How many odd three-digits numbers are there whose all three digits are different












4












$begingroup$


I faced this problem on one test. I wrote my solution but then I found out that my solution is wrong, I still cannot find where my mistake is.



The problem says: How many three-digits numbers are there such that they are odd and their digits are all different.



Here is my approach:



We have three digits. Since the number should be odd, the last digit should be one of those numbers $1, 3, 5, 7, 9$. Now the second digits can be one of the digits: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ There are 10 different digits for the second digits, but since the digits should be different we cannot place 10 digits, but we can place 9 digits. And for the first digits we can place digits in the range $1...9$ but we cannot place the digits that are used in the two other digits and we can place only 7 digits.



So my result is $7cdot9cdot5 = 315$



However the result is not correct, because there are $320$ odd three-digits numbers with different digits.



Can you point me where is my mistake, thanks in advance.






combinatorics number-theory







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The mistake is you might have used 0 for the middle digit. So there are two cases.
    $endgroup$
    – quasi
    Jul 9 '17 at 8:36












  • $begingroup$
    I don't see why we cant use the 0 for the middle digit, number 503, 305, 307 are all valid for counting.
    $endgroup$
    – someone123123
    Jul 9 '17 at 8:39






  • 2




    $begingroup$
    @someone123123 if you use 0 for the middle digit, you have 8 options available for the first digit. This gives 5 additional cases.
    $endgroup$
    – IanF1
    Jul 9 '17 at 15:34










  • $begingroup$
    To really pinpoint the problem: "For the first digits we can place digits in the range 1...9 [good so far] but we cannot place the digits that are used in the two other digits [still good] and we can place only 7 digits [this is not correct!].".
    $endgroup$
    – Daniel Wagner
    Jul 9 '17 at 23:19
















4












$begingroup$


I faced this problem on one test. I wrote my solution but then I found out that my solution is wrong, I still cannot find where my mistake is.



The problem says: How many three-digits numbers are there such that they are odd and their digits are all different.



Here is my approach:



We have three digits. Since the number should be odd, the last digit should be one of those numbers $1, 3, 5, 7, 9$. Now the second digits can be one of the digits: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ There are 10 different digits for the second digits, but since the digits should be different we cannot place 10 digits, but we can place 9 digits. And for the first digits we can place digits in the range $1...9$ but we cannot place the digits that are used in the two other digits and we can place only 7 digits.



So my result is $7cdot9cdot5 = 315$



However the result is not correct, because there are $320$ odd three-digits numbers with different digits.



Can you point me where is my mistake, thanks in advance.






combinatorics number-theory







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The mistake is you might have used 0 for the middle digit. So there are two cases.
    $endgroup$
    – quasi
    Jul 9 '17 at 8:36












  • $begingroup$
    I don't see why we cant use the 0 for the middle digit, number 503, 305, 307 are all valid for counting.
    $endgroup$
    – someone123123
    Jul 9 '17 at 8:39






  • 2




    $begingroup$
    @someone123123 if you use 0 for the middle digit, you have 8 options available for the first digit. This gives 5 additional cases.
    $endgroup$
    – IanF1
    Jul 9 '17 at 15:34










  • $begingroup$
    To really pinpoint the problem: "For the first digits we can place digits in the range 1...9 [good so far] but we cannot place the digits that are used in the two other digits [still good] and we can place only 7 digits [this is not correct!].".
    $endgroup$
    – Daniel Wagner
    Jul 9 '17 at 23:19














4












4








4


2



$begingroup$


I faced this problem on one test. I wrote my solution but then I found out that my solution is wrong, I still cannot find where my mistake is.



The problem says: How many three-digits numbers are there such that they are odd and their digits are all different.



Here is my approach:



We have three digits. Since the number should be odd, the last digit should be one of those numbers $1, 3, 5, 7, 9$. Now the second digits can be one of the digits: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ There are 10 different digits for the second digits, but since the digits should be different we cannot place 10 digits, but we can place 9 digits. And for the first digits we can place digits in the range $1...9$ but we cannot place the digits that are used in the two other digits and we can place only 7 digits.



