How can I argue that for a number to be divisible by 144 it has to be divisible by 36?











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Suppose some number $n in mathbb{N}$ is divisible by $144$.



$$implies frac{n}{144}=k, space space space k in mathbb{Z} \ iff frac{n}{36cdot4}=k iff frac{n}{36}=4k$$



Since any whole number times a whole number is still a whole number, it follows that $n$ must also be divisible by $36$. However, what I think I have just shown is:



$$text{A number }n space text{is divisble by} space 144 implies n space text{is divisible by} space 36 space (1)$$



Is that the same as saying: $$text{For a number to be divisible by 144 it has to be divisible by 36} space (2)$$



In other words, are statements (1) and (2) equivalent?










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  • 14




    Yes, absolutely.
    – Bernard
    2 days ago






  • 1




    Pedantically, the first sentence could have a different meaning if it somehow isn't obvious from context that $n$ is universally quantified. But in general they're the same.
    – Ian
    2 days ago










  • You mean this sentence "Since any whole number times a whole number is still a whole number, it follows that n must also be divisible by 36. ",right?
    – Nullspace
    2 days ago










  • In general, if $a$ is divisible by $b$, and $b$ is divisible by $c$, then $a$ is divisible by $c$.
    – AlexanderJ93
    2 days ago















up vote
19
down vote

favorite
3












Suppose some number $n in mathbb{N}$ is divisible by $144$.



$$implies frac{n}{144}=k, space space space k in mathbb{Z} \ iff frac{n}{36cdot4}=k iff frac{n}{36}=4k$$



Since any whole number times a whole number is still a whole number, it follows that $n$ must also be divisible by $36$. However, what I think I have just shown is:



$$text{A number }n space text{is divisble by} space 144 implies n space text{is divisible by} space 36 space (1)$$



Is that the same as saying: $$text{For a number to be divisible by 144 it has to be divisible by 36} space (2)$$



In other words, are statements (1) and (2) equivalent?










share|cite|improve this question


















  • 14




    Yes, absolutely.
    – Bernard
    2 days ago






  • 1




    Pedantically, the first sentence could have a different meaning if it somehow isn't obvious from context that $n$ is universally quantified. But in general they're the same.
    – Ian
    2 days ago










  • You mean this sentence "Since any whole number times a whole number is still a whole number, it follows that n must also be divisible by 36. ",right?
    – Nullspace
    2 days ago










  • In general, if $a$ is divisible by $b$, and $b$ is divisible by $c$, then $a$ is divisible by $c$.
    – AlexanderJ93
    2 days ago













up vote
19
down vote

favorite
3









up vote
19
down vote

favorite
3






3





Suppose some number $n in mathbb{N}$ is divisible by $144$.



$$implies frac{n}{144}=k, space space space k in mathbb{Z} \ iff frac{n}{36cdot4}=k iff frac{n}{36}=4k$$



Since any whole number times a whole number is still a whole number, it follows that $n$ must also be divisible by $36$. However, what I think I have just shown is:



$$text{A number }n space text{is divisble by} space 144 implies n space text{is divisible by} space 36 space (1)$$



Is that the same as saying: $$text{For a number to be divisible by 144 it has to be divisible by 36} space (2)$$



In other words, are statements (1) and (2) equivalent?










share|cite|improve this question













Suppose some number $n in mathbb{N}$ is divisible by $144$.



$$implies frac{n}{144}=k, space space space k in mathbb{Z} \ iff frac{n}{36cdot4}=k iff frac{n}{36}=4k$$



Since any whole number times a whole number is still a whole number, it follows that $n$ must also be divisible by $36$. However, what I think I have just shown is:



$$text{A number }n space text{is divisble by} space 144 implies n space text{is divisible by} space 36 space (1)$$



Is that the same as saying: $$text{For a number to be divisible by 144 it has to be divisible by 36} space (2)$$



In other words, are statements (1) and (2) equivalent?







proof-verification logic






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share|cite|improve this question











