Finding largest 4 digit number divisible by certain numbers.












2












$begingroup$


Okay I am weak in mathematics, I have no shame in saying that.



I am preparing for my internals and the concept of the GCD and LCM is quite confusing to me.



There is this question which is:



Find the largest 4 digit number that is exactly divisible by 12,15,18 and 27.



The solution is by finding out the LCM of these numbers. I mean why ?



Why is it not GCD? What is the difference between GCD and LCM?



I would really appreciate someone expalining it thoroughly. Please try to help me with this.



Thank you.






greatest-common-divisor least-common-multiple







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Any integer divisible by those four is divisible by their LCM, so that's a useful concept here. But you don't need to follow that route. You could just work from the prime factorization. Since $(12,15,18,27)=(2^2times 3, 3times 5, 2times 3^2,3^3)$ you know that any number divisible by each is divisible by $2^2times 3^3times 5=540$ (this is the LCM, by the way).
    $endgroup$
    – lulu
    Sep 2 '17 at 21:05






  • 1




    $begingroup$
    then you find the largest four-digit number that is a multiple of 540.
    $endgroup$
    – steven gregory
    Sep 2 '17 at 21:07










  • $begingroup$
    But why dont we find the GCD ? I dont understand the concept
    $endgroup$
    – nban
    Sep 2 '17 at 21:08












  • $begingroup$
    Hint $ ,a,b,c,dmid niff {rm lcm}(a,b,c,d)mid n $ That is the "concept" of LCM.
    $endgroup$
    – Bill Dubuque
    Sep 2 '17 at 21:21








  • 2




    $begingroup$
    The greatest common denominator here is $3$. Knowing that something is divisible by $3$ however doesn't tell us anything about whether or not it is divisible by any or all of $12,15,18,27$. Knowing that something is divisible by $540$ on the other hand does tell us that it is divisible by all of $12,15,18,27$.
    $endgroup$
    – JMoravitz
    Sep 2 '17 at 22:09
















2












$begingroup$


Okay I am weak in mathematics, I have no shame in saying that.



I am preparing for my internals and the concept of the GCD and LCM is quite confusing to me.



There is this question which is:



Find the largest 4 digit number that is exactly divisible by 12,15,18 and 27.



The solution is by finding out the LCM of these numbers. I mean why ?



Why is it not GCD? What is the difference between GCD and LCM?



I would really appreciate someone expalining it thoroughly. Please try to help me with this.



Thank you.






greatest-common-divisor least-common-multiple







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Any integer divisible by those four is divisible by their LCM, so that's a useful concept here. But you don't need to follow that route. You could just work from the prime factorization. Since $(12,15,18,27)=(2^2times 3, 3times 5, 2times 3^2,3^3)$ you know that any number divisible by each is divisible by $2^2times 3^3times 5=540$ (this is the LCM, by the way).
    $endgroup$
    – lulu
    Sep 2 '17 at 21:05






  • 1




    $begingroup$
    then you find the largest four-digit number that is a multiple of 540.
    $endgroup$
    – steven gregory
    Sep 2 '17 at 21:07










  • $begingroup$
    But why dont we find the GCD ? I dont understand the concept
    $endgroup$
    – nban
    Sep 2 '17 at 21:08












  • $begingroup$
    Hint $ ,a,b,c,dmid niff {rm lcm}(a,b,c,d)mid n $ That is the "concept" of LCM.
    $endgroup$
    – Bill Dubuque
    Sep 2 '17 at 21:21








  • 2




    $begingroup$
    The greatest common denominator here is $3$. Knowing that something is divisible by $3$ however doesn't tell us anything about whether or not it is divisible by any or all of $12,15,18,27$. Knowing that something is divisible by $540$ on the other hand does tell us that it is divisible by all of $12,15,18,27$.
    $endgroup$
    – JMoravitz
    Sep 2 '17 at 22:09














2












2








2





$begingroup$


Okay I am weak in mathematics, I have no shame in saying that.



I am preparing for my internals and the concept of the GCD and LCM is quite confusing to me.



There is this question which is:



Find the largest 4 digit number that is exactly divisible by 12,15,18 and 27.



The solution is by finding out the LCM of these numbers. I mean why ?



Why is it not GCD? What is the difference between GCD and LCM?



I would really appreciate someone expalining it thoroughly. Please try to help me with this.



Thank you.






greatest-common-divisor least-common-multiple







share|cite|improve this question









$endgroup$




Okay I am weak in mathematics, I have no shame in saying that.



I am preparing for my internals and the concept of the GCD and LCM is quite confusing to me.



There is this question which is:



Find the largest 4 digit number that is exactly divisible by 12,15,18 and 27.



The solution is by finding out the LCM of these numbers. I mean why ?



Why is it not GCD? What is the difference between GCD and LCM?



I would really appreciate someone expalining it thoroughly. Please try to help me with this.