So my result is $7cdot9cdot5 = 315$



However the result is not correct, because there are $320$ odd three-digits numbers with different digits.



Can you point me where is my mistake, thanks in advance.






combinatorics number-theory







share|cite|improve this question









$endgroup$




I faced this problem on one test. I wrote my solution but then I found out that my solution is wrong, I still cannot find where my mistake is.



The problem says: How many three-digits numbers are there such that they are odd and their digits are all different.



Here is my approach:



We have three digits. Since the number should be odd, the last digit should be one of those numbers $1, 3, 5, 7, 9$. Now the second digits can be one of the digits: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ There are 10 different digits for the second digits, but since the digits should be different we cannot place 10 digits, but we can place 9 digits. And for the first digits we can place digits in the range $1...9$ but we cannot place the digits that are used in the two other digits and we can place only 7 digits.



So my result is $7cdot9cdot5 = 315$



However the result is not correct, because there are $320$ odd three-digits numbers with different digits.



Can you point me where is my mistake, thanks in advance.






combinatorics number-theory




combinatorics number-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jul 9 '17 at 8:33









someone123123someone123123

428314




428314








  • 1




    $begingroup$
    The mistake is you might have used 0 for the middle digit. So there are two cases.
    $endgroup$
    – quasi
    Jul 9 '17 at 8:36












  • $begingroup$
    I don't see why we cant use the 0 for the middle digit, number 503, 305, 307 are all valid for counting.
    $endgroup$
    – someone123123
    Jul 9 '17 at 8:39






  • 2




    $begingroup$
    @someone123123 if you use 0 for the middle digit, you have 8 options available for the first digit. This gives 5 additional cases.
    $endgroup$
    – IanF1
    Jul 9 '17 at 15:34










  • $begingroup$
    To really pinpoint the problem: "For the first digits we can place digits in the range 1...9 [good so far] but we cannot place the digits that are used in the two other digits [still good] and we can place only 7 digits [this is not correct!].".
    $endgroup$
    – Daniel Wagner
    Jul 9 '17 at 23:19














  • 1




    $begingroup$
    The mistake is you might have used 0 for the middle digit. So there are two cases.
    $endgroup$
    – quasi
    Jul 9 '17 at 8:36












  • $begingroup$
    I don't see why we cant use the 0 for the middle digit, number 503, 305, 307 are all valid for counting.
    $endgroup$
    – someone123123
    Jul 9 '17 at 8:39






  • 2




    $begingroup$
    @someone123123 if you use 0 for the middle digit, you have 8 options available for the first digit. This gives 5 additional cases.
    $endgroup$
    – IanF1
    Jul 9 '17 at 15:34










  • $begingroup$
    To really pinpoint the problem: "For the first digits we can place digits in the range 1...9 [good so far] but we cannot place the digits that are used in the two other digits [still good] and we can place only 7 digits [this is not correct!].".
    $endgroup$
    – Daniel Wagner
    Jul 9 '17 at 23:19








1




1




$begingroup$
The mistake is you might have used 0 for the middle digit. So there are two cases.
$endgroup$
– quasi
Jul 9 '17 at 8:36






$begingroup$
The mistake is you might have used 0 for the middle digit. So there are two cases.
$endgroup$
– quasi
Jul 9 '17 at 8:36














$begingroup$
I don't see why we cant use the 0 for the middle digit, number 503, 305, 307 are all valid for counting.
$endgroup$
– someone123123
Jul 9 '17 at 8:39




$begingroup$
I don't see why we cant use the 0 for the middle digit, number 503, 305, 307 are all valid for counting.
$endgroup$
– someone123123
Jul 9 '17 at 8:39




2




2




$begingroup$
@someone123123 if you use 0 for the middle digit, you have 8 options available for the first digit. This gives 5 additional cases.
$endgroup$
– IanF1
Jul 9 '17 at 15:34