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share|cite|improve this question










asked 2 days ago









Nullspace

12417




12417








  • 14




    Yes, absolutely.
    – Bernard
    2 days ago






  • 1




    Pedantically, the first sentence could have a different meaning if it somehow isn't obvious from context that $n$ is universally quantified. But in general they're the same.
    – Ian
    2 days ago










  • You mean this sentence "Since any whole number times a whole number is still a whole number, it follows that n must also be divisible by 36. ",right?
    – Nullspace
    2 days ago










  • In general, if $a$ is divisible by $b$, and $b$ is divisible by $c$, then $a$ is divisible by $c$.
    – AlexanderJ93
    2 days ago














  • 14




    Yes, absolutely.
    – Bernard
    2 days ago






  • 1




    Pedantically, the first sentence could have a different meaning if it somehow isn't obvious from context that $n$ is universally quantified. But in general they're the same.
    – Ian
    2 days ago










  • You mean this sentence "Since any whole number times a whole number is still a whole number, it follows that n must also be divisible by 36. ",right?
    – Nullspace
    2 days ago










  • In general, if $a$ is divisible by $b$, and $b$ is divisible by $c$, then $a$ is divisible by $c$.
    – AlexanderJ93
    2 days ago








14




14




Yes, absolutely.
– Bernard
2 days ago




Yes, absolutely.
– Bernard
2 days ago




1




1




Pedantically, the first sentence could have a different meaning if it somehow isn't obvious from context that $n$ is universally quantified. But in general they're the same.
– Ian
2 days ago




Pedantically, the first sentence could have a different meaning if it somehow isn't obvious from context that $n$ is universally quantified. But in general they're the same.
– Ian
2 days ago












You mean this sentence "Since any whole number times a whole number is still a whole number, it follows that n must also be divisible by 36. ",right?
– Nullspace
2 days ago




You mean this sentence "Since any whole number times a whole number is still a whole number, it follows that n must also be divisible by 36. ",right?
– Nullspace
2 days ago












In general, if $a$ is divisible by $b$, and $b$ is divisible by $c$, then $a$ is divisible by $c$.
– AlexanderJ93
2 days ago




In general, if $a$ is divisible by $b$, and $b$ is divisible by $c$, then $a$ is divisible by $c$.
– AlexanderJ93
2 days ago










6 Answers
6






active

oldest

votes

















up vote
44
down vote



accepted










Yes, it's the same. $Aimplies B$ is equivalent to "if we have $A$, we must have $B$".



And your proof looks fine. Good job.



If I were to offer some constructive criticism, it would be of the general kind: In number theory, even though we call the property "divisible", we usually avoid division whenever possible. Of the four basic arithmetic operations it is the only one which makes integers into non-integers. And number theory is all about integers.



Therefore, "$n$ is divisible by $144$", or "$144$ divides $n$" as it's also called, is defined a bit backwards:




There is an integer $k$ such that $n=144k$




(This is defined for any number in place of $144$, except $0$.)



Using that definition, your proof becomes something like this:



If $n$ is divisible by $144$, then there is an integer $k$ such that $n=144k$. This gives
$$
n=144k=(36cdot4)k=36(4k)
$$

Since $4k$ is an integer, this means $n$ is also divisible by $36$.






share|cite|improve this answer



















  • 2




    Programmers might may say mod() = 0 rather than "divides", if n modulus 144 is zero then n modulus 36 is zero etc. Non zero values are integers.
    – mckenzm
    2 days ago










  • Every definition I ever see of divisibility specifically excludes $0$ as a potential divisor (checked a few abstract algebra books here, etc.). If that were not the case, it would wind up violating the Division Algorithm, leave greatest common divisors not well-defined, etc.
    – Daniel R. Collins
    22 hours ago


















up vote
5
down vote













Yes that's correct or simply note that



$$n=144cdot k= 36cdot (4cdot k)$$



but $n=36$ is not divisible by $144$.






share|cite|improve this answer




























    up vote
    4
    down vote













    The $implies$ symbol is defined as follows:



    If $p implies q$ then if $p$ is true, then $q$ must also be true. So when you say $144 mid n implies 36 mid n$ it's the same thing as saying that if $144 mid n$, then it must also be true that $36 mid n$.






    share|cite|improve this answer




























      up vote
      4
      down vote













      There are often multiple approaches to proofs. Is usually good to be familiar with multiple techniques.



