Let M and N be single-digit integers. If the product 2M5 x 13N is divisible by 36, how many ordered pairs (M,N) are possible? [closed]Unique Licence PlatesThe Trickster's GameThree mathematicians are forever in PrisonThe Calculator with Misbehaving `+` and `*`Megan and the parcels of tilesPFG: A pretty spiral!sum and gcd june19 challenge!How many codes are possible?How to solve the input output puzzle?

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Let M and N be single-digit integers. If the product 2M5 x 13N is divisible by 36, how many ordered pairs (M,N) are possible? [closed]


Unique Licence PlatesThe Trickster's GameThree mathematicians are forever in PrisonThe Calculator with Misbehaving `+` and `*`Megan and the parcels of tilesPFG: A pretty spiral!sum and gcd june19 challenge!How many codes are possible?How to solve the input output puzzle?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;

.everyonelovesstackoverflowposition:absolute;height:1px;width:1px;opacity:0;top:0;left:0;pointer-events:none;








3














$begingroup$



Let M and N be single-digit integers. If the product 2M5 x 13N is divisible by 36, how many ordered pairs (M,N) are possible?



-- source




I tried it by reducing 36 into its positive factors (1,2,3,4,6,9,18,36) and then solving, but I got way too many pairs.
Can somebody help?










share|improve this question












$endgroup$





closed as off-topic by elias, gabbo1092, Glorfindel, Rupert Morrish, athin Jun 9 at 22:38


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – elias, gabbo1092, Glorfindel, Rupert Morrish, athin
If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    What do you mean by $2M5$ and $13N$? Are these concatenated numbers (two hundred something and five, one hundred and thirty something) or products (two times M times five, thirteen times N)?
    $endgroup$
    – Rand al'Thor
    Jun 6 at 9:32






  • 1




    $begingroup$
    Also, what is your source for these questions?
    $endgroup$
    – Rand al'Thor
    Jun 6 at 9:33






  • 1




    $begingroup$
    Or power $2M^5$?
    $endgroup$
    – Weather Vane
    Jun 6 at 9:38










  • $begingroup$
    To the close-voters: as the first answerer here, I think this is mathematically "interesting" enough not to be closed. It requires some knowledge of basic number theory (prime factorisations) if you want to do it in a quick neat way rather than brute-forcing all possibilities.
    $endgroup$
    – Rand al'Thor
    Jun 6 at 15:36











  • $begingroup$
    Sorry for answering so late. Yes, @Randal'Thor, 2M5 and 13N are concatenated numbers. And as for the source of these questions, I have a book.
    $endgroup$
    – Siddharth Garg
    Jun 6 at 16:37

















3














$begingroup$



Let M and N be single-digit integers. If the product 2M5 x 13N is divisible by 36, how many ordered pairs (M,N) are possible?



-- source




I tried it by reducing 36 into its positive factors (1,2,3,4,6,9,18,36) and then solving, but I got way too many pairs.
Can somebody help?










share|improve this question












$endgroup$





closed as off-topic by elias, gabbo1092, Glorfindel, Rupert Morrish, athin Jun 9 at 22:38


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – elias, gabbo1092, Glorfindel, Rupert Morrish, athin
If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    What do you mean by $2M5$ and $13N$? Are these concatenated numbers (two hundred something and five, one hundred and thirty something) or products (two times M times five, thirteen times N)?
    $endgroup$
    – Rand al'Thor
    Jun 6 at 9:32






  • 1




    $begingroup$
    Also, what is your source for these questions?
    $endgroup$
    – Rand al'Thor
    Jun 6 at 9:33






  • 1




    $begingroup$
    Or power $2M^5$?
    $endgroup$
    – Weather Vane
    Jun 6 at 9:38










  • $begingroup$
    To the close-voters: as the first answerer here, I think this is mathematically "interesting" enough not to be closed. It requires some knowledge of basic number theory (prime factorisations) if you want to do it in a quick neat way rather than brute-forcing all possibilities.
    $endgroup$
    – Rand al'Thor
    Jun 6 at 15:36











  • $begingroup$
    Sorry for answering so late. Yes, @Randal'Thor, 2M5 and 13N are concatenated numbers. And as for the source of these questions, I have a book.
    $endgroup$
    – Siddharth Garg
    Jun 6 at 16:37













3












3








3


0



$begingroup$



Let M and N be single-digit integers. If the product 2M5 x 13N is divisible by 36, how many ordered pairs (M,N) are possible?