Thank you.






greatest-common-divisor least-common-multiple




greatest-common-divisor least-common-multiple






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 2 '17 at 21:01









nbannban

194




194








  • 2




    $begingroup$
    Any integer divisible by those four is divisible by their LCM, so that's a useful concept here. But you don't need to follow that route. You could just work from the prime factorization. Since $(12,15,18,27)=(2^2times 3, 3times 5, 2times 3^2,3^3)$ you know that any number divisible by each is divisible by $2^2times 3^3times 5=540$ (this is the LCM, by the way).
    $endgroup$
    – lulu
    Sep 2 '17 at 21:05






  • 1




    $begingroup$
    then you find the largest four-digit number that is a multiple of 540.
    $endgroup$
    – steven gregory
    Sep 2 '17 at 21:07










  • $begingroup$
    But why dont we find the GCD ? I dont understand the concept
    $endgroup$
    – nban
    Sep 2 '17 at 21:08












  • $begingroup$
    Hint $ ,a,b,c,dmid niff {rm lcm}(a,b,c,d)mid n $ That is the "concept" of LCM.
    $endgroup$
    – Bill Dubuque
    Sep 2 '17 at 21:21








  • 2




    $begingroup$
    The greatest common denominator here is $3$. Knowing that something is divisible by $3$ however doesn't tell us anything about whether or not it is divisible by any or all of $12,15,18,27$. Knowing that something is divisible by $540$ on the other hand does tell us that it is divisible by all of $12,15,18,27$.
    $endgroup$
    – JMoravitz
    Sep 2 '17 at 22:09














  • 2




    $begingroup$
    Any integer divisible by those four is divisible by their LCM, so that's a useful concept here. But you don't need to follow that route. You could just work from the prime factorization. Since $(12,15,18,27)=(2^2times 3, 3times 5, 2times 3^2,3^3)$ you know that any number divisible by each is divisible by $2^2times 3^3times 5=540$ (this is the LCM, by the way).
    $endgroup$
    – lulu
    Sep 2 '17 at 21:05






  • 1




    $begingroup$
    then you find the largest four-digit number that is a multiple of 540.
    $endgroup$
    – steven gregory
    Sep 2 '17 at 21:07










  • $begingroup$
    But why dont we find the GCD ? I dont understand the concept
    $endgroup$
    – nban
    Sep 2 '17 at 21:08












  • $begingroup$
    Hint $ ,a,b,c,dmid niff {rm lcm}(a,b,c,d)mid n $ That is the "concept" of LCM.
    $endgroup$
    – Bill Dubuque
    Sep 2 '17 at 21:21








  • 2




    $begingroup$
    The greatest common denominator here is $3$. Knowing that something is divisible by $3$ however doesn't tell us anything about whether or not it is divisible by any or all of $12,15,18,27$. Knowing that something is divisible by $540$ on the other hand does tell us that it is divisible by all of $12,15,18,27$.
    $endgroup$
    – JMoravitz
    Sep 2 '17 at 22:09








2




2




$begingroup$
Any integer divisible by those four is divisible by their LCM, so that's a useful concept here. But you don't need to follow that route. You could just work from the prime factorization. Since $(12,15,18,27)=(2^2times 3, 3times 5, 2times 3^2,3^3)$ you know that any number divisible by each is divisible by $2^2times 3^3times 5=540$ (this is the LCM, by the way).
$endgroup$
– lulu
Sep 2 '17 at 21:05




$begingroup$
Any integer divisible by those four is divisible by their LCM, so that's a useful concept here. But you don't need to follow that route. You could just work from the prime factorization. Since $(12,15,18,27)=(2^2times 3, 3times 5, 2times 3^2,3^3)$ you know that any number divisible by each is divisible by $2^2times 3^3times 5=540$ (this is the LCM, by the way).
$endgroup$
– lulu
Sep 2 '17 at 21:05




1




1




$begingroup$
then you find the largest four-digit number that is a multiple of 540.
$endgroup$
– steven gregory
Sep 2 '17 at 21:07




$begingroup$
then you find the largest four-digit number that is a multiple of 540.
$endgroup$
– steven gregory
Sep 2 '17 at 21:07












$begingroup$
But why dont we find the GCD ? I dont understand the concept
$endgroup$
– nban
Sep 2 '17 at 21:08






$begingroup$
But why dont we find the GCD ? I dont understand the concept
$endgroup$
– nban
Sep 2 '17 at 21:08














$begingroup$
Hint $ ,a,b,c,dmid niff {rm lcm}(a,b,c,d)mid n $ That is the "concept" of LCM.
$endgroup$
– Bill Dubuque
Sep 2 '17 at 21:21






$begingroup$
Hint $ ,a,b,c,dmid niff {rm lcm}(a,b,c,d)mid n $ That is the "concept" of LCM.
$endgroup$
– Bill Dubuque
Sep 2 '17 at 21:21






2




2




$begingroup$
The greatest common denominator here is $3$. Knowing that something is divisible by $3$ however doesn't tell us anything about whether or not it is divisible by any or all of $12,15,18,27$. Knowing that something is divisible by $540$ on the other hand does tell us that it is divisible by all of $12,15,18,27$.
$endgroup$
– JMoravitz
Sep 2 '17 at 22:09




$begingroup$
The greatest common denominator here is $3$. Knowing that something is divisible by $3$ however doesn't tell us anything about whether or not it is divisible by any or all of $12,15,18,27$. Knowing that something is divisible by $540$ on the other hand does tell us that it is divisible by all of $12,15,18,27$.
$endgroup$
– JMoravitz
Sep 2 '17 at 22:09










2 Answers
2






active

oldest

votes


















0












$begingroup$

You want a number that is divisible by all of $( 12,15,18, 27)$, which means that your target number $N$ must be a multiple of each of those.