      You have a very good approach. Parsimonious and references only the particular entities at hand.



      Some times, for the sake of illustrating the use of additional concepts, you might want to deviate from parsimony.



      In that spirit, here's an additional proof.



      According to The Fundamental Theorem of Arithmetic, if something is divisible by 144, then it is divisible by at least the same primes raised to the powers you need to yield 144. In other words, $144=2^43^2$. For something to be divisible by 144, the primes 2 and 3 must appear in its prime factorization. The powers of 2 and 3 must be at least 4 and 2, respectively. Now $36=2^23^2$. So any number who's prime factorization incluedes the primes 2 and 3, and has them raised at least to the power of 2, then it is also divisible by 36. If something is divisible by 144, we are guaranteed that 2 and 3 appear in its prime factorization. We area also guaranteed that the exponents on 2 and 3 are 4 and 2 respectively. So divisibility by 144 implies divisibility by 36 since the exponents satisfy the established criteria.






      share|cite|improve this answer




























        up vote
        0
        down vote













        The prime factorization of 144 is 2 * 2 * 2 * 2 * 3 * 3.
        The prime factorization of 36 is 2 * 2 * 3 * 3.



        If X is divisible by Y, then X's prime factorization contains all of the factors in Y's prime factorization.



        Since 144's prime factorization contains all of the factors in 36's prime factorization, 144 is divisible by 36.



        The prime factorization of any number that is divisible by 144 contains all of the prime factors of 144, which contains all of the prime factors of 36. Hence any number that is divisible by 144 has a prime factorization that contains all the prime factors of 36, and therefore is divisible by 36.






        share|cite|improve this answer








        New contributor




        CCC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          up vote
          0
          down vote













          If $n$ is a multiple of $144$, it must automatically be a multiple of any divisor $d$ of $144$. The prime factorization of $144$ is $(2^4)(3^2)$, and so $36 = (2^2)(3^2)$ must be a divisor of $144$ and hence of $n$.






          share|cite|improve this answer





















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            6 Answers
            6






            active

            oldest

            votes








            6 Answers
            6






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            44
            down vote



            accepted










            Yes, it's the same. $Aimplies B$ is equivalent to "if we have $A$, we must have $B$".



            And your proof looks fine. Good job.



            If I were to offer some constructive criticism, it would be of the general kind: In number theory, even though we call the property "divisible", we usually avoid division whenever possible. Of the four basic arithmetic operations it is the only one which makes integers into non-integers. And number theory is all about integers.



            Therefore, "$n$ is divisible by $144$", or "$144$ divides $n$" as it's also called, is defined a bit backwards:




            There is an integer $k$ such that $n=144k$




            (This is defined for any number in place of $144$, except $0$.)



            Using that definition, your proof becomes something like this:



            If $n$ is divisible by $144$, then there is an integer $k$ such that $n=144k$. This gives
            $$
            n=144k=(36cdot4)k=36(4k)
            $$

            Since $4k$ is an integer, this means $n$ is also divisible by $36$.






            share|cite|improve this answer



















            • 2




              Programmers might may say mod() = 0 rather than "divides", if n modulus 144 is zero then n modulus 36 is zero etc. Non zero values are integers.
              – mckenzm
              2 days ago










            • Every definition I ever see of divisibility specifically excludes $0$ as a potential divisor (checked a few abstract algebra books here, etc.). If that were not the case, it would wind up violating the Division Algorithm, leave greatest common divisors not well-defined, etc.
              – Daniel R. Collins
              22 hours ago















            up vote
            44
            down vote



            accepted










            Yes, it's the same. $Aimplies B$ is equivalent to "if we have $A$, we must have $B$".