-- source




I tried it by reducing 36 into its positive factors (1,2,3,4,6,9,18,36) and then solving, but I got way too many pairs.
Can somebody help?










share|improve this question












$endgroup$





Let M and N be single-digit integers. If the product 2M5 x 13N is divisible by 36, how many ordered pairs (M,N) are possible?



-- source




I tried it by reducing 36 into its positive factors (1,2,3,4,6,9,18,36) and then solving, but I got way too many pairs.
Can somebody help?







mathematics number-theory






share|improve this question
















share|improve this question













share|improve this question




share|improve this question








edited Jun 9 at 19:27









Rand al'Thor

76.7k15 gold badges252 silver badges506 bronze badges




76.7k15 gold badges252 silver badges506 bronze badges










asked Jun 6 at 9:28









Siddharth GargSiddharth Garg

516 bronze badges




516 bronze badges





closed as off-topic by elias, gabbo1092, Glorfindel, Rupert Morrish, athin Jun 9 at 22:38


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – elias, gabbo1092, Glorfindel, Rupert Morrish, athin
If this question can be reworded to fit the rules in the help center, please edit the question.









closed as off-topic by elias, gabbo1092, Glorfindel, Rupert Morrish, athin Jun 9 at 22:38


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – elias, gabbo1092, Glorfindel, Rupert Morrish, athin
If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by elias, gabbo1092, Glorfindel, Rupert Morrish, athin Jun 9 at 22:38


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – elias, gabbo1092, Glorfindel, Rupert Morrish, athin
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 1




    $begingroup$
    What do you mean by $2M5$ and $13N$? Are these concatenated numbers (two hundred something and five, one hundred and thirty something) or products (two times M times five, thirteen times N)?
    $endgroup$
    – Rand al'Thor
    Jun 6 at 9:32






  • 1




    $begingroup$
    Also, what is your source for these questions?
    $endgroup$
    – Rand al'Thor
    Jun 6 at 9:33






  • 1




    $begingroup$
    Or power $2M^5$?
    $endgroup$
    – Weather Vane
    Jun 6 at 9:38










  • $begingroup$
    To the close-voters: as the first answerer here, I think this is mathematically "interesting" enough not to be closed. It requires some knowledge of basic number theory (prime factorisations) if you want to do it in a quick neat way rather than brute-forcing all possibilities.
    $endgroup$
    – Rand al'Thor
    Jun 6 at 15:36











  • $begingroup$
    Sorry for answering so late. Yes, @Randal'Thor, 2M5 and 13N are concatenated numbers. And as for the source of these questions, I have a book.
    $endgroup$
    – Siddharth Garg
    Jun 6 at 16:37












  • 1




    $begingroup$
    What do you mean by $2M5$ and $13N$? Are these concatenated numbers (two hundred something and five, one hundred and thirty something) or products (two times M times five, thirteen times N)?
    $endgroup$
    – Rand al'Thor
    Jun 6 at 9:32






  • 1




    $begingroup$
    Also, what is your source for these questions?
    $endgroup$
    – Rand al'Thor
    Jun 6 at 9:33






  • 1




    $begingroup$
    Or power $2M^5$?
    $endgroup$
    – Weather Vane
    Jun 6 at 9:38










  • $begingroup$
    To the close-voters: as the first answerer here, I think this is mathematically "interesting" enough not to be closed. It requires some knowledge of basic number theory (prime factorisations) if you want to do it in a quick neat way rather than brute-forcing all possibilities.
    $endgroup$
    – Rand al'Thor
    Jun 6 at 15:36











  • $begingroup$
    Sorry for answering so late. Yes, @Randal'Thor, 2M5 and 13N are concatenated numbers. And as for the source of these questions, I have a book.
    $endgroup$
    – Siddharth Garg
    Jun 6 at 16:37







