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I got

$CNS(945)=15$ and $CNS(947)=1$.

Short way:

From playing around with other numbers, I noticed that $CNS(n)=o-1$, where $o$ is the number of odd divisors of $n$. I'm not sure how to prove this, but this does give us $CNS(945)=15$ and $CNS(947)=1$ as shown below, since there are sixteen odd divisors of $945$ and only two for $947$ ($1$ and $947$, since it's prime).

Long way:

For $CNS(945)$:

For there to be an odd number of consecutive integers, the average of them (i.e. the number in the middle) must be odd. Thus, we can find all the odd factors of $945$, and what they multiply with to get $945$:

$945 = 3\times315 = 314+315+316$
$945 = 5\times189=187+188+189+190+191$
$945 = 7\times135=132+133+134+135+136+137+138$
$945 = 9\times105=101+102+\cdots+109$
$945 = 15 \times 63=56+57+\cdots+70$
$945 = 21\times 45=35+36+\cdots+55$
$945 = 27\times 35=22+23+\cdots+48$
$945 = 35\times 27=10+11+\cdots+44$

All products after $35\times27$ (i.e. $45\times21$, $63\times15$, etc.) do not work, since there aren't enough positive numbers available on either side of the average number to reach $945$.

For an even number of integers, the average will be $\textrm{something}.5$. Thus, we need to find integer solutions for $m$ and $n$ such that $\frac{945}{m}=n+0.5$. Solving for $m$ and $n$ here, we get the following sums:

$945 = 2\times 472.5=472+473$
$945 = 6\times 157.5=155+156+157+158+159+160$
$945 = 10\times 94.5=90+91+\cdots+99$
$945 = 14\times 67.5=61+62+\cdots+74$
$945 = 18\times 52.5=44+45+\cdots+61$
$945 = 30\times 31.5=17+18+\cdots+46$
$945 = 42\times 22.5=2+3+\cdots+43$

Similar to above, all products after $42\times22.5$ (i.e. $54\times17.5$, $70\times17.5$, etc.) do not work, since there again aren't enough positive numbers available on either side of the average number to reach $945$.

We have eight sums with an odd middle and seven with an even middle, thus $CNS(945)=15$.

For $CNS(947)$:

We can use a similar method to above. First we have to find all the odd factors of $947$, but since $947$ is prime, the only factors are $1$ and $947$. Therefore there are no strings of an odd number of consecutive integers that sum to $947$.
For an even number of integers, we can again use our formula, $\frac{947}{m}=n+0.5$, to find all solutions. Solving this, we find that the only solution that works is $947=2\times473.5=473+474$. This is thus the only way to sum consecutive integers to $947$, and so $CNS(947)=1$.

I got

$CNS(945)=15$ and $CNS(947)=1$.

Short way:

From playing around with other numbers, I noticed that $CNS(n)=o(n)-1$, where $o(n)$ is the number of odd divisors of $n$. I'm not sure how to prove this, but this does give us $CNS(945)=15$ and $CNS(947)=1$ as shown below, since there are sixteen odd divisors of $945$ and only two for $947$ ($1$ and $947$, since it's prime).

Long way:

For $CNS(945)$:

For there to be an odd number of consecutive integers, the average of them (i.e. the number in the middle) must be odd. Thus, we can find all the odd factors of $945$, and what they multiply with to get $945$:

$945 = 3\times315 = 314+315+316$
$945 = 5\times189=187+188+189+190+191$
$945 = 7\times135=132+133+134+135+136+137+138$
$945 = 9\times105=101+102+\cdots+109$
$945 = 15 \times 63=56+57+\cdots+70$
$945 = 21\times 45=35+36+\cdots+55$
$945 = 27\times 35=22+23+\cdots+48$
$945 = 35\times 27=10+11+\cdots+44$

All products after $35\times27$ (i.e. $45\times21$, $63\times15$, etc.) do not work, since there aren't enough positive numbers available on either side of the average number to reach $945$.

For an even number of integers, the average will be $\textrm{something}.5$. Thus, we need to find integer solutions for $m$ and $n$ such that $\frac{945}{m}=n+0.5$. Solving for $m$ and $n$ here, we get the following sums:

$945 = 2\times 472.5=472+473$
$945 = 6\times 157.5=155+156+157+158+159+160$
$945 = 10\times 94.5=90+91+\cdots+99$
$945 = 14\times 67.5=61+62+\cdots+74$
$945 = 18\times 52.5=44+45+\cdots+61$
$945 = 30\times 31.5=17+18+\cdots+46$
$945 = 42\times 22.5=2+3+\cdots+43$

Similar to above, all products after $42\times22.5$ (i.e. $54\times17.5$, $70\times17.5$, etc.) do not work, since there again aren't enough positive numbers available on either side of the average number to reach $945$.

We have eight sums with an odd middle and seven with an even middle, thus $CNS(945)=15$.