            And your proof looks fine. Good job.



            If I were to offer some constructive criticism, it would be of the general kind: In number theory, even though we call the property "divisible", we usually avoid division whenever possible. Of the four basic arithmetic operations it is the only one which makes integers into non-integers. And number theory is all about integers.



            Therefore, "$n$ is divisible by $144$", or "$144$ divides $n$" as it's also called, is defined a bit backwards:




            There is an integer $k$ such that $n=144k$




            (This is defined for any number in place of $144$, except $0$.)



            Using that definition, your proof becomes something like this:



            If $n$ is divisible by $144$, then there is an integer $k$ such that $n=144k$. This gives
            $$
            n=144k=(36cdot4)k=36(4k)
            $$

            Since $4k$ is an integer, this means $n$ is also divisible by $36$.






            share|cite|improve this answer



















            • 2




              Programmers might may say mod() = 0 rather than "divides", if n modulus 144 is zero then n modulus 36 is zero etc. Non zero values are integers.
              – mckenzm
              2 days ago










            • Every definition I ever see of divisibility specifically excludes $0$ as a potential divisor (checked a few abstract algebra books here, etc.). If that were not the case, it would wind up violating the Division Algorithm, leave greatest common divisors not well-defined, etc.
              – Daniel R. Collins
              22 hours ago













            up vote
            44
            down vote



            accepted







            up vote
            44
            down vote



            accepted






            Yes, it's the same. $Aimplies B$ is equivalent to "if we have $A$, we must have $B$".



            And your proof looks fine. Good job.



            If I were to offer some constructive criticism, it would be of the general kind: In number theory, even though we call the property "divisible", we usually avoid division whenever possible. Of the four basic arithmetic operations it is the only one which makes integers into non-integers. And number theory is all about integers.



            Therefore, "$n$ is divisible by $144$", or "$144$ divides $n$" as it's also called, is defined a bit backwards:




            There is an integer $k$ such that $n=144k$




            (This is defined for any number in place of $144$, except $0$.)



            Using that definition, your proof becomes something like this:



            If $n$ is divisible by $144$, then there is an integer $k$ such that $n=144k$. This gives
            $$
            n=144k=(36cdot4)k=36(4k)
            $$

            Since $4k$ is an integer, this means $n$ is also divisible by $36$.






            share|cite|improve this answer














            Yes, it's the same. $Aimplies B$ is equivalent to "if we have $A$, we must have $B$".



            And your proof looks fine. Good job.



            If I were to offer some constructive criticism, it would be of the general kind: In number theory, even though we call the property "divisible", we usually avoid division whenever possible. Of the four basic arithmetic operations it is the only one which makes integers into non-integers. And number theory is all about integers.



            Therefore, "$n$ is divisible by $144$", or "$144$ divides $n$" as it's also called, is defined a bit backwards:




            There is an integer $k$ such that $n=144k$




            (This is defined for any number in place of $144$, except $0$.)



            Using that definition, your proof becomes something like this:



            If $n$ is divisible by $144$, then there is an integer $k$ such that $n=144k$. This gives
            $$
            n=144k=(36cdot4)k=36(4k)
            $$

            Since $4k$ is an integer, this means $n$ is also divisible by $36$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 22 hours ago

























            answered 2 days ago









            Arthur

            107k7103186




            107k7103186








            • 2




              Programmers might may say mod() = 0 rather than "divides", if n modulus 144 is zero then n modulus 36 is zero etc. Non zero values are integers.
              – mckenzm
              2 days ago










            • Every definition I ever see of divisibility specifically excludes $0$ as a potential divisor (checked a few abstract algebra books here, etc.). If that were not the case, it would wind up violating the Division Algorithm, leave greatest common divisors not well-defined, etc.
              – Daniel R. Collins
              22 hours ago














            • 2




              Programmers might may say mod() = 0 rather than "divides", if n modulus 144 is zero then n modulus 36 is zero etc. Non zero values are integers.
              – mckenzm
              2 days ago