In order to be a multiple of all of those, it will also need to be a multiple of their least common multiple (which is perhaps less obvious). Basically the multiples align at zero, then align again each time another "LCM" has been accumulated. For this case, we need the greatest multiple of the LCM which is still less than $10000$, which is not hard once you have determined the LCM.



The greatest common divisor is less than (or equal to) all the numbers being considered, and to find other numbers that divide every one of a set of numbers you would need the divisors of the GCD.



The LCM and GCD are related when you are just referring to two numbers, $a,b$, by $text{lcm}(a,b)cdot gcd(a,b) = ab$ - all the factors of the numbers which are excluded from the GCD must be included in the LCM.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    So can I say that when I need a number that is divisible by a set of numbers I need to find LCM, and when I want to divide all the number in that set I need to find GCD ?
    $endgroup$
    – nban
    Sep 2 '17 at 21:27






  • 1




    $begingroup$
    @nban More precisely, this is what you should say (e.g. for a set of three integers) $$begin{align} a,b,cmid n&iff {rm lcm}(a,b,c)mid n\[.2em] nmid a,b,c &iff nmid gcd(a,b,c) end{align}$$ Those are actually the definition of lcm and gcd for more general number systems.
    $endgroup$
    – Bill Dubuque
    Sep 2 '17 at 21:29












  • $begingroup$
    So this is like, find a number that is divisible by three numbers (LCM) and find a number that divides all three numbers (GCD) ?
    $endgroup$
    – nban
    Sep 2 '17 at 21:35












  • $begingroup$
    @nban Compare to the following universal definitions of union and intersection of sets $$begin{align} A,B,C subset S &iff Acup Bcup C subset S\ S subset A,B,C &iff Ssubset Acap Bcap Cend{align}$$
    $endgroup$
    – Bill Dubuque
    Sep 2 '17 at 21:37








  • 1




    $begingroup$
    @nban Let's consider the above analogy with sets. The set union allows us to replace many common containments $,Asubset S, Bsubset S, Csubset S,$ by one equivalent containment $ Acup Bcup Csubset S.,$ Similarly, LCM replaces many common multiples $, amid n, bmid n, cmid n,$ by one equivalent multiple $,{rm lcm}(a,b,c)mid n,$ statement. You should commit the above equvalences to memory since they are ubiquitous. Then you can easily apply them as rewrite rules whenever need be.
    $endgroup$
    – Bill Dubuque
    Sep 2 '17 at 22:02





















0












$begingroup$

"Find the largest $4$ digit number that is exactly divisible by $12,15,18$ and $27$."



If all those number divide into that 4-digit number than that 4-digit number is a multiple of all those numbers.



So the 4-digit number is a common multiple of all of those.



"The solution is by finding out the LCM of these numbers. I mean why ?"



Well, we don't have to find the least common multiple but we have to find a common multiple. And if we can find the least common multiple that will give us all the infinite number of common multiples by multiply the least common multiple by every possible positive integer.



The question is asking us to find the greatest common multiple that is less than or equal to $9999$. We do that be finding the least common multiple and adding that and adding that until we get the largest possible value less than or equal to $9999$.



"Why is it not GCD? What is the difference between GCD and LCM?"



Because GCD is the "greatest common divisor". It is a number (in this case $3$) that divides into each of the $12,$15,$18$ and $27$. The least common multiple is a number that all the numbers that $12$, $15$, $18$, and $27$ themselves divide into; not what divides into them.



For example $12 $ divides into $24$, $36$, $48$ .etc. But $15$ divides into $30$, $45$ and $60$. The smallest number that both $12$ and $15$ divide into is $60$. But $18$ and $27$ do not divide into $60$. The all divide into $12*15*18*27 = 87480$ that that probably isn't the smallest number they all divide into. Do you know what is. (And if you can find the largest common multiple that is four digits, you have solved the question.)



On the other hand the GCD isn't a number they divide into; it is a number that divides into *them. $6$ divides into $12$ and $18$ but not $15$ or $27$. $9$ divides into $18$ and $27$ but not $12$ and $15$. But $3$ divides into all $4$. Is it the greatest common divisor? Well, yes it is.



====



Remember $12,15$



$n$= LCM = least common MULTIPLE. $12$ and $15$ go into $n$. So it is $60$ that they go into. $12$ and $15$ go into $60$.



$m$ = GCD = greatest common DIVISOR. $m$ goes into $12$ and $15$. So it is $3$ that goes into them. $3$ goes into $12$ and $15$.