For $CNS(947)$:

We can use a similar method to above. First we have to find all the odd factors of $947$, but since $947$ is prime, the only factors are $1$ and $947$. Therefore there are no strings of an odd number of consecutive integers that sum to $947$.
For an even number of integers, we can again use our formula, $\frac{947}{m}=n+0.5$, to find all solutions. Solving this, we find that the only solution that works is $947=2\times473.5=473+474$. This is thus the only way to sum consecutive integers to $947$, and so $CNS(947)=1$.

I got

$CNS(945)=15$ and $CNS(947)=1$.

For $CNS(945)$:

For there to be an odd number of consecutive integers, the average of them (i.e. the number in the middle) must be odd. Thus, we can find all the odd factors of $945$, and what they multiply with to get $945$:

$945 = 3\times315 = 314+315+316$
$945 = 5\times189=187+188+189+190+191$
$945 = 7\times135=132+133+134+135+136+137+138$
$945 = 9\times105=101+102+\cdots+109$
$945 = 15 \times 63=56+57+\cdots+70$
$945 = 21\times 45=35+36+\cdots+55$
$945 = 27\times 35=22+23+\cdots+48$
$945 = 35\times 27=10+11+\cdots+44$

All products after $35\times27$ (i.e. $45\times21$, $63\times15$, etc.) do not work, since there aren't enough positive numbers available on either side of the average number to reach $945$.

For an even number of integers, the average will be $\textrm{something}.5$. Thus, we need to find integer solutions for $m$ and $n$ such that $\frac{945}{m}=n+0.5$. Solving for $m$ and $n$ here, we get the following sums:

$945 = 2\times 472.5=472+473$
$945 = 6\times 157.5=155+156+157+158+159+160$
$945 = 10\times 94.5=90+91+\cdots+99$
$945 = 14\times 67.5=61+62+\cdots+74$
$945 = 18\times 52.5=44+45+\cdots+61$
$945 = 30\times 31.5=17+18+\cdots+46$
$945 = 42\times 22.5=2+3+\cdots+43$

Similar to above, all products after $42\times22.5$ (i.e. $54\times17.5$, $70\times17.5$, etc.) do not work, since there again aren't enough positive numbers available on either side of the average number to reach $945$.

We have eight sums with an odd middle and seven with an even middle, thus $CNS(945)=15$.

For $CNS(947)$:

We can use a similar method to above. First we have to find all the odd factors of $947$, but since $947$ is prime, the only factors are $1$ and $947$. Therefore there are no strings of an odd number of consecutive integers that sum to $947$.
For an even number of integers, we can again use our formula, $\frac{947}{m}=n+0.5$, to find all solutions. Solving this, we find that the only solution that works is $947=2\times473.5=473+474$. This is thus the only way to sum consecutive integers to $947$, and so $CNS(947)=1$.

I got

$CNS(945)=15$ and $CNS(947)=1$.

Short way:

From playing around with other numbers, I noticed that $CNS(n)=o-1$, where $o$ is the number of odd divisors of $n$. I'm not sure how to prove this, but this does give us $CNS(945)=15$ and $CNS(947)=1$ as shown below, since there are sixteen odd divisors of $945$ and only two for $947$ ($1$ and $947$, since it's prime).

Long way:

For $CNS(945)$:

For there to be an odd number of consecutive integers, the average of them (i.e. the number in the middle) must be odd. Thus, we can find all the odd factors of $945$, and what they multiply with to get $945$:

$945 = 3\times315 = 314+315+316$
$945 = 5\times189=187+188+189+190+191$
$945 = 7\times135=132+133+134+135+136+137+138$
$945 = 9\times105=101+102+\cdots+109$
$945 = 15 \times 63=56+57+\cdots+70$
$945 = 21\times 45=35+36+\cdots+55$
$945 = 27\times 35=22+23+\cdots+48$
$945 = 35\times 27=10+11+\cdots+44$

All products after $35\times27$ (i.e. $45\times21$, $63\times15$, etc.) do not work, since there aren't enough positive numbers available on either side of the average number to reach $945$.

For an even number of integers, the average will be $\textrm{something}.5$. Thus, we need to find integer solutions for $m$ and $n$ such that $\frac{945}{m}=n+0.5$. Solving for $m$ and $n$ here, we get the following sums:

$945 = 2\times 472.5=472+473$
$945 = 6\times 157.5=155+156+157+158+159+160$
$945 = 10\times 94.5=90+91+\cdots+99$
$945 = 14\times 67.5=61+62+\cdots+74$
$945 = 18\times 52.5=44+45+\cdots+61$
$945 = 30\times 31.5=17+18+\cdots+46$
$945 = 42\times 22.5=2+3+\cdots+43$

Similar to above, all products after $42\times22.5$ (i.e. $54\times17.5$, $70\times17.5$, etc.) do not work, since there again aren't enough positive numbers available on either side of the average number to reach $945$.