            • Every definition I ever see of divisibility specifically excludes $0$ as a potential divisor (checked a few abstract algebra books here, etc.). If that were not the case, it would wind up violating the Division Algorithm, leave greatest common divisors not well-defined, etc.
              – Daniel R. Collins
              22 hours ago








            2




            2




            Programmers might may say mod() = 0 rather than "divides", if n modulus 144 is zero then n modulus 36 is zero etc. Non zero values are integers.
            – mckenzm
            2 days ago




            Programmers might may say mod() = 0 rather than "divides", if n modulus 144 is zero then n modulus 36 is zero etc. Non zero values are integers.
            – mckenzm
            2 days ago












            Every definition I ever see of divisibility specifically excludes $0$ as a potential divisor (checked a few abstract algebra books here, etc.). If that were not the case, it would wind up violating the Division Algorithm, leave greatest common divisors not well-defined, etc.
            – Daniel R. Collins
            22 hours ago




            Every definition I ever see of divisibility specifically excludes $0$ as a potential divisor (checked a few abstract algebra books here, etc.). If that were not the case, it would wind up violating the Division Algorithm, leave greatest common divisors not well-defined, etc.
            – Daniel R. Collins
            22 hours ago










            up vote
            5
            down vote













            Yes that's correct or simply note that



            $$n=144cdot k= 36cdot (4cdot k)$$



            but $n=36$ is not divisible by $144$.






            share|cite|improve this answer

























              up vote
              5
              down vote













              Yes that's correct or simply note that



              $$n=144cdot k= 36cdot (4cdot k)$$



              but $n=36$ is not divisible by $144$.






              share|cite|improve this answer























                up vote
                5
                down vote










                up vote
                5
                down vote









                Yes that's correct or simply note that



                $$n=144cdot k= 36cdot (4cdot k)$$



                but $n=36$ is not divisible by $144$.






                share|cite|improve this answer












                Yes that's correct or simply note that



                $$n=144cdot k= 36cdot (4cdot k)$$



                but $n=36$ is not divisible by $144$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                gimusi

                84.1k74292




                84.1k74292






















                    up vote
                    4
                    down vote













                    The $implies$ symbol is defined as follows:



                    If $p implies q$ then if $p$ is true, then $q$ must also be true. So when you say $144 mid n implies 36 mid n$ it's the same thing as saying that if $144 mid n$, then it must also be true that $36 mid n$.






                    share|cite|improve this answer

























                      up vote
                      4
                      down vote













                      The $implies$ symbol is defined as follows:



                      If $p implies q$ then if $p$ is true, then $q$ must also be true. So when you say $144 mid n implies 36 mid n$ it's the same thing as saying that if $144 mid n$, then it must also be true that $36 mid n$.






                      share|cite|improve this answer























                        up vote
                        4
                        down vote










                        up vote
                        4
                        down vote









                        The $implies$ symbol is defined as follows:



                        If $p implies q$ then if $p$ is true, then $q$ must also be true. So when you say $144 mid n implies 36 mid n$ it's the same thing as saying that if $144 mid n$, then it must also be true that $36 mid n$.






                        share|cite|improve this answer












                        The $implies$ symbol is defined as follows:



                        If $p implies q$ then if $p$ is true, then $q$ must also be true. So when you say $144 mid n implies 36 mid n$ it's the same thing as saying that if $144 mid n$, then it must also be true that $36 mid n$.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered 2 days ago









                        TrostAft

                        315211




                        315211






















                            up vote
                            4
                            down vote













                            There are often multiple approaches to proofs. Is usually good to be familiar with multiple techniques.



                            You have a very good approach. Parsimonious and references only the particular entities at hand.



                            Some times, for the sake of illustrating the use of additional concepts, you might want to deviate from parsimony.



                            In that spirit, here's an additional proof.