$begingroup$
@someone123123 if you use 0 for the middle digit, you have 8 options available for the first digit. This gives 5 additional cases.
$endgroup$
– IanF1
Jul 9 '17 at 15:34












$begingroup$
To really pinpoint the problem: "For the first digits we can place digits in the range 1...9 [good so far] but we cannot place the digits that are used in the two other digits [still good] and we can place only 7 digits [this is not correct!].".
$endgroup$
– Daniel Wagner
Jul 9 '17 at 23:19




$begingroup$
To really pinpoint the problem: "For the first digits we can place digits in the range 1...9 [good so far] but we cannot place the digits that are used in the two other digits [still good] and we can place only 7 digits [this is not correct!].".
$endgroup$
– Daniel Wagner
Jul 9 '17 at 23:19










4 Answers
4






active

oldest

votes


















13












$begingroup$

Problems of this type are often solved more simply if we place the digits in a suitable order, instead of necessarily left to right.



Here we'll place the digits in the order last-first-second, i.e. in the order $Units-Hundreds-Tens$




  • $5$ choices for the last digit (to make the number odd)


  • $8$ choices left for the first digit (since $0$ can't be used)


  • $8$ choices still left for the second digit (since $0$ is now available for use)



Thus, without breaking up into cases, we get $5cdot8cdot8 = 320$






share|cite|improve this answer











$endgroup$





















    10












    $begingroup$

    The mistake is you might have used $0$ for the middle digit. So there are two cases.



    Case ($1$): Using $0$ for the middle digit:



    $qquad$The count is $5,{times},1,{times},8=40$.



    Case ($2$): Not using $0$ for the middle digit:



    $qquad$The count is $5,{times},8,{times},7=280$.



    So you get a total count of $40+280=320$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for the answer, but can you explain why having 0 for middle digits makes difference.
      $endgroup$
      – someone123123
      Jul 9 '17 at 8:44






    • 6




      $begingroup$
      Because if you use $0$ for the middle digit, the leading digit has only two exclusions, not three.
      $endgroup$
      – quasi
      Jul 9 '17 at 8:46





















    1












    $begingroup$

    Since this problem is only created by one single digit i.e. zero as you can't put zero on hundred's place like 013 which is not 3 digit number so when you put zero on the ten's place i.e. in the middle you will get 8 remaining digits for hundred place. Therefore we have two cases which are:
    Case (1): Using 0 for the middle digit.



    The count is 5×1×8=40.



    Case (2): Not using 0 for the middle digit.



    The count is 5×8×7=280.



    So you get a total count of 40+280=320.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Welcome to MSE. Your answer adds nothing new to the already existing answers.
      $endgroup$
      – José Carlos Santos
      Jan 17 at 17:54



















    -1












    $begingroup$

    Three digit odd numbers implies the numbers would only be made of digits 1 , 3 , 5 , 7 , 9



    With repetition of digits we would have had 5 * 5 * 5 = 125



    But for every hundreds, maximum of 4 tens is possible (avoiding the duplicate of digit used in hundreds)



    And each of these 4 tens, maximum of 3 units is possible (avoiding the duplicate of digits used in tens)




    5 * 4 * 3 = 60 is my answer






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      I think you didn't understand the question, the question says how many three-digits numbers are there such that they are odd and they have three different digits. The first two digits don't have to be odd.
      $endgroup$
      – someone123123
      Jul 12 '17 at 10:13











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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    13












    $begingroup$

    Problems of this type are often solved more simply if we place the digits in a suitable order, instead of necessarily left to right.



    Here we'll place the digits in the order last-first-second, i.e. in the order $Units-Hundreds-Tens$




    • $5$ choices for the last digit (to make the number odd)


    • $8$ choices left for the first digit (since $0$ can't be used)


    • $8$ choices still left for the second digit (since $0$ is now available for use)



    Thus, without breaking up into cases, we get $5cdot8cdot8 = 320$






    share|cite|improve this answer











    $endgroup$


















      13












      $begingroup$

      Problems of this type are often solved more simply if we place the digits in a suitable order, instead of necessarily left to right.