1




1




$begingroup$
What do you mean by $2M5$ and $13N$? Are these concatenated numbers (two hundred something and five, one hundred and thirty something) or products (two times M times five, thirteen times N)?
$endgroup$
– Rand al'Thor
Jun 6 at 9:32




$begingroup$
What do you mean by $2M5$ and $13N$? Are these concatenated numbers (two hundred something and five, one hundred and thirty something) or products (two times M times five, thirteen times N)?
$endgroup$
– Rand al'Thor
Jun 6 at 9:32




1




1




$begingroup$
Also, what is your source for these questions?
$endgroup$
– Rand al'Thor
Jun 6 at 9:33




$begingroup$
Also, what is your source for these questions?
$endgroup$
– Rand al'Thor
Jun 6 at 9:33




1




1




$begingroup$
Or power $2M^5$?
$endgroup$
– Weather Vane
Jun 6 at 9:38




$begingroup$
Or power $2M^5$?
$endgroup$
– Weather Vane
Jun 6 at 9:38












$begingroup$
To the close-voters: as the first answerer here, I think this is mathematically "interesting" enough not to be closed. It requires some knowledge of basic number theory (prime factorisations) if you want to do it in a quick neat way rather than brute-forcing all possibilities.
$endgroup$
– Rand al'Thor
Jun 6 at 15:36





$begingroup$
To the close-voters: as the first answerer here, I think this is mathematically "interesting" enough not to be closed. It requires some knowledge of basic number theory (prime factorisations) if you want to do it in a quick neat way rather than brute-forcing all possibilities.
$endgroup$
– Rand al'Thor
Jun 6 at 15:36













$begingroup$
Sorry for answering so late. Yes, @Randal'Thor, 2M5 and 13N are concatenated numbers. And as for the source of these questions, I have a book.
$endgroup$
– Siddharth Garg
Jun 6 at 16:37




$begingroup$
Sorry for answering so late. Yes, @Randal'Thor, 2M5 and 13N are concatenated numbers. And as for the source of these questions, I have a book.
$endgroup$
– Siddharth Garg
Jun 6 at 16:37










2 Answers
2






active

oldest

votes


















11
















$begingroup$

I'll assume here that the number $2M5$ and $13N$ are concatenated three-digit numbers rather than products, because if they are products then the question is trivial.



Being divisible by 36 is equivalent to




being divisible by both of its prime factors $4=2^2$ and $9=3^2$.





  1. $2M5$ is odd, so both factors of 2 must come from $13N$. That means $3N$ must be a multiple of 4, so either $N=2$ or $N=6$. In either of these cases, the product is definitely a multiple of 4 - it's an "if and only if" condition.





  2. If the product is divisible by 9, then either both factors are multiples of 3 (i.e. their digit sums are multiples of 3) or one of the two factors is a multiple of 9 (i.e. its digit sum is a multiple of 9).




    Let's consider the two cases from the first numbered point above:




    • If $N=2$, then $132$ is divisible by 3 but not 9, so we need $2M5$ divisible by 3, i.e. $2+M+5$ divisible by 3, which means either $M=2$ or $M=5$ or $M=8$. In all of these cases, the product is definitely a multiple of 9.




    • If $N=6$, then $136$ is not divisible by 3, so we need $2M5$ divisible by 9, i.e. $2+M+5$ divisible by 9, which means $M=2$. Again, in this case the product is definitely a multiple of 9.




So the possibilities for the pair $(M,N)$ are:




$(2,2),(5,2),(8,2),(2,6)$ - four possibilities in all.







share|improve this answer










$endgroup$










  • 1




    $begingroup$
    @WeatherVane It's nothing to do with "you" or "me" - don't take it personally. I don't like answers which use computers instead of elegant mathematics, but i don't go around downvoting or deleting them either (unless they're on questions tagged no-computers, which this isn't). On this site, we're commenting/voting on content not people, as you should know :-)
    $endgroup$
    – Rand al'Thor
    Jun 6 at 10:55










  • $begingroup$
    Just to balance the comment you decided to leave, nobody downvoted or deleted a question, or suggested the mutual voting you accused me of requesting. You could have deleted it with the others, but for some reason decided not to.
    $endgroup$
    – Weather Vane
    Jun 6 at 17:42



















5
















$begingroup$

Confirmation of the answer from Rand al'Thor using brute force with C - only 100 perms:






#include <stdio.h>

int main(void)

for(int M = 0; M < 10; M++)
for(int N = 0; N < 10; N++)
if(((205 + 10 * M) * (130 + N)) % 36 == 0)
printf("%d %dn", M, N);

2 2

2 6

5 2

8 2





share|improve this answer












$endgroup$






















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    11
















    $begingroup$

    I'll assume here that the number $2M5$ and $13N$ are concatenated three-digit numbers rather than products, because if they are products then the question is trivial.