                            According to The Fundamental Theorem of Arithmetic, if something is divisible by 144, then it is divisible by at least the same primes raised to the powers you need to yield 144. In other words, $144=2^43^2$. For something to be divisible by 144, the primes 2 and 3 must appear in its prime factorization. The powers of 2 and 3 must be at least 4 and 2, respectively. Now $36=2^23^2$. So any number who's prime factorization incluedes the primes 2 and 3, and has them raised at least to the power of 2, then it is also divisible by 36. If something is divisible by 144, we are guaranteed that 2 and 3 appear in its prime factorization. We area also guaranteed that the exponents on 2 and 3 are 4 and 2 respectively. So divisibility by 144 implies divisibility by 36 since the exponents satisfy the established criteria.






                            share|cite|improve this answer

























                              up vote
                              4
                              down vote













                              There are often multiple approaches to proofs. Is usually good to be familiar with multiple techniques.



                              You have a very good approach. Parsimonious and references only the particular entities at hand.



                              Some times, for the sake of illustrating the use of additional concepts, you might want to deviate from parsimony.



                              In that spirit, here's an additional proof.



                              According to The Fundamental Theorem of Arithmetic, if something is divisible by 144, then it is divisible by at least the same primes raised to the powers you need to yield 144. In other words, $144=2^43^2$. For something to be divisible by 144, the primes 2 and 3 must appear in its prime factorization. The powers of 2 and 3 must be at least 4 and 2, respectively. Now $36=2^23^2$. So any number who's prime factorization incluedes the primes 2 and 3, and has them raised at least to the power of 2, then it is also divisible by 36. If something is divisible by 144, we are guaranteed that 2 and 3 appear in its prime factorization. We area also guaranteed that the exponents on 2 and 3 are 4 and 2 respectively. So divisibility by 144 implies divisibility by 36 since the exponents satisfy the established criteria.






                              share|cite|improve this answer























                                up vote
                                4
                                down vote










                                up vote
                                4
                                down vote









                                There are often multiple approaches to proofs. Is usually good to be familiar with multiple techniques.



                                You have a very good approach. Parsimonious and references only the particular entities at hand.



                                Some times, for the sake of illustrating the use of additional concepts, you might want to deviate from parsimony.



                                In that spirit, here's an additional proof.



                                According to The Fundamental Theorem of Arithmetic, if something is divisible by 144, then it is divisible by at least the same primes raised to the powers you need to yield 144. In other words, $144=2^43^2$. For something to be divisible by 144, the primes 2 and 3 must appear in its prime factorization. The powers of 2 and 3 must be at least 4 and 2, respectively. Now $36=2^23^2$. So any number who's prime factorization incluedes the primes 2 and 3, and has them raised at least to the power of 2, then it is also divisible by 36. If something is divisible by 144, we are guaranteed that 2 and 3 appear in its prime factorization. We area also guaranteed that the exponents on 2 and 3 are 4 and 2 respectively. So divisibility by 144 implies divisibility by 36 since the exponents satisfy the established criteria.






                                share|cite|improve this answer












                                There are often multiple approaches to proofs. Is usually good to be familiar with multiple techniques.



                                You have a very good approach. Parsimonious and references only the particular entities at hand.



                                Some times, for the sake of illustrating the use of additional concepts, you might want to deviate from parsimony.



                                In that spirit, here's an additional proof.



                                According to The Fundamental Theorem of Arithmetic, if something is divisible by 144, then it is divisible by at least the same primes raised to the powers you need to yield 144. In other words, $144=2^43^2$. For something to be divisible by 144, the primes 2 and 3 must appear in its prime factorization. The powers of 2 and 3 must be at least 4 and 2, respectively. Now $36=2^23^2$. So any number who's prime factorization incluedes the primes 2 and 3, and has them raised at least to the power of 2, then it is also divisible by 36. If something is divisible by 144, we are guaranteed that 2 and 3 appear in its prime factorization. We area also guaranteed that the exponents on 2 and 3 are 4 and 2 respectively. So divisibility by 144 implies divisibility by 36 since the exponents satisfy the established criteria.







