The sum of n consecutive numbers is divisible by the greatest prime factor of n.
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I facilitated the following task with pre-service math teachers:
Take the sum of any three consecutive numbers. Do you notice anything special? Write a clear conjecture. Then write a clear proof for your conjecture.
Now, take the sum of any amount of consecutive numbers. Can you broaden your conjecture from problem 1? Prove your conjecture.
I left the task open because I wanted students to create a variety of conjectures and proofs for whole class discussion. For task 2, one student came up with the following conjecture: "The sum of n consecutive integers is divisible by the greatest prime factor of n". I'm curious if anyone has a proof or counterexample for this claim as I do not.
number-theory
asked 41 mins ago
MathGuyMathGuy
908315
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add a comment |
$begingroup$
I facilitated the following task with pre-service math teachers:
Take the sum of any three consecutive numbers. Do you notice anything special? Write a clear conjecture. Then write a clear proof for your conjecture.
Now, take the sum of any amount of consecutive numbers. Can you broaden your conjecture from problem 1? Prove your conjecture.
I left the task open because I wanted students to create a variety of conjectures and proofs for whole class discussion. For task 2, one student came up with the following conjecture: "The sum of n consecutive integers is divisible by the greatest prime factor of n". I'm curious if anyone has a proof or counterexample for this claim as I do not.
number-theory
asked 41 mins ago
MathGuyMathGuy
908315
$endgroup$
add a comment |
$begingroup$
I facilitated the following task with pre-service math teachers:
Take the sum of any three consecutive numbers. Do you notice anything special? Write a clear conjecture. Then write a clear proof for your conjecture.
Now, take the sum of any amount of consecutive numbers. Can you broaden your conjecture from problem 1? Prove your conjecture.
I left the task open because I wanted students to create a variety of conjectures and proofs for whole class discussion. For task 2, one student came up with the following conjecture: "The sum of n consecutive integers is divisible by the greatest prime factor of n". I'm curious if anyone has a proof or counterexample for this claim as I do not.
number-theory
asked 41 mins ago
MathGuyMathGuy
908315
$endgroup$
I facilitated the following task with pre-service math teachers:
Take the sum of any three consecutive numbers. Do you notice anything special? Write a clear conjecture. Then write a clear proof for your conjecture.
Now, take the sum of any amount of consecutive numbers. Can you broaden your conjecture from problem 1? Prove your conjecture.
I left the task open because I wanted students to create a variety of conjectures and proofs for whole class discussion. For task 2, one student came up with the following conjecture: "The sum of n consecutive integers is divisible by the greatest prime factor of n". I'm curious if anyone has a proof or counterexample for this claim as I do not.
number-theory
number-theory
asked 41 mins ago
MathGuyMathGuy
908315
asked 41 mins ago
MathGuyMathGuy
908315
asked 41 mins ago
MathGuyMathGuy
908315
asked 41 mins ago
MathGuyMathGuy
908315
asked 41 mins ago
MathGuyMathGuy
908315
908315
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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This is an excellent conjecture. It is not quite true, as it fails for $n=2$. The sum of two consecutive numbers is odd. We can say more. The sum of $n$ consecutive numbers is divisible by $n$ if $n$ is odd and by $frac n2$ if $n$ is even. This implies the student's conjecture for $n gt 2$.
To see this, reduce all the numbers $bmod n$. We will then have one each congruent to $0,1,2,ldots n-1 bmod n$. The sum of the numbers from $0$ to $n-1$ is $frac 12(n-1)n$, which is divisible by $n$ or $frac n2$ as required.
answered 30 mins ago
Ross MillikanRoss Millikan
296k23198371
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If $n=2$ the statement is false. Let's look at $n>2$.
The sum of $n$ consecutive numbers starting with $a$ is
$$
z=frac{n}{2}(2a+n-1)
$$
If $n$ is even, $n/2$ is an integer containing the largest prime factor of $n$, hence $z$ is divisible by that prime factor.
If $n$ is odd, $2a-1+n$ is even and $(2a+n-1)/2$ is an integer. Therefore $z$ is divisible by $n$ and by all of its prime factors.
answered 22 mins ago
GReyesGReyes
1,23915
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add a comment |
$begingroup$
If the first of the $n$ summands is $m+1$, then the sum is
$$(m+1)+(m+2)+ldots+(m+n)=nm+1+2+ldots+n=nm+frac{n(n+1)}{2}. $$
- If $n$ is odd, say $n=2k-1$, this is even a multiple of $n$, namely $ncdot(m+k)$. Then even more so, it is a multiple of e.g. the largest prime divisor of $n$.
