Let $M$ be a set of numbers such that … , Prove that $M=mathbb{N^*}$
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Let $M$ be a set of numbers such that $Msubsetmathbb{N^*}$ and $2018in M$
And If $min M$ then all the postive divisors of $m$ are in $M$
And If $k,m in M$ and $1 < k < m $ then $km+1 in M$
prove that $M=mathbb{N^*}$
I think we should do induction
Let's suppose that all numbers $le n$ are in $M$
let s take 2 divisors of $n$ , $k$ and $m$
so $k.m+1=n+1in M$
the problem is for prime numbers and perfect squares
number-theory
asked Nov 23 at 20:35
user600785
2610
add a comment |
up vote
0
down vote
favorite
Let $M$ be a set of numbers such that $Msubsetmathbb{N^*}$ and $2018in M$
And If $min M$ then all the postive divisors of $m$ are in $M$
And If $k,m in M$ and $1 < k < m $ then $km+1 in M$
prove that $M=mathbb{N^*}$
I think we should do induction
Let's suppose that all numbers $le n$ are in $M$
let s take 2 divisors of $n$ , $k$ and $m$
so $k.m+1=n+1in M$
the problem is for prime numbers and perfect squares
number-theory
asked Nov 23 at 20:35
user600785
2610
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $M$ be a set of numbers such that $Msubsetmathbb{N^*}$ and $2018in M$
And If $min M$ then all the postive divisors of $m$ are in $M$
And If $k,m in M$ and $1 < k < m $ then $km+1 in M$
prove that $M=mathbb{N^*}$
I think we should do induction
Let's suppose that all numbers $le n$ are in $M$
let s take 2 divisors of $n$ , $k$ and $m$
so $k.m+1=n+1in M$
the problem is for prime numbers and perfect squares
number-theory
asked Nov 23 at 20:35
user600785
2610
Let $M$ be a set of numbers such that $Msubsetmathbb{N^*}$ and $2018in M$
And If $min M$ then all the postive divisors of $m$ are in $M$
And If $k,m in M$ and $1 < k < m $ then $km+1 in M$
prove that $M=mathbb{N^*}$
I think we should do induction
Let's suppose that all numbers $le n$ are in $M$
let s take 2 divisors of $n$ , $k$ and $m$
so $k.m+1=n+1in M$
the problem is for prime numbers and perfect squares
number-theory
number-theory
asked Nov 23 at 20:35
user600785
2610
asked Nov 23 at 20:35
user600785
2610
edited Nov 23 at 20:46
asked Nov 23 at 20:35
user600785
2610
asked Nov 23 at 20:35
user600785
2610
asked Nov 23 at 20:35
user600785
2610
2610
add a comment |
add a comment |
1 Answer
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From $2018$, we get $1,2,1009,2018$. From these, $2cdot 1009+1=2019$ (and then $3$ and $673$) and $2cdot 2018+1=4037$ (and then $11$ and $367$) and a few more. From $2$ and $3$, $7$. From $2$ and $7$, $15$ and then $5$. From $3$ and $5$, $16$ and then $4,8$. From $5$ and $7$, $36$ and then $6,9,12,18$. So far we have the small numbers
$$1,2,3,4,5,6,7,8,9,11,15,16,18 $$
How can we obtain $10$? As a divisor of $km+1$, For suitable $k,m$. Fortunately, $10pm1$ are already in $M$, so we succeed!
We can cast this into an induction proof: Suppose that for some $n>5$, we already know that $1,ldots, n-1in M$. If $n$ is odd, we can write $n=km+1$ with $k=2 $ and $m=frac {n-1}2$ (so $1<k<m<n-1$) and find $nin M$.
If $n$ is even, then $n+1=km+1$ for $k=2 $ and $m=frac{n}2$ (so again $1<k<m<n-1$. Hence $n+1in M$ and then $n^2=(n+1)(n-1)+1in M$ and then also $n$ as a positive divisor thereof.
answered Nov 23 at 20:45
Hagen von Eitzen
275k21268495
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
From $2018$, we get $1,2,1009,2018$. From these, $2cdot 1009+1=2019$ (and then $3$ and $673$) and $2cdot 2018+1=4037$ (and then $11$ and $367$) and a few more. From $2$ and $3$, $7$. From $2$ and $7$, $15$ and then $5$. From $3$ and $5$, $16$ and then $4,8$. From $5$ and $7$, $36$ and then $6,9,12,18$. So far we have the small numbers
$$1,2,3,4,5,6,7,8,9,11,15,16,18 $$
How can we obtain $10$? As a divisor of $km+1$, For suitable $k,m$. Fortunately, $10pm1$ are already in $M$, so we succeed!
