About “interesting” numbers #2

We call a divisor of a positive integer interesting if it is a divisor of a number and differs from 1 and the number itself. We call a number $X$ very interesting if it has at least two interesting divisors and it is divisible by the difference between any two of its interesting divisors.

How can one find the product of all very interesting numbers?

I experimented numerically and came to the conclusion that such numbers never exceed 1000. Is it really so?

edited Dec 24 '18 at 20:57

Jam

5,01921432

asked Feb 8 '18 at 7:50

Victor

1766

$begingroup$
Do you mean by "divided into" that such a very interesting number is divisible by the difference of any two "interesting" (i.e. nontrivial) divisors of $X$ ?
$endgroup$
– coffeemath
Feb 8 '18 at 8:00

3

$begingroup$
Exaple of very inresting number - $12$. It has intresting dividers ${2,3,4,6}$. Number $12$ is divided by the difference $3 - 2$, $4 - 2$, $6 - 2$, $4 - 3$, $6 - 3$, $6 - 4$.
$endgroup$
– Victor
Feb 8 '18 at 8:14

2

$begingroup$
@Victor: have you found interesting numbers other than $6,8,12$?
$endgroup$
– Oleg567
Feb 8 '18 at 8:41

$begingroup$
It seems that very interesting numbers are quite boring. (Proof: every finite set is boring, and there are only three so-called very interesting numbers.)
$endgroup$
– Asaf Karagila♦
Feb 8 '18 at 15:03

add a comment |

How can one find the product of all very interesting numbers?

I experimented numerically and came to the conclusion that such numbers never exceed 1000. Is it really so?

edited Dec 24 '18 at 20:57

Jam

5,01921432

asked Feb 8 '18 at 7:50

Victor

1766

$begingroup$
Do you mean by "divided into" that such a very interesting number is divisible by the difference of any two "interesting" (i.e. nontrivial) divisors of $X$ ?
$endgroup$
– coffeemath
Feb 8 '18 at 8:00

3

$begingroup$
Exaple of very inresting number - $12$. It has intresting dividers ${2,3,4,6}$. Number $12$ is divided by the difference $3 - 2$, $4 - 2$, $6 - 2$, $4 - 3$, $6 - 3$, $6 - 4$.
$endgroup$
– Victor
Feb 8 '18 at 8:14

2

$begingroup$
@Victor: have you found interesting numbers other than $6,8,12$?
$endgroup$
– Oleg567
Feb 8 '18 at 8:41

$begingroup$
It seems that very interesting numbers are quite boring. (Proof: every finite set is boring, and there are only three so-called very interesting numbers.)
$endgroup$
– Asaf Karagila♦
Feb 8 '18 at 15:03

add a comment |

How can one find the product of all very interesting numbers?

I experimented numerically and came to the conclusion that such numbers never exceed 1000. Is it really so?

edited Dec 24 '18 at 20:57

Jam

5,01921432

asked Feb 8 '18 at 7:50

Victor

1766

How can one find the product of all very interesting numbers?

I experimented numerically and came to the conclusion that such numbers never exceed 1000. Is it really so?

number-theory elementary-number-theory

edited Dec 24 '18 at 20:57

Jam

5,01921432

asked Feb 8 '18 at 7:50

Victor

1766

edited Dec 24 '18 at 20:57

Jam

5,01921432

asked Feb 8 '18 at 7:50

Victor

1766

edited Dec 24 '18 at 20:57

Jam

5,01921432

edited Dec 24 '18 at 20:57

Jam

5,01921432

edited Dec 24 '18 at 20:57

Jam

5,01921432

asked Feb 8 '18 at 7:50

Victor

1766

asked Feb 8 '18 at 7:50

Victor

1766

asked Feb 8 '18 at 7:50

Victor

1766

$begingroup$
Do you mean by "divided into" that such a very interesting number is divisible by the difference of any two "interesting" (i.e. nontrivial) divisors of $X$ ?
$endgroup$
– coffeemath
Feb 8 '18 at 8:00

3

$begingroup$
Exaple of very inresting number - $12$. It has intresting dividers ${2,3,4,6}$. Number $12$ is divided by the difference $3 - 2$, $4 - 2$, $6 - 2$, $4 - 3$, $6 - 3$, $6 - 4$.
$endgroup$
– Victor
Feb 8 '18 at 8:14