      Here we'll place the digits in the order last-first-second, i.e. in the order $Units-Hundreds-Tens$




      • $5$ choices for the last digit (to make the number odd)


      • $8$ choices left for the first digit (since $0$ can't be used)


      • $8$ choices still left for the second digit (since $0$ is now available for use)



      Thus, without breaking up into cases, we get $5cdot8cdot8 = 320$






      share|cite|improve this answer











      $endgroup$
















        13












        13








        13





        $begingroup$

        Problems of this type are often solved more simply if we place the digits in a suitable order, instead of necessarily left to right.



        Here we'll place the digits in the order last-first-second, i.e. in the order $Units-Hundreds-Tens$




        • $5$ choices for the last digit (to make the number odd)


        • $8$ choices left for the first digit (since $0$ can't be used)


        • $8$ choices still left for the second digit (since $0$ is now available for use)



        Thus, without breaking up into cases, we get $5cdot8cdot8 = 320$






        share|cite|improve this answer











        $endgroup$



        Problems of this type are often solved more simply if we place the digits in a suitable order, instead of necessarily left to right.



        Here we'll place the digits in the order last-first-second, i.e. in the order $Units-Hundreds-Tens$




        • $5$ choices for the last digit (to make the number odd)


        • $8$ choices left for the first digit (since $0$ can't be used)


        • $8$ choices still left for the second digit (since $0$ is now available for use)



        Thus, without breaking up into cases, we get $5cdot8cdot8 = 320$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jul 9 '17 at 15:15

























        answered Jul 9 '17 at 9:16









        true blue aniltrue blue anil

        20.5k11941




        20.5k11941























            10












            $begingroup$

            The mistake is you might have used $0$ for the middle digit. So there are two cases.



            Case ($1$): Using $0$ for the middle digit:



            $qquad$The count is $5,{times},1,{times},8=40$.



            Case ($2$): Not using $0$ for the middle digit:



            $qquad$The count is $5,{times},8,{times},7=280$.



            So you get a total count of $40+280=320$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thanks for the answer, but can you explain why having 0 for middle digits makes difference.
              $endgroup$
              – someone123123
              Jul 9 '17 at 8:44






            • 6




              $begingroup$
              Because if you use $0$ for the middle digit, the leading digit has only two exclusions, not three.
              $endgroup$
              – quasi
              Jul 9 '17 at 8:46


















            10












            $begingroup$

            The mistake is you might have used $0$ for the middle digit. So there are two cases.



            Case ($1$): Using $0$ for the middle digit:



            $qquad$The count is $5,{times},1,{times},8=40$.



            Case ($2$): Not using $0$ for the middle digit:



            $qquad$The count is $5,{times},8,{times},7=280$.



            So you get a total count of $40+280=320$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thanks for the answer, but can you explain why having 0 for middle digits makes difference.
              $endgroup$
              – someone123123
              Jul 9 '17 at 8:44






            • 6




              $begingroup$
              Because if you use $0$ for the middle digit, the leading digit has only two exclusions, not three.
              $endgroup$
              – quasi
              Jul 9 '17 at 8:46
















            10












            10








            10





            $begingroup$

            The mistake is you might have used $0$ for the middle digit. So there are two cases.



            Case ($1$): Using $0$ for the middle digit:



            $qquad$The count is $5,{times},1,{times},8=40$.



            Case ($2$): Not using $0$ for the middle digit:



            $qquad$The count is $5,{times},8,{times},7=280$.



            So you get a total count of $40+280=320$.






            share|cite|improve this answer











            $endgroup$



            The mistake is you might have used $0$ for the middle digit. So there are two cases.



            Case ($1$): Using $0$ for the middle digit:



            $qquad$The count is $5,{times},1,{times},8=40$.



            Case ($2$): Not using $0$ for the middle digit:



            $qquad$The count is $5,{times},8,{times},7=280$.



