Bsrgvty

What are the applications of the Mean Value Theorem?

Why does Docker hub allow non-official Docker images and why do they exist?

Why combine commands on a single line in Bash script?

Is it possible to have a healthy work-life balance as a professor?

Do animals summoned by the Conjure Animals spell get extra HP from Circle of the Shepherd's Spirit Totem ability?

Travel with Expired Greek Passport from UK to Greece and Return

How can I find the weakness of our "Quality Assurance Process"?

How does an aircraft descend without its nose pointing down?

How would a young girl/boy (about 14) who never gets old survive in the 16th century?

Echo bracket symbol to terminal

Is there a problem using A LOT of locks in a process?

Is velocity a valid measure of team and process improvement?

What is the best focal length for DSLR film negative scan?

Why is the Falcon Heavy center core recovery done at sea?

Run "cd" command as superuser in Linux

Can a stolen Android phone with USB debugging enabled have screen lock bypassed?

How to help my son improve without being discouraging?

Why did Grima shed a tear?

Are effects pedals overkill for acoustic instruments?

Is is likely that my lack of post-secondary education is holding my resume back?

What are the downsides of Hodl Invoices?

18-month-old kicked out of church nursery

What does "speed checked" mean?

Reviewer wants me to do massive amount of work, the result would be a different article. Should I tell that to the editor?

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

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?

mathematics number-theory

share|improve this question

edited Jun 9 at 19:27

Rand al'Thor

76.7k1515 gold badges252252 silver badges506506 bronze badges

asked Jun 6 at 9:28

Siddharth GargSiddharth Garg

5166 bronze badges

$endgroup$

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
  

  
  
  
  

add a comment
 | 

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?

mathematics number-theory

share|improve this question

edited Jun 9 at 19:27

Rand al'Thor

76.7k1515 gold badges252252 silver badges506506 bronze badges

asked Jun 6 at 9:28

Siddharth GargSiddharth Garg

5166 bronze badges

$endgroup$

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
  

  
  
  
  

add a comment
 | 

$begingroup$

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?

mathematics number-theory

share|improve this question

edited Jun 9 at 19:27

Rand al'Thor

76.7k1515 gold badges252252 silver badges506506 bronze badges

asked Jun 6 at 9:28

Siddharth GargSiddharth Garg

5166 bronze badges

$endgroup$

share|improve this question

edited Jun 9 at 19:27

Rand al'Thor

76.7k1515 gold badges252252 silver badges506506 bronze badges

asked Jun 6 at 9:28

Siddharth GargSiddharth Garg

5166 bronze badges

share|improve this question

edited Jun 9 at 19:27

Rand al'Thor

76.7k1515 gold badges252252 silver badges506506 bronze badges

asked Jun 6 at 9:28

Siddharth GargSiddharth Garg

5166 bronze badges

edited Jun 9 at 19:27

Rand al'Thor

76.7k1515 gold badges252252 silver badges506506 bronze badges

edited Jun 9 at 19:27

Rand al'Thor

76.7k1515 gold badges252252 silver badges506506 bronze badges

asked Jun 6 at 9:28

Siddharth GargSiddharth Garg

5166 bronze badges

asked Jun 6 at 9:28

Siddharth GargSiddharth Garg

5166 bronze badges

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 $9=3^2$.

2. 
   

   $2M5$ 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.

   
   



3. 
   

   
   

   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

answered Jun 6 at 9:41

Rand al'ThorRand al'Thor

76.7k1515 gold badges252252 silver badges506506 bronze badges

$endgroup$

• 
  
  
  

  
  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
  
  

  
  
  
  

add a comment
 | 

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

edited Jun 6 at 10:02

answered Jun 6 at 9:57

Weather VaneWeather Vane

6,46611 gold badge44 silver badges2626 bronze badges

$endgroup$

add a comment
 | 

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 $9=3^2$.

2. 
   

   $2M5$ 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.

   
   



3. 
   

   
   

   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

answered Jun 6 at 9:41

Rand al'ThorRand al'Thor

76.7k1515 gold badges252252 silver badges506506 bronze badges

$endgroup$

• 
  
  
  

  
  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
  
  

  
  
  
  

add a comment
 | 

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 $9=3^2$.

2. 
   

   $2M5$ 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.

   
   



3. 
   

   
   

   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

answered Jun 6 at 9:41

Rand al'ThorRand al'Thor

76.7k1515 gold badges252252 silver badges506506 bronze badges

$endgroup$

• 
  
  
  

  
  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
  
  

  
  
  
  

add a comment
 | 

$begingroup$

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 $9=3^2$.

2. 
   

   $2M5$ 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.

   
   



3. 
   

   
   

   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

answered Jun 6 at 9:41

Rand al'ThorRand al'Thor

76.7k1515 gold badges252252 silver badges506506 bronze badges

$endgroup$

share|improve this answer

answered Jun 6 at 9:41

Rand al'ThorRand al'Thor

76.7k1515 gold badges252252 silver badges506506 bronze badges

answered Jun 6 at 9:41

Rand al'ThorRand al'Thor

76.7k1515 gold badges252252 silver badges506506 bronze badges

answered Jun 6 at 9:41

Rand al'ThorRand al'Thor

76.7k1515 gold badges252252 silver badges506506 bronze badges

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

edited Jun 6 at 10:02

answered Jun 6 at 9:57

Weather VaneWeather Vane

6,46611 gold badge44 silver badges2626 bronze badges

$endgroup$

add a comment
 | 

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

edited Jun 6 at 10:02

answered Jun 6 at 9:57

Weather VaneWeather Vane

6,46611 gold badge44 silver badges2626 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

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

edited Jun 6 at 10:02

answered Jun 6 at 9:57

Weather VaneWeather Vane

6,46611 gold badge44 silver badges2626 bronze badges

$endgroup$

share|improve this answer

edited Jun 6 at 10:02

answered Jun 6 at 9:57

Weather VaneWeather Vane

6,46611 gold badge44 silver badges2626 bronze badges

answered Jun 6 at 9:57

Weather VaneWeather Vane

6,46611 gold badge44 silver badges2626 bronze badges

answered Jun 6 at 9:57

Weather VaneWeather Vane

6,46611 gold badge44 silver badges2626 bronze badges