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2414676%2ffinding-largest-4-digit-number-divisible-by-certain-numbers%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    You want a number that is divisible by all of $( 12,15,18, 27)$, which means that your target number $N$ must be a multiple of each of those.



    In order to be a multiple of all of those, it will also need to be a multiple of their least common multiple (which is perhaps less obvious). Basically the multiples align at zero, then align again each time another "LCM" has been accumulated. For this case, we need the greatest multiple of the LCM which is still less than $10000$, which is not hard once you have determined the LCM.



    The greatest common divisor is less than (or equal to) all the numbers being considered, and to find other numbers that divide every one of a set of numbers you would need the divisors of the GCD.



    The LCM and GCD are related when you are just referring to two numbers, $a,b$, by $text{lcm}(a,b)cdot gcd(a,b) = ab$ - all the factors of the numbers which are excluded from the GCD must be included in the LCM.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      So can I say that when I need a number that is divisible by a set of numbers I need to find LCM, and when I want to divide all the number in that set I need to find GCD ?
      $endgroup$
      – nban
      Sep 2 '17 at 21:27






    • 1




      $begingroup$
      @nban More precisely, this is what you should say (e.g. for a set of three integers) $$begin{align} a,b,cmid n&iff {rm lcm}(a,b,c)mid n\[.2em] nmid a,b,c &iff nmid gcd(a,b,c) end{align}$$ Those are actually the definition of lcm and gcd for more general number systems.
      $endgroup$
      – Bill Dubuque
      Sep 2 '17 at 21:29












    • $begingroup$
      So this is like, find a number that is divisible by three numbers (LCM) and find a number that divides all three numbers (GCD) ?
      $endgroup$
      – nban
      Sep 2 '17 at 21:35












    • $begingroup$
      @nban Compare to the following universal definitions of union and intersection of sets $$begin{align} A,B,C subset S &iff Acup Bcup C subset S\ S subset A,B,C &iff Ssubset Acap Bcap Cend{align}$$
      $endgroup$
      – Bill Dubuque
      Sep 2 '17 at 21:37








    • 1




      $begingroup$
      @nban Let's consider the above analogy with sets. The set union allows us to replace many common containments $,Asubset S, Bsubset S, Csubset S,$ by one equivalent containment $ Acup Bcup Csubset S.,$ Similarly, LCM replaces many common multiples $, amid n, bmid n, cmid n,$ by one equivalent multiple $,{rm lcm}(a,b,c)mid n,$ statement. You should commit the above equvalences to memory since they are ubiquitous. Then you can easily apply them as rewrite rules whenever need be.
      $endgroup$
      – Bill Dubuque
      Sep 2 '17 at 22:02


















    0












    $begingroup$

    You want a number that is divisible by all of $( 12,15,18, 27)$, which means that your target number $N$ must be a multiple of each of those.



    In order to be a multiple of all of those, it will also need to be a multiple of their least common multiple (which is perhaps less obvious). Basically the multiples align at zero, then align again each time another "LCM" has been accumulated. For this case, we need the greatest multiple of the LCM which is still less than $10000$, which is not hard once you have determined the LCM.



    The greatest common divisor is less than (or equal to) all the numbers being considered, and to find other numbers that divide every one of a set of numbers you would need the divisors of the GCD.



    The LCM and GCD are related when you are just referring to two numbers, $a,b$, by $text{lcm}(a,b)cdot gcd(a,b) = ab$ - all the factors of the numbers which are excluded from the GCD must be included in the LCM.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      So can I say that when I need a number that is divisible by a set of numbers I need to find LCM, and when I want to divide all the number in that set I need to find GCD ?
      $endgroup$
      – nban
      Sep 2 '17 at 21:27






    • 1




      $begingroup$
      @nban More precisely, this is what you should say (e.g. for a set of three integers) $$begin{align} a,b,cmid n&iff {rm lcm}(a,b,c)mid n\[.2em] nmid a,b,c &iff nmid gcd(a,b,c) end{align}$$ Those are actually the definition of lcm and gcd for more general number systems.
      $endgroup$
      – Bill Dubuque
      Sep 2 '17 at 21:29












    • $begingroup$
      So this is like, find a number that is divisible by three numbers (LCM) and find a number that divides all three numbers (GCD) ?
      $endgroup$
      – nban
      Sep 2 '17 at 21:35












    • $begingroup$
      @nban Compare to the following universal definitions of union and intersection of sets $$begin{align} A,B,C subset S &iff Acup Bcup C subset S\ S subset A,B,C &iff Ssubset Acap Bcap Cend{align}$$
      $endgroup$
      – Bill Dubuque
      Sep 2 '17 at 21:37








    • 1




      $begingroup$
      @nban Let's consider the above analogy with sets. The set union allows us to replace many common containments $,Asubset S, Bsubset S, Csubset S,$ by one equivalent containment $ Acup Bcup Csubset S.,$ Similarly, LCM replaces many common multiples $, amid n, bmid n, cmid n,$ by one equivalent multiple $,{rm lcm}(a,b,c)mid n,$ statement. You should commit the above equvalences to memory since they are ubiquitous. Then you can easily apply them as rewrite rules whenever need be.
      $endgroup$
      – Bill Dubuque
      Sep 2 '17 at 22:02
















    0












    0








    0





    $begingroup$

    You want a number that is divisible by all of $( 12,15,18, 27)$, which means that your target number $N$ must be a multiple of each of those.