2

$begingroup$
@Victor: have you found interesting numbers other than $6,8,12$?
$endgroup$
– Oleg567
Feb 8 '18 at 8:41

$begingroup$
It seems that very interesting numbers are quite boring. (Proof: every finite set is boring, and there are only three so-called very interesting numbers.)
$endgroup$
– Asaf Karagila♦
Feb 8 '18 at 15:03

add a comment |

$begingroup$
Do you mean by "divided into" that such a very interesting number is divisible by the difference of any two "interesting" (i.e. nontrivial) divisors of $X$ ?
$endgroup$
– coffeemath
Feb 8 '18 at 8:00

3

$begingroup$
Exaple of very inresting number - $12$. It has intresting dividers ${2,3,4,6}$. Number $12$ is divided by the difference $3 - 2$, $4 - 2$, $6 - 2$, $4 - 3$, $6 - 3$, $6 - 4$.
$endgroup$
– Victor
Feb 8 '18 at 8:14

2

$begingroup$
@Victor: have you found interesting numbers other than $6,8,12$?
$endgroup$
– Oleg567
Feb 8 '18 at 8:41

$begingroup$
It seems that very interesting numbers are quite boring. (Proof: every finite set is boring, and there are only three so-called very interesting numbers.)
$endgroup$
– Asaf Karagila♦
Feb 8 '18 at 15:03

Do you mean by "divided into" that such a very interesting number is divisible by the difference of any two "interesting" (i.e. nontrivial) divisors of $X$ ?

– coffeemath
Feb 8 '18 at 8:00

Exaple of very inresting number - $12$. It has intresting dividers ${2,3,4,6}$. Number $12$ is divided by the difference $3 - 2$, $4 - 2$, $6 - 2$, $4 - 3$, $6 - 3$, $6 - 4$.

– Victor
Feb 8 '18 at 8:14

@Victor: have you found interesting numbers other than $6,8,12$?

– Oleg567
Feb 8 '18 at 8:41

It seems that very interesting numbers are quite boring. (Proof: every finite set is boring, and there are only three so-called very interesting numbers.)

– Asaf Karagila♦
Feb 8 '18 at 15:03

add a comment |

1 Answer
1

active

oldest

votes

Suppose $n$ is very interesting.

Then $n$ must be composite, hence we can write, $n=ab$, where $a$ is the least prime factor of $n$, and $b > 1$.

If $b le a$, then by minimality of $a$, we must have $b=a$, but then $n=a^2$, which is not possible, since the square of a prime is not a very interesting number. Therefore $b > a$.

If $a$ is odd, then $n$ is odd, hence so is $b$.

But then $b-a$ is even, hence, since $(b-a){,mid,} n$, it follows that $2{,mid,}n$, contradiction.

Hence we must have $a=2$, so $n=2b$.

Then $b-2$ must divide $n$, but then, since
$$n = 2b = 2(b-2) + 4$$
it follows that $(b-2){,mid,}4$, hence $b le 6$, so $n le 12$.

For $1 le n le 12$, the only candidates are $6,8,10,12$, since those are the only even composite numbers which are not the square of a prime.

The number $10$ is not very interesting, since $5-2 = 3$, which is not a divisor of $10$.

It's easily verified that the numbers $6,8,12$ are very interesting, hence those are the only very interesting numbers.

edited Feb 8 '18 at 16:39

answered Feb 8 '18 at 9:33

quasi

36.2k22664

add a comment |

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1 Answer
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1 Answer
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oldest

votes

Suppose $n$ is very interesting.

Then $n$ must be composite, hence we can write, $n=ab$, where $a$ is the least prime factor of $n$, and $b > 1$.

If $b le a$, then by minimality of $a$, we must have $b=a$, but then $n=a^2$, which is not possible, since the square of a prime is not a very interesting number. Therefore $b > a$.

If $a$ is odd, then $n$ is odd, hence so is $b$.

But then $b-a$ is even, hence, since $(b-a){,mid,} n$, it follows that $2{,mid,}n$, contradiction.

Hence we must have $a=2$, so $n=2b$.

Then $b-2$ must divide $n$, but then, since
$$n = 2b = 2(b-2) + 4$$
it follows that $(b-2){,mid,}4$, hence $b le 6$, so $n le 12$.