- If $n$ is even, say $n=2k$, then it is at least a multiple of $k$, namely $kcdot(m+n+1)$. This is still a multiple of the largest prime divisor of $n$, unless $k=1$.
Hence the conjecture fails only for $n=2$ (and is meaningless for $n=1$).
answered 21 mins ago
Hagen von EitzenHagen von Eitzen
279k23271503
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is an excellent conjecture. It is not quite true, as it fails for $n=2$. The sum of two consecutive numbers is odd. We can say more. The sum of $n$ consecutive numbers is divisible by $n$ if $n$ is odd and by $frac n2$ if $n$ is even. This implies the student's conjecture for $n gt 2$.
To see this, reduce all the numbers $bmod n$. We will then have one each congruent to $0,1,2,ldots n-1 bmod n$. The sum of the numbers from $0$ to $n-1$ is $frac 12(n-1)n$, which is divisible by $n$ or $frac n2$ as required.
answered 30 mins ago
Ross MillikanRoss Millikan
296k23198371
$endgroup$
add a comment |
$begingroup$
This is an excellent conjecture. It is not quite true, as it fails for $n=2$. The sum of two consecutive numbers is odd. We can say more. The sum of $n$ consecutive numbers is divisible by $n$ if $n$ is odd and by $frac n2$ if $n$ is even. This implies the student's conjecture for $n gt 2$.
To see this, reduce all the numbers $bmod n$. We will then have one each congruent to $0,1,2,ldots n-1 bmod n$. The sum of the numbers from $0$ to $n-1$ is $frac 12(n-1)n$, which is divisible by $n$ or $frac n2$ as required.
answered 30 mins ago
Ross MillikanRoss Millikan
296k23198371
$endgroup$
add a comment |
$begingroup$
This is an excellent conjecture. It is not quite true, as it fails for $n=2$. The sum of two consecutive numbers is odd. We can say more. The sum of $n$ consecutive numbers is divisible by $n$ if $n$ is odd and by $frac n2$ if $n$ is even. This implies the student's conjecture for $n gt 2$.
To see this, reduce all the numbers $bmod n$. We will then have one each congruent to $0,1,2,ldots n-1 bmod n$. The sum of the numbers from $0$ to $n-1$ is $frac 12(n-1)n$, which is divisible by $n$ or $frac n2$ as required.
answered 30 mins ago
Ross MillikanRoss Millikan
296k23198371
$endgroup$
This is an excellent conjecture. It is not quite true, as it fails for $n=2$. The sum of two consecutive numbers is odd. We can say more. The sum of $n$ consecutive numbers is divisible by $n$ if $n$ is odd and by $frac n2$ if $n$ is even. This implies the student's conjecture for $n gt 2$.
To see this, reduce all the numbers $bmod n$. We will then have one each congruent to $0,1,2,ldots n-1 bmod n$. The sum of the numbers from $0$ to $n-1$ is $frac 12(n-1)n$, which is divisible by $n$ or $frac n2$ as required.
answered 30 mins ago
Ross MillikanRoss Millikan
296k23198371
answered 30 mins ago
Ross MillikanRoss Millikan
296k23198371
answered 30 mins ago
Ross MillikanRoss Millikan
296k23198371
answered 30 mins ago
Ross MillikanRoss Millikan
296k23198371
296k23198371
add a comment |
add a comment |
$begingroup$
If $n=2$ the statement is false. Let's look at $n>2$.
The sum of $n$ consecutive numbers starting with $a$ is
$$
z=frac{n}{2}(2a+n-1)
$$
If $n$ is even, $n/2$ is an integer containing the largest prime factor of $n$, hence $z$ is divisible by that prime factor.
If $n$ is odd, $2a-1+n$ is even and $(2a+n-1)/2$ is an integer. Therefore $z$ is divisible by $n$ and by all of its prime factors.
answered 22 mins ago
GReyesGReyes
1,23915
$endgroup$
add a comment |
$begingroup$
If $n=2$ the statement is false. Let's look at $n>2$.
The sum of $n$ consecutive numbers starting with $a$ is
$$
z=frac{n}{2}(2a+n-1)
$$
If $n$ is even, $n/2$ is an integer containing the largest prime factor of $n$, hence $z$ is divisible by that prime factor.
If $n$ is odd, $2a-1+n$ is even and $(2a+n-1)/2$ is an integer. Therefore $z$ is divisible by $n$ and by all of its prime factors.
answered 22 mins ago
GReyesGReyes
1,23915
$endgroup$
add a comment |
$begingroup$
If $n=2$ the statement is false. Let's look at $n>2$.