We have eight sums with an odd middle and seven with an even middle, thus $CNS(945)=15$.

For $CNS(947)$:

We can use a similar method to above. First we have to find all the odd factors of $947$, but since $947$ is prime, the only factors are $1$ and $947$. Therefore there are no strings of an odd number of consecutive integers that sum to $947$.
For an even number of integers, we can again use our formula, $\frac{947}{m}=n+0.5$, to find all solutions. Solving this, we find that the only solution that works is $947=2\times473.5=473+474$. This is thus the only way to sum consecutive integers to $947$, and so $CNS(947)=1$.

I got

$CNS(945)=15$ and $CNS(947)=1$.

For $CNS(945)$:

For there to be an odd number of consecutive integers, the average of them (i.e. the number in the middle) must be odd. Thus, we can find all the odd factors of $945$:

$945 = 3\times315 = 314+315+316$
$945 = 5\times189=187+188+189+190+191$
$945 = 7\times135=132+133+134+135+136+137+138$
$945 = 9\times105=101+102+\cdots+109$
$945 = 15 \times 63=56+57+\cdots+70$
$945 = 21\times 45=35+36+\cdots+55$
$945 = 27\times 35=22+23+\cdots+48$
$945 = 35\times 27=10+11+\cdots+44$

All products after $35\times27$ (i.e. $45\times21$, $63\times15$, etc.) do not work, since there aren't enough positive numbers available on either side of the average number to reach $945$.

For an even number of integers, the average will be $\textrm{something}.5$. Thus, we need to find integer solutions for $m$ and $n$ such that $\frac{945}{m}=n+0.5$. Solving for $m$ and $n$ here, we get the following sums:

$945 = 2\times 472.5=472+473$
$945 = 6\times 157.5=155+156+157+158+159+160$
$945 = 10\times 94.5=90+91+\cdots+99$
$945 = 14\times 67.5=61+62+\cdots+74$
$945 = 18\times 52.5=44+45+\cdots+61$
$945 = 30\times 31.5=17+18+\cdots+46$
$945 = 42\times 22.5=2+3+\cdots+43$

Similar to above, all products after $42\times22.5$ (i.e. $54\times17.5$, $70\times17.5$, etc.) do not work, since there again aren't enough positive numbers available on either side of the average number to reach $945$.

We have eight sums with an odd middle and seven with an even middle, thus $CNS(945)=15$.

For $CNS(947)$:

explanation incoming

I got

$CNS(945)=15$ and $CNS(947)=1$.

For $CNS(945)$:

For there to be an odd number of consecutive integers, the average of them (i.e. the number in the middle) must be odd. Thus, we can find all the odd factors of $945$, and what they multiply with to get $945$:

$945 = 3\times315 = 314+315+316$
$945 = 5\times189=187+188+189+190+191$
$945 = 7\times135=132+133+134+135+136+137+138$
$945 = 9\times105=101+102+\cdots+109$
$945 = 15 \times 63=56+57+\cdots+70$
$945 = 21\times 45=35+36+\cdots+55$
$945 = 27\times 35=22+23+\cdots+48$
$945 = 35\times 27=10+11+\cdots+44$

All products after $35\times27$ (i.e. $45\times21$, $63\times15$, etc.) do not work, since there aren't enough positive numbers available on either side of the average number to reach $945$.

For an even number of integers, the average will be $\textrm{something}.5$. Thus, we need to find integer solutions for $m$ and $n$ such that $\frac{945}{m}=n+0.5$. Solving for $m$ and $n$ here, we get the following sums:

$945 = 2\times 472.5=472+473$
$945 = 6\times 157.5=155+156+157+158+159+160$
$945 = 10\times 94.5=90+91+\cdots+99$
$945 = 14\times 67.5=61+62+\cdots+74$
$945 = 18\times 52.5=44+45+\cdots+61$
$945 = 30\times 31.5=17+18+\cdots+46$
$945 = 42\times 22.5=2+3+\cdots+43$

Similar to above, all products after $42\times22.5$ (i.e. $54\times17.5$, $70\times17.5$, etc.) do not work, since there again aren't enough positive numbers available on either side of the average number to reach $945$.

We have eight sums with an odd middle and seven with an even middle, thus $CNS(945)=15$.

For $CNS(947)$:

We can use a similar method to above. First we have to find all the odd factors of $947$, but since $947$ is prime, the only factors are $1$ and $947$. Therefore there are no strings of an odd number of consecutive integers that sum to $947$.
For an even number of integers, we can again use our formula, $\frac{947}{m}=n+0.5$, to find all solutions. Solving this, we find that the only solution that works is $947=2\times473.5=473+474$. This is thus the only way to sum consecutive integers to $947$, and so $CNS(947)=1$.