N has all divisors up to 31 except two The Next CEO of Stack OverflowNumber theory problem divides“Quadly” numbers with just 4 factorsTrue or false division algorithm problemProve that $1$ has only one divisorA Question Related to Zsigmondy's Theoremnumber system and divisibilityWritten as sum of positive divisorsUnder what conditions does an integer dividing the square of n imply that the integer must divide n?divisibility proofProve that if $3n^2 + 2n$ is even, then $n$ is even
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N has all divisors up to 31 except two
The Next CEO of Stack OverflowNumber theory problem divides“Quadly” numbers with just 4 factorsTrue or false division algorithm problemProve that $1$ has only one divisorA Question Related to Zsigmondy's Theoremnumber system and divisibilityWritten as sum of positive divisorsUnder what conditions does an integer dividing the square of n imply that the integer must divide n?divisibility proofProve that if $3n^2 + 2n$ is even, then $n$ is even
$begingroup$
Let $N$ be a positive integer. It is true that $1|N, 2|N, 3|N, dots, 31|N$, except two. Which are false?
It seems like the problem is impossible. For example, any two consecutive integers must have an even integer E, and this even integer is divisible by 2 and an integer less than it, and since 2 and that integer divide N, E must divide $N$, which is a contradiction. For example, if we chose 31 and 30, then since 2 and 15 divide N, it must be the case that 30 divides $N$, which is a contradiction. So I'm not sure where to go with this problem.
divisibility
$endgroup$
add a comment |
$begingroup$
Let $N$ be a positive integer. It is true that $1|N, 2|N, 3|N, dots, 31|N$, except two. Which are false?
It seems like the problem is impossible. For example, any two consecutive integers must have an even integer E, and this even integer is divisible by 2 and an integer less than it, and since 2 and that integer divide N, E must divide $N$, which is a contradiction. For example, if we chose 31 and 30, then since 2 and 15 divide N, it must be the case that 30 divides $N$, which is a contradiction. So I'm not sure where to go with this problem.
divisibility
$endgroup$
$begingroup$
I understand the problem as follows: Of the 31 statements $i|N, i=1,2,ldots,31$ 29 are true and 2 are false. Which are the 2 that are false? As your example shows, an even divisor is likely not among them.
$endgroup$
– Ingix
Mar 19 at 21:42
$begingroup$
It seems like there is more than one solution. We could have $$N= 31!over 17cdot19$$ but we could choose any two primes $17leq p,qleq31$ in the denominator.
$endgroup$
– saulspatz
Mar 19 at 22:10
add a comment |
$begingroup$
Let $N$ be a positive integer. It is true that $1|N, 2|N, 3|N, dots, 31|N$, except two. Which are false?
It seems like the problem is impossible. For example, any two consecutive integers must have an even integer E, and this even integer is divisible by 2 and an integer less than it, and since 2 and that integer divide N, E must divide $N$, which is a contradiction. For example, if we chose 31 and 30, then since 2 and 15 divide N, it must be the case that 30 divides $N$, which is a contradiction. So I'm not sure where to go with this problem.
divisibility
$endgroup$
Let $N$ be a positive integer. It is true that $1|N, 2|N, 3|N, dots, 31|N$, except two. Which are false?
It seems like the problem is impossible. For example, any two consecutive integers must have an even integer E, and this even integer is divisible by 2 and an integer less than it, and since 2 and that integer divide N, E must divide $N$, which is a contradiction. For example, if we chose 31 and 30, then since 2 and 15 divide N, it must be the case that 30 divides $N$, which is a contradiction. So I'm not sure where to go with this problem.
divisibility
divisibility
asked Mar 19 at 21:28
WesleyWesley
609413
609413
$begingroup$
I understand the problem as follows: Of the 31 statements $i|N, i=1,2,ldots,31$ 29 are true and 2 are false. Which are the 2 that are false? As your example shows, an even divisor is likely not among them.
$endgroup$
– Ingix
Mar 19 at 21:42
$begingroup$
It seems like there is more than one solution. We could have $$N= 31!over 17cdot19$$ but we could choose any two primes $17leq p,qleq31$ in the denominator.
$endgroup$
– saulspatz
Mar 19 at 22:10
add a comment |
$begingroup$
I understand the problem as follows: Of the 31 statements $i|N, i=1,2,ldots,31$ 29 are true and 2 are false. Which are the 2 that are false? As your example shows, an even divisor is likely not among them.
$endgroup$
– Ingix
Mar 19 at 21:42
$begingroup$
It seems like there is more than one solution. We could have $$N= 31!over 17cdot19$$ but we could choose any two primes $17leq p,qleq31$ in the denominator.
$endgroup$
– saulspatz
Mar 19 at 22:10
$begingroup$
I understand the problem as follows: Of the 31 statements $i|N, i=1,2,ldots,31$ 29 are true and 2 are false. Which are the 2 that are false? As your example shows, an even divisor is likely not among them.
$endgroup$
– Ingix
Mar 19 at 21:42
$begingroup$
I understand the problem as follows: Of the 31 statements $i|N, i=1,2,ldots,31$ 29 are true and 2 are false. Which are the 2 that are false? As your example shows, an even divisor is likely not among them.
$endgroup$
– Ingix
Mar 19 at 21:42
$begingroup$
It seems like there is more than one solution. We could have $$N= 31!over 17cdot19$$ but we could choose any two primes $17leq p,qleq31$ in the denominator.
$endgroup$
– saulspatz
Mar 19 at 22:10
$begingroup$
It seems like there is more than one solution. We could have $$N= 31!over 17cdot19$$ but we could choose any two primes $17leq p,qleq31$ in the denominator.
$endgroup$
– saulspatz
Mar 19 at 22:10
add a comment |
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$begingroup$
I understand the problem as follows: Of the 31 statements $i|N, i=1,2,ldots,31$ 29 are true and 2 are false. Which are the 2 that are false? As your example shows, an even divisor is likely not among them.
$endgroup$
– Ingix
Mar 19 at 21:42
$begingroup$
It seems like there is more than one solution. We could have $$N= 31!over 17cdot19$$ but we could choose any two primes $17leq p,qleq31$ in the denominator.
$endgroup$
– saulspatz
Mar 19 at 22:10