The sum of $n$ consecutive numbers starting with $a$ is
$$
z=frac{n}{2}(2a+n-1)
$$
If $n$ is even, $n/2$ is an integer containing the largest prime factor of $n$, hence $z$ is divisible by that prime factor.
If $n$ is odd, $2a-1+n$ is even and $(2a+n-1)/2$ is an integer. Therefore $z$ is divisible by $n$ and by all of its prime factors.
answered 22 mins ago
GReyesGReyes
1,23915
$endgroup$
If $n=2$ the statement is false. Let's look at $n>2$.
The sum of $n$ consecutive numbers starting with $a$ is
$$
z=frac{n}{2}(2a+n-1)
$$
If $n$ is even, $n/2$ is an integer containing the largest prime factor of $n$, hence $z$ is divisible by that prime factor.
If $n$ is odd, $2a-1+n$ is even and $(2a+n-1)/2$ is an integer. Therefore $z$ is divisible by $n$ and by all of its prime factors.
answered 22 mins ago
GReyesGReyes
1,23915
answered 22 mins ago
GReyesGReyes
1,23915
answered 22 mins ago
GReyesGReyes
1,23915
answered 22 mins ago
GReyesGReyes
1,23915
1,23915
add a comment |
add a comment |
$begingroup$
If the first of the $n$ summands is $m+1$, then the sum is
$$(m+1)+(m+2)+ldots+(m+n)=nm+1+2+ldots+n=nm+frac{n(n+1)}{2}. $$
- If $n$ is odd, say $n=2k-1$, this is even a multiple of $n$, namely $ncdot(m+k)$. Then even more so, it is a multiple of e.g. the largest prime divisor of $n$.
- If $n$ is even, say $n=2k$, then it is at least a multiple of $k$, namely $kcdot(m+n+1)$. This is still a multiple of the largest prime divisor of $n$, unless $k=1$.
Hence the conjecture fails only for $n=2$ (and is meaningless for $n=1$).
answered 21 mins ago
Hagen von EitzenHagen von Eitzen
279k23271503
$endgroup$
add a comment |
$begingroup$
If the first of the $n$ summands is $m+1$, then the sum is
$$(m+1)+(m+2)+ldots+(m+n)=nm+1+2+ldots+n=nm+frac{n(n+1)}{2}. $$
- If $n$ is odd, say $n=2k-1$, this is even a multiple of $n$, namely $ncdot(m+k)$. Then even more so, it is a multiple of e.g. the largest prime divisor of $n$.
- If $n$ is even, say $n=2k$, then it is at least a multiple of $k$, namely $kcdot(m+n+1)$. This is still a multiple of the largest prime divisor of $n$, unless $k=1$.
Hence the conjecture fails only for $n=2$ (and is meaningless for $n=1$).
answered 21 mins ago
Hagen von EitzenHagen von Eitzen
279k23271503
$endgroup$
add a comment |
$begingroup$
If the first of the $n$ summands is $m+1$, then the sum is
$$(m+1)+(m+2)+ldots+(m+n)=nm+1+2+ldots+n=nm+frac{n(n+1)}{2}. $$
- If $n$ is odd, say $n=2k-1$, this is even a multiple of $n$, namely $ncdot(m+k)$. Then even more so, it is a multiple of e.g. the largest prime divisor of $n$.
- If $n$ is even, say $n=2k$, then it is at least a multiple of $k$, namely $kcdot(m+n+1)$. This is still a multiple of the largest prime divisor of $n$, unless $k=1$.
Hence the conjecture fails only for $n=2$ (and is meaningless for $n=1$).
answered 21 mins ago
Hagen von EitzenHagen von Eitzen
279k23271503
$endgroup$
If the first of the $n$ summands is $m+1$, then the sum is
$$(m+1)+(m+2)+ldots+(m+n)=nm+1+2+ldots+n=nm+frac{n(n+1)}{2}. $$
- If $n$ is odd, say $n=2k-1$, this is even a multiple of $n$, namely $ncdot(m+k)$. Then even more so, it is a multiple of e.g. the largest prime divisor of $n$.
- If $n$ is even, say $n=2k$, then it is at least a multiple of $k$, namely $kcdot(m+n+1)$. This is still a multiple of the largest prime divisor of $n$, unless $k=1$.
Hence the conjecture fails only for $n=2$ (and is meaningless for $n=1$).
answered 21 mins ago
Hagen von EitzenHagen von Eitzen
279k23271503
answered 21 mins ago
Hagen von EitzenHagen von Eitzen
279k23271503
answered 21 mins ago
Hagen von EitzenHagen von Eitzen
279k23271503
answered 21 mins ago
Hagen von EitzenHagen von Eitzen
279k23271503
279k23271503
add a comment |
add a comment |
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