Mixing
How do I solve this problem involving the LCM of 200 numbers?
$begingroup$
Evaluate $x$ if:
$$xcdotoperatorname{lcm}{(102ldots 200)}=operatorname{lcm}{(1,2,ldots 200)}$$
Here's what I have so far,
LEMMA 1: In any set of $n$ consecutive positive integers, there must be atleast one number divisible by $n$.
LEMMA 2:
$operatorname{lcm}{(a_1,a_2ldots)}=operatorname{lcm}{(operatorname{lcm}{(a_1,a_2)},a_3ldots)}$
LEMMA 3:If $amid b$ then,
$operatorname{lcm}{(a,b)}=b$.
Let
$$A={1,2 ldots 200}$$
$$B={102,103ldots 200}$$
Now, $B$ contains 99 integers.
So, there must be subsets of $B$ with $k$ consecutive integers for all $1leq kleq 99$.
So for each such $k$, using Lemma 1, there is a
$$l: kmid l$$
Therefore, using Lemma 2:
$$operatorname{lcm}{(A)}=operatorname{lcm}{(A-{k,l},operatorname{lcm}{(k,l)})}$$
Now, using Lemma 3,
$$operatorname{lcm}{(A)}=operatorname{lcm}{(A-{k,l},l)}=operatorname{lcm}{(A-{k})}$$
So, doing this with all the $k$, we can conclude that,
$$operatorname{lcm}{(1,2ldots 200)}=operatorname{lcm}{(100,101,102ldots 200)}$$
Trivially, we can remove the 100 as 200 is divisible by it.
So, the original equation becomes:
$$ x=dfrac{operatorname{lcm}{(101,102ldots 200)}}{operatorname{lcm}{(102,103ldots 200)}}$$
Thus, I conclude $boxed{x=101}$.
Is this proof correct?
(Any proof writing tips are also appreciated. I have no experience writing number theoretic proofs)
elementary-number-theory least-common-multiple
asked May 27 '17 at 15:57
ArchimedesprincipleArchimedesprinciple
34418
$endgroup$
add a comment |
$begingroup$
Evaluate $x$ if:
$$xcdotoperatorname{lcm}{(102ldots 200)}=operatorname{lcm}{(1,2,ldots 200)}$$
Here's what I have so far,
LEMMA 1: In any set of $n$ consecutive positive integers, there must be atleast one number divisible by $n$.
LEMMA 2:
$operatorname{lcm}{(a_1,a_2ldots)}=operatorname{lcm}{(operatorname{lcm}{(a_1,a_2)},a_3ldots)}$
LEMMA 3:If $amid b$ then,
$operatorname{lcm}{(a,b)}=b$.
Let
$$A={1,2 ldots 200}$$
$$B={102,103ldots 200}$$
Now, $B$ contains 99 integers.
So, there must be subsets of $B$ with $k$ consecutive integers for all $1leq kleq 99$.
So for each such $k$, using Lemma 1, there is a
$$l: kmid l$$
Therefore, using Lemma 2:
$$operatorname{lcm}{(A)}=operatorname{lcm}{(A-{k,l},operatorname{lcm}{(k,l)})}$$
Now, using Lemma 3,
$$operatorname{lcm}{(A)}=operatorname{lcm}{(A-{k,l},l)}=operatorname{lcm}{(A-{k})}$$
So, doing this with all the $k$, we can conclude that,
$$operatorname{lcm}{(1,2ldots 200)}=operatorname{lcm}{(100,101,102ldots 200)}$$
Trivially, we can remove the 100 as 200 is divisible by it.
So, the original equation becomes:
$$ x=dfrac{operatorname{lcm}{(101,102ldots 200)}}{operatorname{lcm}{(102,103ldots 200)}}$$
Thus, I conclude $boxed{x=101}$.
Is this proof correct?
(Any proof writing tips are also appreciated. I have no experience writing number theoretic proofs)
elementary-number-theory least-common-multiple
asked May 27 '17 at 15:57
ArchimedesprincipleArchimedesprinciple
34418
$endgroup$
$begingroup$
Looks good to me. The only minor thing is that most people prefer to use setminus ($Asetminus B$) to - ($A-B$) for the set difference in $LaTeX$.
$endgroup$
– fedja
May 27 '17 at 16:17
add a comment |
$begingroup$
Evaluate $x$ if:
$$xcdotoperatorname{lcm}{(102ldots 200)}=operatorname{lcm}{(1,2,ldots 200)}$$
Here's what I have so far,
LEMMA 1: In any set of $n$ consecutive positive integers, there must be atleast one number divisible by $n$.
LEMMA 2:
$operatorname{lcm}{(a_1,a_2ldots)}=operatorname{lcm}{(operatorname{lcm}{(a_1,a_2)},a_3ldots)}$
LEMMA 3:If $amid b$ then,
$operatorname{lcm}{(a,b)}=b$.
Let
$$A={1,2 ldots 200}$$
$$B={102,103ldots 200}$$
Now, $B$ contains 99 integers.
So, there must be subsets of $B$ with $k$ consecutive integers for all $1leq kleq 99$.
So for each such $k$, using Lemma 1, there is a
$$l: kmid l$$
Therefore, using Lemma 2:
$$operatorname{lcm}{(A)}=operatorname{lcm}{(A-{k,l},operatorname{lcm}{(k,l)})}$$
Now, using Lemma 3,
$$operatorname{lcm}{(A)}=operatorname{lcm}{(A-{k,l},l)}=operatorname{lcm}{(A-{k})}$$
So, doing this with all the $k$, we can conclude that,
$$operatorname{lcm}{(1,2ldots 200)}=operatorname{lcm}{(100,101,102ldots 200)}$$
Trivially, we can remove the 100 as 200 is divisible by it.