    Being divisible by 36 is equivalent to




    being divisible by both of its prime factors $4=2^2$ and $9=3^2$.





    1. $2M5$ is odd, so both factors of 2 must come from $13N$. That means $3N$ must be a multiple of 4, so either $N=2$ or $N=6$. In either of these cases, the product is definitely a multiple of 4 - it's an "if and only if" condition.





    2. If the product is divisible by 9, then either both factors are multiples of 3 (i.e. their digit sums are multiples of 3) or one of the two factors is a multiple of 9 (i.e. its digit sum is a multiple of 9).




      Let's consider the two cases from the first numbered point above:




      • If $N=2$, then $132$ is divisible by 3 but not 9, so we need $2M5$ divisible by 3, i.e. $2+M+5$ divisible by 3, which means either $M=2$ or $M=5$ or $M=8$. In all of these cases, the product is definitely a multiple of 9.




      • If $N=6$, then $136$ is not divisible by 3, so we need $2M5$ divisible by 9, i.e. $2+M+5$ divisible by 9, which means $M=2$. Again, in this case the product is definitely a multiple of 9.




    So the possibilities for the pair $(M,N)$ are:




    $(2,2),(5,2),(8,2),(2,6)$ - four possibilities in all.







    share|improve this answer










    $endgroup$










    • 1




      $begingroup$
      @WeatherVane It's nothing to do with "you" or "me" - don't take it personally. I don't like answers which use computers instead of elegant mathematics, but i don't go around downvoting or deleting them either (unless they're on questions tagged no-computers, which this isn't). On this site, we're commenting/voting on content not people, as you should know :-)
      $endgroup$
      – Rand al'Thor
      Jun 6 at 10:55










    • $begingroup$
      Just to balance the comment you decided to leave, nobody downvoted or deleted a question, or suggested the mutual voting you accused me of requesting. You could have deleted it with the others, but for some reason decided not to.
      $endgroup$
      – Weather Vane
      Jun 6 at 17:42
















    11
















    $begingroup$

    I'll assume here that the number $2M5$ and $13N$ are concatenated three-digit numbers rather than products, because if they are products then the question is trivial.



    Being divisible by 36 is equivalent to




    being divisible by both of its prime factors $4=2^2$ and $9=3^2$.





    1. $2M5$ is odd, so both factors of 2 must come from $13N$. That means $3N$ must be a multiple of 4, so either $N=2$ or $N=6$. In either of these cases, the product is definitely a multiple of 4 - it's an "if and only if" condition.





    2. If the product is divisible by 9, then either both factors are multiples of 3 (i.e. their digit sums are multiples of 3) or one of the two factors is a multiple of 9 (i.e. its digit sum is a multiple of 9).




      Let's consider the two cases from the first numbered point above:




      • If $N=2$, then $132$ is divisible by 3 but not 9, so we need $2M5$ divisible by 3, i.e. $2+M+5$ divisible by 3, which means either $M=2$ or $M=5$ or $M=8$. In all of these cases, the product is definitely a multiple of 9.




      • If $N=6$, then $136$ is not divisible by 3, so we need $2M5$ divisible by 9, i.e. $2+M+5$ divisible by 9, which means $M=2$. Again, in this case the product is definitely a multiple of 9.