            So you get a total count of $40+280=320$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jul 9 '17 at 8:45

























            answered Jul 9 '17 at 8:41









            quasiquasi

            36k22663




            36k22663












            • $begingroup$
              Thanks for the answer, but can you explain why having 0 for middle digits makes difference.
              $endgroup$
              – someone123123
              Jul 9 '17 at 8:44






            • 6




              $begingroup$
              Because if you use $0$ for the middle digit, the leading digit has only two exclusions, not three.
              $endgroup$
              – quasi
              Jul 9 '17 at 8:46




















            • $begingroup$
              Thanks for the answer, but can you explain why having 0 for middle digits makes difference.
              $endgroup$
              – someone123123
              Jul 9 '17 at 8:44






            • 6




              $begingroup$
              Because if you use $0$ for the middle digit, the leading digit has only two exclusions, not three.
              $endgroup$
              – quasi
              Jul 9 '17 at 8:46


















            $begingroup$
            Thanks for the answer, but can you explain why having 0 for middle digits makes difference.
            $endgroup$
            – someone123123
            Jul 9 '17 at 8:44




            $begingroup$
            Thanks for the answer, but can you explain why having 0 for middle digits makes difference.
            $endgroup$
            – someone123123
            Jul 9 '17 at 8:44




            6




            6




            $begingroup$
            Because if you use $0$ for the middle digit, the leading digit has only two exclusions, not three.
            $endgroup$
            – quasi
            Jul 9 '17 at 8:46






            $begingroup$
            Because if you use $0$ for the middle digit, the leading digit has only two exclusions, not three.
            $endgroup$
            – quasi
            Jul 9 '17 at 8:46













            1












            $begingroup$

            Since this problem is only created by one single digit i.e. zero as you can't put zero on hundred's place like 013 which is not 3 digit number so when you put zero on the ten's place i.e. in the middle you will get 8 remaining digits for hundred place. Therefore we have two cases which are:
            Case (1): Using 0 for the middle digit.



            The count is 5×1×8=40.



            Case (2): Not using 0 for the middle digit.



            The count is 5×8×7=280.



            So you get a total count of 40+280=320.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Welcome to MSE. Your answer adds nothing new to the already existing answers.
              $endgroup$
              – José Carlos Santos
              Jan 17 at 17:54
















            1












            $begingroup$

            Since this problem is only created by one single digit i.e. zero as you can't put zero on hundred's place like 013 which is not 3 digit number so when you put zero on the ten's place i.e. in the middle you will get 8 remaining digits for hundred place. Therefore we have two cases which are:
            Case (1): Using 0 for the middle digit.



            The count is 5×1×8=40.



            Case (2): Not using 0 for the middle digit.



            The count is 5×8×7=280.



            So you get a total count of 40+280=320.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Welcome to MSE. Your answer adds nothing new to the already existing answers.
              $endgroup$
              – José Carlos Santos
              Jan 17 at 17:54














            1












            1








            1





            $begingroup$

            Since this problem is only created by one single digit i.e. zero as you can't put zero on hundred's place like 013 which is not 3 digit number so when you put zero on the ten's place i.e. in the middle you will get 8 remaining digits for hundred place. Therefore we have two cases which are:
            Case (1): Using 0 for the middle digit.



            The count is 5×1×8=40.



            Case (2): Not using 0 for the middle digit.



            The count is 5×8×7=280.



            So you get a total count of 40+280=320.






            share|cite|improve this answer









            $endgroup$



            Since this problem is only created by one single digit i.e. zero as you can't put zero on hundred's place like 013 which is not 3 digit number so when you put zero on the ten's place i.e. in the middle you will get 8 remaining digits for hundred place. Therefore we have two cases which are:
            Case (1): Using 0 for the middle digit.



            The count is 5×1×8=40.



            Case (2): Not using 0 for the middle digit.



            The count is 5×8×7=280.