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                                TurlocTheRed

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                                    The prime factorization of 144 is 2 * 2 * 2 * 2 * 3 * 3.
                                    The prime factorization of 36 is 2 * 2 * 3 * 3.



                                    If X is divisible by Y, then X's prime factorization contains all of the factors in Y's prime factorization.



                                    Since 144's prime factorization contains all of the factors in 36's prime factorization, 144 is divisible by 36.



                                    The prime factorization of any number that is divisible by 144 contains all of the prime factors of 144, which contains all of the prime factors of 36. Hence any number that is divisible by 144 has a prime factorization that contains all the prime factors of 36, and therefore is divisible by 36.






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                                      up vote
                                      0
                                      down vote













                                      The prime factorization of 144 is 2 * 2 * 2 * 2 * 3 * 3.
                                      The prime factorization of 36 is 2 * 2 * 3 * 3.



                                      If X is divisible by Y, then X's prime factorization contains all of the factors in Y's prime factorization.



                                      Since 144's prime factorization contains all of the factors in 36's prime factorization, 144 is divisible by 36.



                                      The prime factorization of any number that is divisible by 144 contains all of the prime factors of 144, which contains all of the prime factors of 36. Hence any number that is divisible by 144 has a prime factorization that contains all the prime factors of 36, and therefore is divisible by 36.






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                                        up vote
                                        0
                                        down vote










                                        up vote
                                        0
                                        down vote









                                        The prime factorization of 144 is 2 * 2 * 2 * 2 * 3 * 3.
                                        The prime factorization of 36 is 2 * 2 * 3 * 3.



                                        If X is divisible by Y, then X's prime factorization contains all of the factors in Y's prime factorization.



                                        Since 144's prime factorization contains all of the factors in 36's prime factorization, 144 is divisible by 36.



                                        The prime factorization of any number that is divisible by 144 contains all of the prime factors of 144, which contains all of the prime factors of 36. Hence any number that is divisible by 144 has a prime factorization that contains all the prime factors of 36, and therefore is divisible by 36.






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                                        New contributor




                                        CCC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                                        The prime factorization of 144 is 2 * 2 * 2 * 2 * 3 * 3.
                                        The prime factorization of 36 is 2 * 2 * 3 * 3.



                                        If X is divisible by Y, then X's prime factorization contains all of the factors in Y's prime factorization.



                                        Since 144's prime factorization contains all of the factors in 36's prime factorization, 144 is divisible by 36.



                                        The prime factorization of any number that is divisible by 144 contains all of the prime factors of 144, which contains all of the prime factors of 36. Hence any number that is divisible by 144 has a prime factorization that contains all the prime factors of 36, and therefore is divisible by 36.







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                                        answered yesterday









                                        CCC

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                                            If $n$ is a multiple of $144$, it must automatically be a multiple of any divisor $d$ of $144$. The prime factorization of $144$ is $(2^4)(3^2)$, and so $36 = (2^2)(3^2)$ must be a divisor of $144$ and hence of $n$.






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                                              If $n$ is a multiple of $144$, it must automatically be a multiple of any divisor $d$ of $144$. The prime factorization of $144$ is $(2^4)(3^2)$, and so $36 = (2^2)(3^2)$ must be a divisor of $144$ and hence of $n$.






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                                                up vote
                                                0
                                                down vote









                                                If $n$ is a multiple of $144$, it must automatically be a multiple of any divisor $d$ of $144$. The prime factorization of $144$ is $(2^4)(3^2)$, and so $36 = (2^2)(3^2)$ must be a divisor of $144$ and hence of $n$.






                                                share|cite|improve this answer












                                                If $n$ is a multiple of $144$, it must automatically be a multiple of any divisor $d$ of $144$. The prime factorization of $144$ is $(2^4)(3^2)$, and so $36 = (2^2)(3^2)$ must be a divisor of $144$ and hence of $n$.







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                                                answered 5 hours ago









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