    So the possibilities for the pair $(M,N)$ are:




    $(2,2),(5,2),(8,2),(2,6)$ - four possibilities in all.







    share|improve this answer










    $endgroup$










    • 1




      $begingroup$
      @WeatherVane It's nothing to do with "you" or "me" - don't take it personally. I don't like answers which use computers instead of elegant mathematics, but i don't go around downvoting or deleting them either (unless they're on questions tagged no-computers, which this isn't). On this site, we're commenting/voting on content not people, as you should know :-)
      $endgroup$
      – Rand al'Thor
      Jun 6 at 10:55










    • $begingroup$
      Just to balance the comment you decided to leave, nobody downvoted or deleted a question, or suggested the mutual voting you accused me of requesting. You could have deleted it with the others, but for some reason decided not to.
      $endgroup$
      – Weather Vane
      Jun 6 at 17:42














    11














    11










    11







    $begingroup$

    I'll assume here that the number $2M5$ and $13N$ are concatenated three-digit numbers rather than products, because if they are products then the question is trivial.



    Being divisible by 36 is equivalent to




    being divisible by both of its prime factors $4=2^2$ and $9=3^2$.





    1. $2M5$ is odd, so both factors of 2 must come from $13N$. That means $3N$ must be a multiple of 4, so either $N=2$ or $N=6$. In either of these cases, the product is definitely a multiple of 4 - it's an "if and only if" condition.





    2. If the product is divisible by 9, then either both factors are multiples of 3 (i.e. their digit sums are multiples of 3) or one of the two factors is a multiple of 9 (i.e. its digit sum is a multiple of 9).




      Let's consider the two cases from the first numbered point above:




      • If $N=2$, then $132$ is divisible by 3 but not 9, so we need $2M5$ divisible by 3, i.e. $2+M+5$ divisible by 3, which means either $M=2$ or $M=5$ or $M=8$. In all of these cases, the product is definitely a multiple of 9.




      • If $N=6$, then $136$ is not divisible by 3, so we need $2M5$ divisible by 9, i.e. $2+M+5$ divisible by 9, which means $M=2$. Again, in this case the product is definitely a multiple of 9.




    So the possibilities for the pair $(M,N)$ are:




    $(2,2),(5,2),(8,2),(2,6)$ - four possibilities in all.







    share|improve this answer










    $endgroup$



    I'll assume here that the number $2M5$ and $13N$ are concatenated three-digit numbers rather than products, because if they are products then the question is trivial.



    Being divisible by 36 is equivalent to




    being divisible by both of its prime factors $4=2^2$ and $9=3^2$.





    1. $2M5$ is odd, so both factors of 2 must come from $13N$. That means $3N$ must be a multiple of 4, so either $N=2$ or $N=6$. In either of these cases, the product is definitely a multiple of 4 - it's an "if and only if" condition.





    2. If the product is divisible by 9, then either both factors are multiples of 3 (i.e. their digit sums are multiples of 3) or one of the two factors is a multiple of 9 (i.e. its digit sum is a multiple of 9).




      Let's consider the two cases from the first numbered point above:




      • If $N=2$, then $132$ is divisible by 3 but not 9, so we need $2M5$ divisible by 3, i.e. $2+M+5$ divisible by 3, which means either $M=2$ or $M=5$ or $M=8$. In all of these cases, the product is definitely a multiple of 9.




      • If $N=6$, then $136$ is not divisible by 3, so we need $2M5$ divisible by 9, i.e. $2+M+5$ divisible by 9, which means $M=2$. Again, in this case the product is definitely a multiple of 9.




    So the possibilities for the pair $(M,N)$ are:




    $(2,2),(5,2),(8,2),(2,6)$ - four possibilities in all.








    share|improve this answer













    share|improve this answer




    share|improve this answer










    answered Jun 6 at 9:41









    Rand al'ThorRand al'Thor

    76.7k15 gold badges252 silver badges506 bronze badges




    76.7k15 gold badges252 silver badges506 bronze badges










    • 1




      $begingroup$
      @WeatherVane It's nothing to do with "you" or "me" - don't take it personally. I don't like answers which use computers instead of elegant mathematics, but i don't go around downvoting or deleting them either (unless they're on questions tagged no-computers, which this isn't). On this site, we're commenting/voting on content not people, as you should know :-)
      $endgroup$
      – Rand al'Thor
      Jun 6 at 10:55










    • $begingroup$
      Just to balance the comment you decided to leave, nobody downvoted or deleted a question, or suggested the mutual voting you accused me of requesting. You could have deleted it with the others, but for some reason decided not to.
      $endgroup$
      – Weather Vane
      Jun 6 at 17:42