            So you get a total count of 40+280=320.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 17 at 17:26









            AJ WARRIORAJ WARRIOR

            112




            112












            • $begingroup$
              Welcome to MSE. Your answer adds nothing new to the already existing answers.
              $endgroup$
              – José Carlos Santos
              Jan 17 at 17:54


















            • $begingroup$
              Welcome to MSE. Your answer adds nothing new to the already existing answers.
              $endgroup$
              – José Carlos Santos
              Jan 17 at 17:54
















            $begingroup$
            Welcome to MSE. Your answer adds nothing new to the already existing answers.
            $endgroup$
            – José Carlos Santos
            Jan 17 at 17:54




            $begingroup$
            Welcome to MSE. Your answer adds nothing new to the already existing answers.
            $endgroup$
            – José Carlos Santos
            Jan 17 at 17:54











            -1












            $begingroup$

            Three digit odd numbers implies the numbers would only be made of digits 1 , 3 , 5 , 7 , 9



            With repetition of digits we would have had 5 * 5 * 5 = 125



            But for every hundreds, maximum of 4 tens is possible (avoiding the duplicate of digit used in hundreds)



            And each of these 4 tens, maximum of 3 units is possible (avoiding the duplicate of digits used in tens)




            5 * 4 * 3 = 60 is my answer






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              I think you didn't understand the question, the question says how many three-digits numbers are there such that they are odd and they have three different digits. The first two digits don't have to be odd.
              $endgroup$
              – someone123123
              Jul 12 '17 at 10:13
















            -1












            $begingroup$

            Three digit odd numbers implies the numbers would only be made of digits 1 , 3 , 5 , 7 , 9



            With repetition of digits we would have had 5 * 5 * 5 = 125



            But for every hundreds, maximum of 4 tens is possible (avoiding the duplicate of digit used in hundreds)



            And each of these 4 tens, maximum of 3 units is possible (avoiding the duplicate of digits used in tens)




            5 * 4 * 3 = 60 is my answer






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              I think you didn't understand the question, the question says how many three-digits numbers are there such that they are odd and they have three different digits. The first two digits don't have to be odd.
              $endgroup$
              – someone123123
              Jul 12 '17 at 10:13














            -1












            -1








            -1





            $begingroup$

            Three digit odd numbers implies the numbers would only be made of digits 1 , 3 , 5 , 7 , 9



            With repetition of digits we would have had 5 * 5 * 5 = 125



            But for every hundreds, maximum of 4 tens is possible (avoiding the duplicate of digit used in hundreds)



            And each of these 4 tens, maximum of 3 units is possible (avoiding the duplicate of digits used in tens)




            5 * 4 * 3 = 60 is my answer






            share|cite|improve this answer









            $endgroup$



            Three digit odd numbers implies the numbers would only be made of digits 1 , 3 , 5 , 7 , 9



            With repetition of digits we would have had 5 * 5 * 5 = 125



            But for every hundreds, maximum of 4 tens is possible (avoiding the duplicate of digit used in hundreds)



            And each of these 4 tens, maximum of 3 units is possible (avoiding the duplicate of digits used in tens)




            5 * 4 * 3 = 60 is my answer







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jul 11 '17 at 21:57









            korwalskiykorwalskiy

            1091




            1091








            • 1




              $begingroup$
              I think you didn't understand the question, the question says how many three-digits numbers are there such that they are odd and they have three different digits. The first two digits don't have to be odd.
              $endgroup$
              – someone123123
              Jul 12 '17 at 10:13














            • 1




              $begingroup$
              I think you didn't understand the question, the question says how many three-digits numbers are there such that they are odd and they have three different digits. The first two digits don't have to be odd.
              $endgroup$
              – someone123123
              Jul 12 '17 at 10:13








            1




            1




            $begingroup$
            I think you didn't understand the question, the question says how many three-digits numbers are there such that they are odd and they have three different digits. The first two digits don't have to be odd.
            $endgroup$
            – someone123123
            Jul 12 '17 at 10:13




            $begingroup$
            I think you didn't understand the question, the question says how many three-digits numbers are there such that they are odd and they have three different digits. The first two digits don't have to be odd.
            $endgroup$
            – someone123123
            Jul 12 '17 at 10:13


















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