For $1 le n le 12$, the only candidates are $6,8,10,12$, since those are the only even composite numbers which are not the square of a prime.

The number $10$ is not very interesting, since $5-2 = 3$, which is not a divisor of $10$.

It's easily verified that the numbers $6,8,12$ are very interesting, hence those are the only very interesting numbers.

edited Feb 8 '18 at 16:39

answered Feb 8 '18 at 9:33

quasi

36.2k22664

add a comment |

Suppose $n$ is very interesting.

Then $n$ must be composite, hence we can write, $n=ab$, where $a$ is the least prime factor of $n$, and $b > 1$.

If $b le a$, then by minimality of $a$, we must have $b=a$, but then $n=a^2$, which is not possible, since the square of a prime is not a very interesting number. Therefore $b > a$.

If $a$ is odd, then $n$ is odd, hence so is $b$.

But then $b-a$ is even, hence, since $(b-a){,mid,} n$, it follows that $2{,mid,}n$, contradiction.

Hence we must have $a=2$, so $n=2b$.

Then $b-2$ must divide $n$, but then, since
$$n = 2b = 2(b-2) + 4$$
it follows that $(b-2){,mid,}4$, hence $b le 6$, so $n le 12$.

For $1 le n le 12$, the only candidates are $6,8,10,12$, since those are the only even composite numbers which are not the square of a prime.

The number $10$ is not very interesting, since $5-2 = 3$, which is not a divisor of $10$.

It's easily verified that the numbers $6,8,12$ are very interesting, hence those are the only very interesting numbers.

edited Feb 8 '18 at 16:39

answered Feb 8 '18 at 9:33

quasi

36.2k22664

add a comment |

Suppose $n$ is very interesting.

Then $n$ must be composite, hence we can write, $n=ab$, where $a$ is the least prime factor of $n$, and $b > 1$.

If $b le a$, then by minimality of $a$, we must have $b=a$, but then $n=a^2$, which is not possible, since the square of a prime is not a very interesting number. Therefore $b > a$.

If $a$ is odd, then $n$ is odd, hence so is $b$.

But then $b-a$ is even, hence, since $(b-a){,mid,} n$, it follows that $2{,mid,}n$, contradiction.

Hence we must have $a=2$, so $n=2b$.

Then $b-2$ must divide $n$, but then, since
$$n = 2b = 2(b-2) + 4$$
it follows that $(b-2){,mid,}4$, hence $b le 6$, so $n le 12$.

For $1 le n le 12$, the only candidates are $6,8,10,12$, since those are the only even composite numbers which are not the square of a prime.

The number $10$ is not very interesting, since $5-2 = 3$, which is not a divisor of $10$.

It's easily verified that the numbers $6,8,12$ are very interesting, hence those are the only very interesting numbers.

edited Feb 8 '18 at 16:39

answered Feb 8 '18 at 9:33

quasi

36.2k22664

Suppose $n$ is very interesting.

Then $n$ must be composite, hence we can write, $n=ab$, where $a$ is the least prime factor of $n$, and $b > 1$.

If $b le a$, then by minimality of $a$, we must have $b=a$, but then $n=a^2$, which is not possible, since the square of a prime is not a very interesting number. Therefore $b > a$.

If $a$ is odd, then $n$ is odd, hence so is $b$.

But then $b-a$ is even, hence, since $(b-a){,mid,} n$, it follows that $2{,mid,}n$, contradiction.

Hence we must have $a=2$, so $n=2b$.

Then $b-2$ must divide $n$, but then, since
$$n = 2b = 2(b-2) + 4$$
it follows that $(b-2){,mid,}4$, hence $b le 6$, so $n le 12$.

For $1 le n le 12$, the only candidates are $6,8,10,12$, since those are the only even composite numbers which are not the square of a prime.

The number $10$ is not very interesting, since $5-2 = 3$, which is not a divisor of $10$.

It's easily verified that the numbers $6,8,12$ are very interesting, hence those are the only very interesting numbers.

edited Feb 8 '18 at 16:39

answered Feb 8 '18 at 9:33

quasi

36.2k22664

edited Feb 8 '18 at 16:39

answered Feb 8 '18 at 9:33

quasi

36.2k22664

answered Feb 8 '18 at 9:33

quasi

36.2k22664

answered Feb 8 '18 at 9:33

quasi

36.2k22664

add a comment |

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