    In order to be a multiple of all of those, it will also need to be a multiple of their least common multiple (which is perhaps less obvious). Basically the multiples align at zero, then align again each time another "LCM" has been accumulated. For this case, we need the greatest multiple of the LCM which is still less than $10000$, which is not hard once you have determined the LCM.



    The greatest common divisor is less than (or equal to) all the numbers being considered, and to find other numbers that divide every one of a set of numbers you would need the divisors of the GCD.



    The LCM and GCD are related when you are just referring to two numbers, $a,b$, by $text{lcm}(a,b)cdot gcd(a,b) = ab$ - all the factors of the numbers which are excluded from the GCD must be included in the LCM.






    share|cite|improve this answer









    $endgroup$



    You want a number that is divisible by all of $( 12,15,18, 27)$, which means that your target number $N$ must be a multiple of each of those.



    In order to be a multiple of all of those, it will also need to be a multiple of their least common multiple (which is perhaps less obvious). Basically the multiples align at zero, then align again each time another "LCM" has been accumulated. For this case, we need the greatest multiple of the LCM which is still less than $10000$, which is not hard once you have determined the LCM.



    The greatest common divisor is less than (or equal to) all the numbers being considered, and to find other numbers that divide every one of a set of numbers you would need the divisors of the GCD.



    The LCM and GCD are related when you are just referring to two numbers, $a,b$, by $text{lcm}(a,b)cdot gcd(a,b) = ab$ - all the factors of the numbers which are excluded from the GCD must be included in the LCM.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Sep 2 '17 at 21:19









    JoffanJoffan

    32.6k43269




    32.6k43269








    • 1




      $begingroup$
      So can I say that when I need a number that is divisible by a set of numbers I need to find LCM, and when I want to divide all the number in that set I need to find GCD ?
      $endgroup$
      – nban
      Sep 2 '17 at 21:27






    • 1




      $begingroup$
      @nban More precisely, this is what you should say (e.g. for a set of three integers) $$begin{align} a,b,cmid n&iff {rm lcm}(a,b,c)mid n\[.2em] nmid a,b,c &iff nmid gcd(a,b,c) end{align}$$ Those are actually the definition of lcm and gcd for more general number systems.
      $endgroup$
      – Bill Dubuque
      Sep 2 '17 at 21:29












    • $begingroup$
      So this is like, find a number that is divisible by three numbers (LCM) and find a number that divides all three numbers (GCD) ?
      $endgroup$
      – nban
      Sep 2 '17 at 21:35












    • $begingroup$
      @nban Compare to the following universal definitions of union and intersection of sets $$begin{align} A,B,C subset S &iff Acup Bcup C subset S\ S subset A,B,C &iff Ssubset Acap Bcap Cend{align}$$
      $endgroup$
      – Bill Dubuque
      Sep 2 '17 at 21:37








    • 1




      $begingroup$
      @nban Let's consider the above analogy with sets. The set union allows us to replace many common containments $,Asubset S, Bsubset S, Csubset S,$ by one equivalent containment $ Acup Bcup Csubset S.,$ Similarly, LCM replaces many common multiples $, amid n, bmid n, cmid n,$ by one equivalent multiple $,{rm lcm}(a,b,c)mid n,$ statement. You should commit the above equvalences to memory since they are ubiquitous. Then you can easily apply them as rewrite rules whenever need be.
      $endgroup$
      – Bill Dubuque
      Sep 2 '17 at 22:02
















    • 1




      $begingroup$
      So can I say that when I need a number that is divisible by a set of numbers I need to find LCM, and when I want to divide all the number in that set I need to find GCD ?
      $endgroup$
      – nban
      Sep 2 '17 at 21:27






    • 1




      $begingroup$
      @nban More precisely, this is what you should say (e.g. for a set of three integers) $$begin{align} a,b,cmid n&iff {rm lcm}(a,b,c)mid n\[.2em] nmid a,b,c &iff nmid gcd(a,b,c) end{align}$$ Those are actually the definition of lcm and gcd for more general number systems.
      $endgroup$
      – Bill Dubuque
      Sep 2 '17 at 21:29












    • $begingroup$
      So this is like, find a number that is divisible by three numbers (LCM) and find a number that divides all three numbers (GCD) ?
      $endgroup$
      – nban
      Sep 2 '17 at 21:35












    • $begingroup$
      @nban Compare to the following universal definitions of union and intersection of sets $$begin{align} A,B,C subset S &iff Acup Bcup C subset S\ S subset A,B,C &iff Ssubset Acap Bcap Cend{align}$$
      $endgroup$
      – Bill Dubuque
      Sep 2 '17 at 21:37