So, the original equation becomes:
$$ x=dfrac{operatorname{lcm}{(101,102ldots 200)}}{operatorname{lcm}{(102,103ldots 200)}}$$
Thus, I conclude $boxed{x=101}$.
Is this proof correct?
(Any proof writing tips are also appreciated. I have no experience writing number theoretic proofs)
elementary-number-theory least-common-multiple
asked May 27 '17 at 15:57
ArchimedesprincipleArchimedesprinciple
34418
$endgroup$
Evaluate $x$ if:
$$xcdotoperatorname{lcm}{(102ldots 200)}=operatorname{lcm}{(1,2,ldots 200)}$$
Here's what I have so far,
LEMMA 1: In any set of $n$ consecutive positive integers, there must be atleast one number divisible by $n$.
LEMMA 2:
$operatorname{lcm}{(a_1,a_2ldots)}=operatorname{lcm}{(operatorname{lcm}{(a_1,a_2)},a_3ldots)}$
LEMMA 3:If $amid b$ then,
$operatorname{lcm}{(a,b)}=b$.
Let
$$A={1,2 ldots 200}$$
$$B={102,103ldots 200}$$
Now, $B$ contains 99 integers.
So, there must be subsets of $B$ with $k$ consecutive integers for all $1leq kleq 99$.
So for each such $k$, using Lemma 1, there is a
$$l: kmid l$$
Therefore, using Lemma 2:
$$operatorname{lcm}{(A)}=operatorname{lcm}{(A-{k,l},operatorname{lcm}{(k,l)})}$$
Now, using Lemma 3,
$$operatorname{lcm}{(A)}=operatorname{lcm}{(A-{k,l},l)}=operatorname{lcm}{(A-{k})}$$
So, doing this with all the $k$, we can conclude that,
$$operatorname{lcm}{(1,2ldots 200)}=operatorname{lcm}{(100,101,102ldots 200)}$$
Trivially, we can remove the 100 as 200 is divisible by it.
So, the original equation becomes:
$$ x=dfrac{operatorname{lcm}{(101,102ldots 200)}}{operatorname{lcm}{(102,103ldots 200)}}$$
Thus, I conclude $boxed{x=101}$.
Is this proof correct?
(Any proof writing tips are also appreciated. I have no experience writing number theoretic proofs)
elementary-number-theory least-common-multiple
elementary-number-theory least-common-multiple
asked May 27 '17 at 15:57
ArchimedesprincipleArchimedesprinciple
34418
asked May 27 '17 at 15:57
ArchimedesprincipleArchimedesprinciple
34418
edited May 27 '17 at 16:08
Archimedesprinciple
asked May 27 '17 at 15:57
ArchimedesprincipleArchimedesprinciple
34418
asked May 27 '17 at 15:57
ArchimedesprincipleArchimedesprinciple
34418
asked May 27 '17 at 15:57
ArchimedesprincipleArchimedesprinciple
34418
34418
$begingroup$
Looks good to me. The only minor thing is that most people prefer to use setminus ($Asetminus B$) to - ($A-B$) for the set difference in $LaTeX$.
$endgroup$
– fedja
May 27 '17 at 16:17
add a comment |
$begingroup$
Looks good to me. The only minor thing is that most people prefer to use setminus ($Asetminus B$) to - ($A-B$) for the set difference in $LaTeX$.
$endgroup$
– fedja
May 27 '17 at 16:17
$begingroup$
Looks good to me. The only minor thing is that most people prefer to use setminus ($Asetminus B$) to - ($A-B$) for the set difference in $LaTeX$.
$endgroup$
– fedja
May 27 '17 at 16:17
$begingroup$
Looks good to me. The only minor thing is that most people prefer to use setminus ($Asetminus B$) to - ($A-B$) for the set difference in $LaTeX$.
$endgroup$
– fedja
May 27 '17 at 16:17
add a comment |
1 Answer
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$begingroup$
This is correct. It is easier to note that all the numbers from $1$ to $100$ have a multiple in the range $102-200$ so they don't contribute to the LCM
answered May 27 '17 at 17:56
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
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$begingroup$
This is correct. It is easier to note that all the numbers from $1$ to $100$ have a multiple in the range $102-200$ so they don't contribute to the LCM
answered May 27 '17 at 17:56
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
This is correct. It is easier to note that all the numbers from $1$ to $100$ have a multiple in the range $102-200$ so they don't contribute to the LCM
answered May 27 '17 at 17:56
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
This is correct. It is easier to note that all the numbers from $1$ to $100$ have a multiple in the range $102-200$ so they don't contribute to the LCM
answered May 27 '17 at 17:56
Ross MillikanRoss Millikan
301k24200375
$endgroup$
This is correct. It is easier to note that all the numbers from $1$ to $100$ have a multiple in the range $102-200$ so they don't contribute to the LCM
answered May 27 '17 at 17:56
Ross MillikanRoss Millikan
301k24200375
answered May 27 '17 at 17:56
Ross MillikanRoss Millikan
301k24200375
answered May 27 '17 at 17:56
Ross MillikanRoss Millikan
301k24200375
answered May 27 '17 at 17:56
Ross MillikanRoss Millikan
301k24200375
301k24200375
add a comment |
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$begingroup$
Looks good to me. The only minor thing is that most people prefer to use setminus ($Asetminus B$) to - ($A-B$) for the set difference in $LaTeX$.
$endgroup$
– fedja
May 27 '17 at 16:17