We can cast this into an induction proof: Suppose that for some $n>5$, we already know that $1,ldots, n-1in M$. If $n$ is odd, we can write $n=km+1$ with $k=2 $ and $m=frac {n-1}2$ (so $1<k<m<n-1$) and find $nin M$.
If $n$ is even, then $n+1=km+1$ for $k=2 $ and $m=frac{n}2$ (so again $1<k<m<n-1$. Hence $n+1in M$ and then $n^2=(n+1)(n-1)+1in M$ and then also $n$ as a positive divisor thereof.
answered Nov 23 at 20:45
Hagen von Eitzen
275k21268495
add a comment |
up vote
2
down vote
accepted
From $2018$, we get $1,2,1009,2018$. From these, $2cdot 1009+1=2019$ (and then $3$ and $673$) and $2cdot 2018+1=4037$ (and then $11$ and $367$) and a few more. From $2$ and $3$, $7$. From $2$ and $7$, $15$ and then $5$. From $3$ and $5$, $16$ and then $4,8$. From $5$ and $7$, $36$ and then $6,9,12,18$. So far we have the small numbers
$$1,2,3,4,5,6,7,8,9,11,15,16,18 $$
How can we obtain $10$? As a divisor of $km+1$, For suitable $k,m$. Fortunately, $10pm1$ are already in $M$, so we succeed!
We can cast this into an induction proof: Suppose that for some $n>5$, we already know that $1,ldots, n-1in M$. If $n$ is odd, we can write $n=km+1$ with $k=2 $ and $m=frac {n-1}2$ (so $1<k<m<n-1$) and find $nin M$.
If $n$ is even, then $n+1=km+1$ for $k=2 $ and $m=frac{n}2$ (so again $1<k<m<n-1$. Hence $n+1in M$ and then $n^2=(n+1)(n-1)+1in M$ and then also $n$ as a positive divisor thereof.
answered Nov 23 at 20:45
Hagen von Eitzen
275k21268495
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
From $2018$, we get $1,2,1009,2018$. From these, $2cdot 1009+1=2019$ (and then $3$ and $673$) and $2cdot 2018+1=4037$ (and then $11$ and $367$) and a few more. From $2$ and $3$, $7$. From $2$ and $7$, $15$ and then $5$. From $3$ and $5$, $16$ and then $4,8$. From $5$ and $7$, $36$ and then $6,9,12,18$. So far we have the small numbers
$$1,2,3,4,5,6,7,8,9,11,15,16,18 $$
How can we obtain $10$? As a divisor of $km+1$, For suitable $k,m$. Fortunately, $10pm1$ are already in $M$, so we succeed!
We can cast this into an induction proof: Suppose that for some $n>5$, we already know that $1,ldots, n-1in M$. If $n$ is odd, we can write $n=km+1$ with $k=2 $ and $m=frac {n-1}2$ (so $1<k<m<n-1$) and find $nin M$.
If $n$ is even, then $n+1=km+1$ for $k=2 $ and $m=frac{n}2$ (so again $1<k<m<n-1$. Hence $n+1in M$ and then $n^2=(n+1)(n-1)+1in M$ and then also $n$ as a positive divisor thereof.
answered Nov 23 at 20:45
Hagen von Eitzen
275k21268495
From $2018$, we get $1,2,1009,2018$. From these, $2cdot 1009+1=2019$ (and then $3$ and $673$) and $2cdot 2018+1=4037$ (and then $11$ and $367$) and a few more. From $2$ and $3$, $7$. From $2$ and $7$, $15$ and then $5$. From $3$ and $5$, $16$ and then $4,8$. From $5$ and $7$, $36$ and then $6,9,12,18$. So far we have the small numbers
$$1,2,3,4,5,6,7,8,9,11,15,16,18 $$
How can we obtain $10$? As a divisor of $km+1$, For suitable $k,m$. Fortunately, $10pm1$ are already in $M$, so we succeed!
We can cast this into an induction proof: Suppose that for some $n>5$, we already know that $1,ldots, n-1in M$. If $n$ is odd, we can write $n=km+1$ with $k=2 $ and $m=frac {n-1}2$ (so $1<k<m<n-1$) and find $nin M$.
If $n$ is even, then $n+1=km+1$ for $k=2 $ and $m=frac{n}2$ (so again $1<k<m<n-1$. Hence $n+1in M$ and then $n^2=(n+1)(n-1)+1in M$ and then also $n$ as a positive divisor thereof.
answered Nov 23 at 20:45
Hagen von Eitzen
275k21268495
edited Nov 23 at 21:04
answered Nov 23 at 20:45
Hagen von Eitzen
275k21268495
answered Nov 23 at 20:45
Hagen von Eitzen
275k21268495
answered Nov 23 at 20:45
Hagen von Eitzen
275k21268495
275k21268495
add a comment |
add a comment |
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