    • 1




      $begingroup$
      @nban Let's consider the above analogy with sets. The set union allows us to replace many common containments $,Asubset S, Bsubset S, Csubset S,$ by one equivalent containment $ Acup Bcup Csubset S.,$ Similarly, LCM replaces many common multiples $, amid n, bmid n, cmid n,$ by one equivalent multiple $,{rm lcm}(a,b,c)mid n,$ statement. You should commit the above equvalences to memory since they are ubiquitous. Then you can easily apply them as rewrite rules whenever need be.
      $endgroup$
      – Bill Dubuque
      Sep 2 '17 at 22:02










    1




    1




    $begingroup$
    So can I say that when I need a number that is divisible by a set of numbers I need to find LCM, and when I want to divide all the number in that set I need to find GCD ?
    $endgroup$
    – nban
    Sep 2 '17 at 21:27




    $begingroup$
    So can I say that when I need a number that is divisible by a set of numbers I need to find LCM, and when I want to divide all the number in that set I need to find GCD ?
    $endgroup$
    – nban
    Sep 2 '17 at 21:27




    1




    1




    $begingroup$
    @nban More precisely, this is what you should say (e.g. for a set of three integers) $$begin{align} a,b,cmid n&iff {rm lcm}(a,b,c)mid n\[.2em] nmid a,b,c &iff nmid gcd(a,b,c) end{align}$$ Those are actually the definition of lcm and gcd for more general number systems.
    $endgroup$
    – Bill Dubuque
    Sep 2 '17 at 21:29






    $begingroup$
    @nban More precisely, this is what you should say (e.g. for a set of three integers) $$begin{align} a,b,cmid n&iff {rm lcm}(a,b,c)mid n\[.2em] nmid a,b,c &iff nmid gcd(a,b,c) end{align}$$ Those are actually the definition of lcm and gcd for more general number systems.
    $endgroup$
    – Bill Dubuque
    Sep 2 '17 at 21:29














    $begingroup$
    So this is like, find a number that is divisible by three numbers (LCM) and find a number that divides all three numbers (GCD) ?
    $endgroup$
    – nban
    Sep 2 '17 at 21:35






    $begingroup$
    So this is like, find a number that is divisible by three numbers (LCM) and find a number that divides all three numbers (GCD) ?
    $endgroup$
    – nban
    Sep 2 '17 at 21:35














    $begingroup$
    @nban Compare to the following universal definitions of union and intersection of sets $$begin{align} A,B,C subset S &iff Acup Bcup C subset S\ S subset A,B,C &iff Ssubset Acap Bcap Cend{align}$$
    $endgroup$
    – Bill Dubuque
    Sep 2 '17 at 21:37






    $begingroup$
    @nban Compare to the following universal definitions of union and intersection of sets $$begin{align} A,B,C subset S &iff Acup Bcup C subset S\ S subset A,B,C &iff Ssubset Acap Bcap Cend{align}$$
    $endgroup$
    – Bill Dubuque
    Sep 2 '17 at 21:37






    1




    1




    $begingroup$
    @nban Let's consider the above analogy with sets. The set union allows us to replace many common containments $,Asubset S, Bsubset S, Csubset S,$ by one equivalent containment $ Acup Bcup Csubset S.,$ Similarly, LCM replaces many common multiples $, amid n, bmid n, cmid n,$ by one equivalent multiple $,{rm lcm}(a,b,c)mid n,$ statement. You should commit the above equvalences to memory since they are ubiquitous. Then you can easily apply them as rewrite rules whenever need be.
    $endgroup$
    – Bill Dubuque
    Sep 2 '17 at 22:02






    $begingroup$
    @nban Let's consider the above analogy with sets. The set union allows us to replace many common containments $,Asubset S, Bsubset S, Csubset S,$ by one equivalent containment $ Acup Bcup Csubset S.,$ Similarly, LCM replaces many common multiples $, amid n, bmid n, cmid n,$ by one equivalent multiple $,{rm lcm}(a,b,c)mid n,$ statement. You should commit the above equvalences to memory since they are ubiquitous. Then you can easily apply them as rewrite rules whenever need be.
    $endgroup$
    – Bill Dubuque
    Sep 2 '17 at 22:02













    0












    $begingroup$

    "Find the largest $4$ digit number that is exactly divisible by $12,15,18$ and $27$."



    If all those number divide into that 4-digit number than that 4-digit number is a multiple of all those numbers.



    So the 4-digit number is a common multiple of all of those.



    "The solution is by finding out the LCM of these numbers. I mean why ?"



    Well, we don't have to find the least common multiple but we have to find a common multiple. And if we can find the least common multiple that will give us all the infinite number of common multiples by multiply the least common multiple by every possible positive integer.



    The question is asking us to find the greatest common multiple that is less than or equal to $9999$. We do that be finding the least common multiple and adding that and adding that until we get the largest possible value less than or equal to $9999$.



    "Why is it not GCD? What is the difference between GCD and LCM?"



    Because GCD is the "greatest common divisor". It is a number (in this case $3$) that divides into each of the $12,$15,$18$ and $27$. The least common multiple is a number that all the numbers that $12$, $15$, $18$, and $27$ themselves divide into; not what divides into them.



