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 GameLock OptimisationThree mathematicians are forever in PrisonThe Calculator with Misbehaving `+` and `*`Megan and the parcels of tilesPFG: A pretty spiral!Help me visit my friend through his new digital key lockHow many trees and birds are there?How many ordered pairs (a,b) satisfy a^2=b^3+1, where a and b are integers?
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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 GameLock OptimisationThree mathematicians are forever in PrisonThe Calculator with Misbehaving `+` and `*`Megan and the parcels of tilesPFG: A pretty spiral!Help me visit my friend through his new digital key lockHow many trees and birds are there?How many ordered pairs (a,b) satisfy a^2=b^3+1, where a and b are integers?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom: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
edited Jun 9 at 19:27
Rand al'Thor
73.7k15 gold badges244 silver badges490 bronze badges
asked Jun 6 at 9:28
Siddharth GargSiddharth Garg
516 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
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
edited Jun 9 at 19:27
Rand al'Thor
73.7k15 gold badges244 silver badges490 bronze badges
asked Jun 6 at 9:28
Siddharth GargSiddharth Garg
516 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
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
edited Jun 9 at 19:27
Rand al'Thor
73.7k15 gold badges244 silver badges490 bronze badges
asked Jun 6 at 9:28
Siddharth GargSiddharth Garg
516 bronze badges
$endgroup$
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
mathematics number-theory
edited Jun 9 at 19:27
Rand al'Thor
73.7k15 gold badges244 silver badges490 bronze badges
asked Jun 6 at 9:28
Siddharth GargSiddharth Garg
516 bronze badges
edited Jun 9 at 19:27
Rand al'Thor
73.7k15 gold badges244 silver badges490 bronze badges
asked Jun 6 at 9:28
Siddharth GargSiddharth Garg
516 bronze badges
edited Jun 9 at 19:27
Rand al'Thor
73.7k15 gold badges244 silver badges490 bronze badges
edited Jun 9 at 19:27
Rand al'Thor
73.7k15 gold badges244 silver badges490 bronze badges
edited Jun 9 at 19:27
Rand al'Thor
73.7k15 gold badges244 silver badges490 bronze badges
73.7k15 gold badges244 silver badges490 bronze badges
asked Jun 6 at 9:28
Siddharth GargSiddharth Garg
516 bronze badges
asked Jun 6 at 9:28
Siddharth GargSiddharth Garg
516 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
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
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 |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $9=3^2$.
$2M5$ 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.
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.
answered Jun 6 at 9:41
Rand al'ThorRand al'Thor
73.7k15 gold badges244 silver badges490 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 |
$begingroup$
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
answered Jun 6 at 9:57
Weather VaneWeather Vane
4,1321 gold badge1 silver badge19 bronze badges
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $9=3^2$.
$2M5$ 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.
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.
answered Jun 6 at 9:41
Rand al'ThorRand al'Thor
73.7k15 gold badges244 silver badges490 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 $9=3^2$.
$2M5$ 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.
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.
answered Jun 6 at 9:41
Rand al'ThorRand al'Thor
73.7k15 gold badges244 silver badges490 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 $9=3^2$.
$2M5$ 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.
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.
answered Jun 6 at 9:41
Rand al'ThorRand al'Thor
73.7k15 gold badges244 silver badges490 bronze badges
$endgroup$
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 $9=3^2$.
$2M5$ 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.
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.
answered Jun 6 at 9:41
Rand al'ThorRand al'Thor
73.7k15 gold badges244 silver badges490 bronze badges
answered Jun 6 at 9:41
Rand al'ThorRand al'Thor
73.7k15 gold badges244 silver badges490 bronze badges
answered Jun 6 at 9:41
Rand al'ThorRand al'Thor
73.7k15 gold badges244 silver badges490 bronze badges
answered Jun 6 at 9:41
Rand al'ThorRand al'Thor
73.7k15 gold badges244 silver badges490 bronze badges
73.7k15 gold badges244 silver badges490 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
add a comment |
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
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
answered Jun 6 at 9:57
Weather VaneWeather Vane
4,1321 gold badge1 silver badge19 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
answered Jun 6 at 9:57
Weather VaneWeather Vane
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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
answered Jun 6 at 9:57
Weather VaneWeather Vane
4,1321 gold badge1 silver badge19 bronze badges
$endgroup$
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
answered Jun 6 at 9:57
Weather VaneWeather Vane
4,1321 gold badge1 silver badge19 bronze badges
edited Jun 6 at 10:02
answered Jun 6 at 9:57
Weather VaneWeather Vane
4,1321 gold badge1 silver badge19 bronze badges
answered Jun 6 at 9:57
Weather VaneWeather Vane
4,1321 gold badge1 silver badge19 bronze badges
answered Jun 6 at 9:57
Weather VaneWeather Vane
4,1321 gold badge1 silver badge19 bronze badges
4,1321 gold badge1 silver badge19 bronze badges
add a comment |
add a comment |
1
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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)?
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– Rand al'Thor
Jun 6 at 9:32
1
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Also, what is your source for these questions?
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– Rand al'Thor
Jun 6 at 9:33
1
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Or power $2M^5$?
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– Weather Vane
Jun 6 at 9:38
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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.
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– Rand al'Thor
Jun 6 at 15:36
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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.
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– Siddharth Garg
Jun 6 at 16:37