    • 1




      $begingroup$
      @WeatherVane It's nothing to do with "you" or "me" - don't take it personally. I don't like answers which use computers instead of elegant mathematics, but i don't go around downvoting or deleting them either (unless they're on questions tagged no-computers, which this isn't). On this site, we're commenting/voting on content not people, as you should know :-)
      $endgroup$
      – Rand al'Thor
      Jun 6 at 10:55










    • $begingroup$
      Just to balance the comment you decided to leave, nobody downvoted or deleted a question, or suggested the mutual voting you accused me of requesting. You could have deleted it with the others, but for some reason decided not to.
      $endgroup$
      – Weather Vane
      Jun 6 at 17:42








    1




    1




    $begingroup$
    @WeatherVane It's nothing to do with "you" or "me" - don't take it personally. I don't like answers which use computers instead of elegant mathematics, but i don't go around downvoting or deleting them either (unless they're on questions tagged no-computers, which this isn't). On this site, we're commenting/voting on content not people, as you should know :-)
    $endgroup$
    – Rand al'Thor
    Jun 6 at 10:55




    $begingroup$
    @WeatherVane It's nothing to do with "you" or "me" - don't take it personally. I don't like answers which use computers instead of elegant mathematics, but i don't go around downvoting or deleting them either (unless they're on questions tagged no-computers, which this isn't). On this site, we're commenting/voting on content not people, as you should know :-)
    $endgroup$
    – Rand al'Thor
    Jun 6 at 10:55












    $begingroup$
    Just to balance the comment you decided to leave, nobody downvoted or deleted a question, or suggested the mutual voting you accused me of requesting. You could have deleted it with the others, but for some reason decided not to.
    $endgroup$
    – Weather Vane
    Jun 6 at 17:42





    $begingroup$
    Just to balance the comment you decided to leave, nobody downvoted or deleted a question, or suggested the mutual voting you accused me of requesting. You could have deleted it with the others, but for some reason decided not to.
    $endgroup$
    – Weather Vane
    Jun 6 at 17:42














    5
















    $begingroup$

    Confirmation of the answer from Rand al'Thor using brute force with C - only 100 perms:






    #include <stdio.h>

    int main(void)

    for(int M = 0; M < 10; M++)
    for(int N = 0; N < 10; N++)
    if(((205 + 10 * M) * (130 + N)) % 36 == 0)
    printf("%d %dn", M, N);

    2 2

    2 6

    5 2

    8 2





    share|improve this answer












    $endgroup$



















      5
















      $begingroup$

      Confirmation of the answer from Rand al'Thor using brute force with C - only 100 perms:






      #include <stdio.h>

      int main(void)

      for(int M = 0; M < 10; M++)
      for(int N = 0; N < 10; N++)
      if(((205 + 10 * M) * (130 + N)) % 36 == 0)
      printf("%d %dn", M, N);

      2 2

      2 6

      5 2

      8 2





      share|improve this answer












      $endgroup$

















        5














        5










        5







        $begingroup$

        Confirmation of the answer from Rand al'Thor using brute force with C - only 100 perms:






        #include <stdio.h>

        int main(void)

        for(int M = 0; M < 10; M++)
        for(int N = 0; N < 10; N++)
        if(((205 + 10 * M) * (130 + N)) % 36 == 0)
        printf("%d %dn", M, N);

        2 2

        2 6

        5 2

        8 2





        share|improve this answer












        $endgroup$



        Confirmation of the answer from Rand al'Thor using brute force with C - only 100 perms:






        #include <stdio.h>

        int main(void)

        for(int M = 0; M < 10; M++)
        for(int N = 0; N < 10; N++)
        if(((205 + 10 * M) * (130 + N)) % 36 == 0)
        printf("%d %dn", M, N);

        2 2

        2 6

        5 2

        8 2






        share|improve this answer















        share|improve this answer




        share|improve this answer








        edited Jun 6 at 10:02

























        answered Jun 6 at 9:57









        Weather VaneWeather Vane

        6,4661 gold badge4 silver badges26 bronze badges




        6,4661 gold badge4 silver badges26 bronze badges
















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