    For example $12 $ divides into $24$, $36$, $48$ .etc. But $15$ divides into $30$, $45$ and $60$. The smallest number that both $12$ and $15$ divide into is $60$. But $18$ and $27$ do not divide into $60$. The all divide into $12*15*18*27 = 87480$ that that probably isn't the smallest number they all divide into. Do you know what is. (And if you can find the largest common multiple that is four digits, you have solved the question.)



    On the other hand the GCD isn't a number they divide into; it is a number that divides into *them. $6$ divides into $12$ and $18$ but not $15$ or $27$. $9$ divides into $18$ and $27$ but not $12$ and $15$. But $3$ divides into all $4$. Is it the greatest common divisor? Well, yes it is.



    ====



    Remember $12,15$



    $n$= LCM = least common MULTIPLE. $12$ and $15$ go into $n$. So it is $60$ that they go into. $12$ and $15$ go into $60$.



    $m$ = GCD = greatest common DIVISOR. $m$ goes into $12$ and $15$. So it is $3$ that goes into them. $3$ goes into $12$ and $15$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      "Find the largest $4$ digit number that is exactly divisible by $12,15,18$ and $27$."



      If all those number divide into that 4-digit number than that 4-digit number is a multiple of all those numbers.



      So the 4-digit number is a common multiple of all of those.



      "The solution is by finding out the LCM of these numbers. I mean why ?"



      Well, we don't have to find the least common multiple but we have to find a common multiple. And if we can find the least common multiple that will give us all the infinite number of common multiples by multiply the least common multiple by every possible positive integer.



      The question is asking us to find the greatest common multiple that is less than or equal to $9999$. We do that be finding the least common multiple and adding that and adding that until we get the largest possible value less than or equal to $9999$.



      "Why is it not GCD? What is the difference between GCD and LCM?"



      Because GCD is the "greatest common divisor". It is a number (in this case $3$) that divides into each of the $12,$15,$18$ and $27$. The least common multiple is a number that all the numbers that $12$, $15$, $18$, and $27$ themselves divide into; not what divides into them.



      For example $12 $ divides into $24$, $36$, $48$ .etc. But $15$ divides into $30$, $45$ and $60$. The smallest number that both $12$ and $15$ divide into is $60$. But $18$ and $27$ do not divide into $60$. The all divide into $12*15*18*27 = 87480$ that that probably isn't the smallest number they all divide into. Do you know what is. (And if you can find the largest common multiple that is four digits, you have solved the question.)



      On the other hand the GCD isn't a number they divide into; it is a number that divides into *them. $6$ divides into $12$ and $18$ but not $15$ or $27$. $9$ divides into $18$ and $27$ but not $12$ and $15$. But $3$ divides into all $4$. Is it the greatest common divisor? Well, yes it is.



      ====



      Remember $12,15$



      $n$= LCM = least common MULTIPLE. $12$ and $15$ go into $n$. So it is $60$ that they go into. $12$ and $15$ go into $60$.



      $m$ = GCD = greatest common DIVISOR. $m$ goes into $12$ and $15$. So it is $3$ that goes into them. $3$ goes into $12$ and $15$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        "Find the largest $4$ digit number that is exactly divisible by $12,15,18$ and $27$."



        If all those number divide into that 4-digit number than that 4-digit number is a multiple of all those numbers.



        So the 4-digit number is a common multiple of all of those.



        "The solution is by finding out the LCM of these numbers. I mean why ?"



        Well, we don't have to find the least common multiple but we have to find a common multiple. And if we can find the least common multiple that will give us all the infinite number of common multiples by multiply the least common multiple by every possible positive integer.



        The question is asking us to find the greatest common multiple that is less than or equal to $9999$. We do that be finding the least common multiple and adding that and adding that until we get the largest possible value less than or equal to $9999$.



        "Why is it not GCD? What is the difference between GCD and LCM?"



        Because GCD is the "greatest common divisor". It is a number (in this case $3$) that divides into each of the $12,$15,$18$ and $27$. The least common multiple is a number that all the numbers that $12$, $15$, $18$, and $27$ themselves divide into; not what divides into them.



        For example $12 $ divides into $24$, $36$, $48$ .etc. But $15$ divides into $30$, $45$ and $60$. The smallest number that both $12$ and $15$ divide into is $60$. But $18$ and $27$ do not divide into $60$. The all divide into $12*15*18*27 = 87480$ that that probably isn't the smallest number they all divide into. Do you know what is. (And if you can find the largest common multiple that is four digits, you have solved the question.)



        On the other hand the GCD isn't a number they divide into; it is a number that divides into *them. $6$ divides into $12$ and $18$ but not $15$ or $27$. $9$ divides into $18$ and $27$ but not $12$ and $15$. But $3$ divides into all $4$. Is it the greatest common divisor? Well, yes it is.



        ====



        Remember $12,15$



        $n$= LCM = least common MULTIPLE. $12$ and $15$ go into $n$. So it is $60$ that they go into. $12$ and $15$ go into $60$.



