Mixing
Prove or disprove: If the least common multiple of $a_1, a_2, …, a_n$ is odd, then none of the $a_i$, $i =...
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Here is what I imagine is a very simple question regarding divisibility, which is nevertheless giving me quite a bit of trouble (as is usually the case).
Let $a_1, a_2, ..., a_n$ be positive integers.
Prove or disprove: If the least common multiple of $a_1, a_2, ..., a_n$ is odd, then none of the $a_i$, $i = 1, ..., n$, are even.
elementary-number-theory
asked Jan 3 at 1:45
thisisourconcerndudethisisourconcerndude
1,0961022
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add a comment |
$begingroup$
Here is what I imagine is a very simple question regarding divisibility, which is nevertheless giving me quite a bit of trouble (as is usually the case).
Let $a_1, a_2, ..., a_n$ be positive integers.
Prove or disprove: If the least common multiple of $a_1, a_2, ..., a_n$ is odd, then none of the $a_i$, $i = 1, ..., n$, are even.
elementary-number-theory
asked Jan 3 at 1:45
thisisourconcerndudethisisourconcerndude
1,0961022
$endgroup$
5
$begingroup$
Hint. Each $a_i$ divides the least common multiple (that's what "common multiple" means"). What happens if one of the $a_i$ is even?
$endgroup$
– Ethan Bolker
Jan 3 at 1:49
1
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@EthanBolker Yes, of course. That would mean an even number divides an odd number—a contradiction. Thanks.
$endgroup$
– thisisourconcerndude
Jan 3 at 2:00
add a comment |
$begingroup$
Here is what I imagine is a very simple question regarding divisibility, which is nevertheless giving me quite a bit of trouble (as is usually the case).
Let $a_1, a_2, ..., a_n$ be positive integers.
Prove or disprove: If the least common multiple of $a_1, a_2, ..., a_n$ is odd, then none of the $a_i$, $i = 1, ..., n$, are even.
elementary-number-theory
asked Jan 3 at 1:45
thisisourconcerndudethisisourconcerndude
1,0961022
$endgroup$
Here is what I imagine is a very simple question regarding divisibility, which is nevertheless giving me quite a bit of trouble (as is usually the case).
Let $a_1, a_2, ..., a_n$ be positive integers.
Prove or disprove: If the least common multiple of $a_1, a_2, ..., a_n$ is odd, then none of the $a_i$, $i = 1, ..., n$, are even.
elementary-number-theory
elementary-number-theory
asked Jan 3 at 1:45
thisisourconcerndudethisisourconcerndude
1,0961022
asked Jan 3 at 1:45
thisisourconcerndudethisisourconcerndude
1,0961022
asked Jan 3 at 1:45
thisisourconcerndudethisisourconcerndude
1,0961022
asked Jan 3 at 1:45
thisisourconcerndudethisisourconcerndude
1,0961022
asked Jan 3 at 1:45
thisisourconcerndudethisisourconcerndude
1,0961022
1,0961022
5
$begingroup$
Hint. Each $a_i$ divides the least common multiple (that's what "common multiple" means"). What happens if one of the $a_i$ is even?
$endgroup$
– Ethan Bolker
Jan 3 at 1:49
1
$begingroup$
@EthanBolker Yes, of course. That would mean an even number divides an odd number—a contradiction. Thanks.
$endgroup$
– thisisourconcerndude
Jan 3 at 2:00
add a comment |
5
$begingroup$
Hint. Each $a_i$ divides the least common multiple (that's what "common multiple" means"). What happens if one of the $a_i$ is even?
$endgroup$
– Ethan Bolker
Jan 3 at 1:49
1
$begingroup$
@EthanBolker Yes, of course. That would mean an even number divides an odd number—a contradiction. Thanks.
$endgroup$
– thisisourconcerndude
Jan 3 at 2:00
5
5
$begingroup$
Hint. Each $a_i$ divides the least common multiple (that's what "common multiple" means"). What happens if one of the $a_i$ is even?
$endgroup$
– Ethan Bolker
Jan 3 at 1:49
$begingroup$
Hint. Each $a_i$ divides the least common multiple (that's what "common multiple" means"). What happens if one of the $a_i$ is even?
$endgroup$
– Ethan Bolker
Jan 3 at 1:49
1
1
$begingroup$
@EthanBolker Yes, of course. That would mean an even number divides an odd number—a contradiction. Thanks.
$endgroup$
– thisisourconcerndude
Jan 3 at 2:00
$begingroup$
@EthanBolker Yes, of course. That would mean an even number divides an odd number—a contradiction. Thanks.
$endgroup$
– thisisourconcerndude
Jan 3 at 2:00
add a comment |
1 Answer
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Just to formalize the answer implicit in the comments, if $M$ is the least common multiple of $a_1,a_2,cdots ,a_n$, then $pmid a_i Rightarrow pmid M$ and $pmid M iff p|a_i$ for some $i$. In particular, this holds for $p=2$. Thus $2notmid M Rightarrow 2not mid a_i$ for any $i$.
answered Jan 3 at 2:13
Keith BackmanKeith Backman
1,1141712
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1 Answer
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$begingroup$
Just to formalize the answer implicit in the comments, if $M$ is the least common multiple of $a_1,a_2,cdots ,a_n$, then $pmid a_i Rightarrow pmid M$ and $pmid M iff p|a_i$ for some $i$. In particular, this holds for $p=2$. Thus $2notmid M Rightarrow 2not mid a_i$ for any $i$.
answered Jan 3 at 2:13
Keith BackmanKeith Backman
1,1141712
$endgroup$
add a comment |
$begingroup$
Just to formalize the answer implicit in the comments, if $M$ is the least common multiple of $a_1,a_2,cdots ,a_n$, then $pmid a_i Rightarrow pmid M$ and $pmid M iff p|a_i$ for some $i$. In particular, this holds for $p=2$. Thus $2notmid M Rightarrow 2not mid a_i$ for any $i$.
answered Jan 3 at 2:13
Keith BackmanKeith Backman
1,1141712
$endgroup$
add a comment |
$begingroup$
Just to formalize the answer implicit in the comments, if $M$ is the least common multiple of $a_1,a_2,cdots ,a_n$, then $pmid a_i Rightarrow pmid M$ and $pmid M iff p|a_i$ for some $i$. In particular, this holds for $p=2$. Thus $2notmid M Rightarrow 2not mid a_i$ for any $i$.
answered Jan 3 at 2:13
Keith BackmanKeith Backman
1,1141712
$endgroup$
Just to formalize the answer implicit in the comments, if $M$ is the least common multiple of $a_1,a_2,cdots ,a_n$, then $pmid a_i Rightarrow pmid M$ and $pmid M iff p|a_i$ for some $i$. In particular, this holds for $p=2$. Thus $2notmid M Rightarrow 2not mid a_i$ for any $i$.
answered Jan 3 at 2:13
Keith BackmanKeith Backman
1,1141712
answered Jan 3 at 2:13
Keith BackmanKeith Backman
1,1141712
answered Jan 3 at 2:13
Keith BackmanKeith Backman
1,1141712
answered Jan 3 at 2:13
Keith BackmanKeith Backman
1,1141712
1,1141712
add a comment |
add a comment |
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5
$begingroup$
Hint. Each $a_i$ divides the least common multiple (that's what "common multiple" means"). What happens if one of the $a_i$ is even?
$endgroup$
– Ethan Bolker
Jan 3 at 1:49
1
$begingroup$
@EthanBolker Yes, of course. That would mean an even number divides an odd number—a contradiction. Thanks.
$endgroup$
– thisisourconcerndude
Jan 3 at 2:00