(2*3) ≠ o(2)*o(3)
$endgroup$
– Chris.
Jan 20 at 11:03
$begingroup$
This is not right... Why is o(n) multiplicative? If you substitute 2 and 3 in, it is not right. o(2*3) ≠ o(2)*o(3)
$endgroup$
– Chris.
Jan 20 at 11:03
$begingroup$
@Chris: yes it is. $varphi(2)=1+2=3, varphi(3)=1+3=4, varphi=1+2+3+6=12=3cdot 4$
$endgroup$
– Ross Millikan
Jan 20 at 14:57
$begingroup$
@Chris: yes it is. $varphi(2)=1+2=3, varphi(3)=1+3=4, varphi=1+2+3+6=12=3cdot 4$
$endgroup$
– Ross Millikan
Jan 20 at 14:57
add a comment |
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$begingroup$
Hello, and welcome to MSE. Although your title says you have no clue, please tell us what have your tried so far? Here is a hint that might help you: The probability would be the # of values of $n$ where $fleft(nright) - phileft(nright)$ is even divided by the total # of values being checked, this being $100$ in your case.
$endgroup$
– John Omielan
Jan 20 at 5:11