        $m$ = GCD = greatest common DIVISOR. $m$ goes into $12$ and $15$. So it is $3$ that goes into them. $3$ goes into $12$ and $15$.






        share|cite|improve this answer









        $endgroup$



        "Find the largest $4$ digit number that is exactly divisible by $12,15,18$ and $27$."



        If all those number divide into that 4-digit number than that 4-digit number is a multiple of all those numbers.



        So the 4-digit number is a common multiple of all of those.



        "The solution is by finding out the LCM of these numbers. I mean why ?"



        Well, we don't have to find the least common multiple but we have to find a common multiple. And if we can find the least common multiple that will give us all the infinite number of common multiples by multiply the least common multiple by every possible positive integer.



        The question is asking us to find the greatest common multiple that is less than or equal to $9999$. We do that be finding the least common multiple and adding that and adding that until we get the largest possible value less than or equal to $9999$.



        "Why is it not GCD? What is the difference between GCD and LCM?"



        Because GCD is the "greatest common divisor". It is a number (in this case $3$) that divides into each of the $12,$15,$18$ and $27$. The least common multiple is a number that all the numbers that $12$, $15$, $18$, and $27$ themselves divide into; not what divides into them.



        For example $12 $ divides into $24$, $36$, $48$ .etc. But $15$ divides into $30$, $45$ and $60$. The smallest number that both $12$ and $15$ divide into is $60$. But $18$ and $27$ do not divide into $60$. The all divide into $12*15*18*27 = 87480$ that that probably isn't the smallest number they all divide into. Do you know what is. (And if you can find the largest common multiple that is four digits, you have solved the question.)



        On the other hand the GCD isn't a number they divide into; it is a number that divides into *them. $6$ divides into $12$ and $18$ but not $15$ or $27$. $9$ divides into $18$ and $27$ but not $12$ and $15$. But $3$ divides into all $4$. Is it the greatest common divisor? Well, yes it is.



        ====



        Remember $12,15$



        $n$= LCM = least common MULTIPLE. $12$ and $15$ go into $n$. So it is $60$ that they go into. $12$ and $15$ go into $60$.



        $m$ = GCD = greatest common DIVISOR. $m$ goes into $12$ and $15$. So it is $3$ that goes into them. $3$ goes into $12$ and $15$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 3 '17 at 0:01









        fleabloodfleablood

        73.4k22791




        73.4k22791






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2414676%2ffinding-largest-4-digit-number-divisible-by-certain-numbers%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Willebadessen

            Image
            Wappen Deutschlandkarte 51.633055555556 9.0330555555556 295 Koordinaten: 51° 38′  N , 9° 2′  O Basisdaten Bundesland: Nordrhein-Westfalen Regierungsbezirk: Detmold Kreis: Höxter Höhe: 295 m ü. NHN Fläche: 128,41 km 2 Einwohner: 8227 (31. Dez. 2017) [1] Bevölkerungsdichte: 64 Einwohner je km 2 Postleitzahlen: 34435, 34437, 34439 Vorwahlen: 05644, 05646, 05642 Kfz-Kennzeichen: HX Gemeindeschlüssel: 05 7 62 040 Stadtgliederung: 13 Stadtteile Adresse der Stadtverwaltung: Abdinghofweg 1 34439 Willebadessen Website: www.willebadessen.de Bürgermeister: Hans Hermann Bluhm (CDU) Lage der Stadt Willebadessen im Kreis Höxter Willebadessen ist eine Kleinstadt in Nordrhein-Westfalen und gehört zum Kreis Höxter im Regierungsbezirk Detmold. Die Stadt liegt im Osten des Regierungsbezirkes mitten im Naturpark Teutoburger Wald/Eggegebirge. Für den heu...
            Read more

            Ida-Boy-Ed-Garten

            Image
            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
            Read more

            Residenzschloss Arolsen

            Image
            Residenzschloss Arolsen Residenzschloss Arolsen Entstehungszeit: 1710–1728 Erhaltungszustand: Vollständig Ständische Stellung: Grafen Ort: Bad Arolsen Geographische Lage 51° 22′ 51″  N , 9° 1′ 19″  O 51.380833333333 9.0219444444444 293 Koordinaten: 51° 22′ 51″  N , 9° 1′ 19″  O Höhe: 293  m ü. NHN Das Residenzschloss Arolsen ist ein barockes Schloss in Bad Arolsen im Kreis Waldeck-Frankenberg in Nordhessen (Deutschland). Das Schloss wurde als dreiflügelige Anlage gebaut, an die sich ein englischer Garten anschließt. Zentrales landschaftsarchitektonisches Gestaltungselement ist ein ausgedehntes Rondell. Inhaltsverzeichnis 1 Geschichte 1.1 Vorgängerbau 1.2 Bau und Geschichte des neuen Schlosses 1.3 Bibliothek 1.4 Finanzielle und staatsrechtliche Konsequenzen 1.5 Heutige Nutzung 2 Innenausstattung 3 Umgebung 3.1 Park 3.2 Wirtschaftsgebäude 4 Literatur 5 Weblinks 6...
            Read more