How do I solve this problem using Permutation and Combination? [on hold]

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The sum of proper divisors of 72 (1 and 72 excluded) is
i. 195
ii.122
iii.194
iv. None of these

I have already solved it by adding the divisors (which was easy to do and the only approach I could think of), also the question was mentioned in a book whose only concern is Permutation and Combination so I am wondering if there is some another approach to this problem.

permutations combinations

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edited 14 hours ago

asked 15 hours ago

Ryu

82

put on hold as off-topic by Arnaud Mortier, Ethan Bolker, José Carlos Santos, amWhy, Strants 15 hours ago

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

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Arnaud Mortier, José Carlos Santos, amWhy, Strants

If this question can be reworded to fit the rules in the help center, please edit the question.

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  1
  

  
  
  
  
  
  

  
  This does not require permut/combinations. You only need to list the proper divisors of $72=2^3cdot3^2$ and add them up.
  – TheSimpliFire
  15 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The approach that does not depend on hand counting is given by the divisor function. It depends on the prime factorization of the number
  – Ross Millikan
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Welcome to stackexchange. That said, I'm voting to close this question because it's simple enough for you to have worked out the answer by listing all the divisors. When you ask a question here it's good practice to include what you've done so far and where you are stuck.
  – Ethan Bolker
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  +TheSimpliFire I need an approach that solves this using permutation and combination
  – Ryu
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  +Ethan The question is too easy to solve by listing all divisors but I want to know how this can be related with permutation and combination
  – Ryu
  15 hours ago
  
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

The sum of proper divisors of 72 (1 and 72 excluded) is
i. 195
ii.122
iii.194
iv. None of these

I have already solved it by adding the divisors (which was easy to do and the only approach I could think of), also the question was mentioned in a book whose only concern is Permutation and Combination so I am wondering if there is some another approach to this problem.

permutations combinations

share|cite|improve this question

edited 14 hours ago

asked 15 hours ago

Ryu

82

put on hold as off-topic by Arnaud Mortier, Ethan Bolker, José Carlos Santos, amWhy, Strants 15 hours ago

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

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Arnaud Mortier, José Carlos Santos, amWhy, Strants

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This does not require permut/combinations. You only need to list the proper divisors of $72=2^3cdot3^2$ and add them up.
  – TheSimpliFire
  15 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The approach that does not depend on hand counting is given by the divisor function. It depends on the prime factorization of the number
  – Ross Millikan
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Welcome to stackexchange. That said, I'm voting to close this question because it's simple enough for you to have worked out the answer by listing all the divisors. When you ask a question here it's good practice to include what you've done so far and where you are stuck.
  – Ethan Bolker
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  +TheSimpliFire I need an approach that solves this using permutation and combination
  – Ryu
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  +Ethan The question is too easy to solve by listing all divisors but I want to know how this can be related with permutation and combination
  – Ryu
  15 hours ago
  
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

up vote
-1
down vote

favorite

The sum of proper divisors of 72 (1 and 72 excluded) is
i. 195
ii.122
iii.194
iv. None of these

I have already solved it by adding the divisors (which was easy to do and the only approach I could think of), also the question was mentioned in a book whose only concern is Permutation and Combination so I am wondering if there is some another approach to this problem.

permutations combinations

share|cite|improve this question

edited 14 hours ago

asked 15 hours ago

Ryu

82

The sum of proper divisors of 72 (1 and 72 excluded) is
i. 195
ii.122
iii.194
iv. None of these

I have already solved it by adding the divisors (which was easy to do and the only approach I could think of), also the question was mentioned in a book whose only concern is Permutation and Combination so I am wondering if there is some another approach to this problem.

permutations combinations

share|cite|improve this question

edited 14 hours ago

asked 15 hours ago

Ryu

82

share|cite|improve this question

share|cite|improve this question

edited 14 hours ago

edited 14 hours ago

edited 14 hours ago

asked 15 hours ago

Ryu

82

asked 15 hours ago

Ryu

82

asked 15 hours ago

Ryu

82

82

put on hold as off-topic by Arnaud Mortier, Ethan Bolker, José Carlos Santos, amWhy, Strants 15 hours ago

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

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Arnaud Mortier, José Carlos Santos, amWhy, Strants

If this question can be reworded to fit the rules in the help center, please edit the question.

put on hold as off-topic by Arnaud Mortier, Ethan Bolker, José Carlos Santos, amWhy, Strants 15 hours ago

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

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Arnaud Mortier, José Carlos Santos, amWhy, Strants

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This does not require permut/combinations. You only need to list the proper divisors of $72=2^3cdot3^2$ and add them up.
  – TheSimpliFire
  15 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The approach that does not depend on hand counting is given by the divisor function. It depends on the prime factorization of the number
  – Ross Millikan
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Welcome to stackexchange. That said, I'm voting to close this question because it's simple enough for you to have worked out the answer by listing all the divisors. When you ask a question here it's good practice to include what you've done so far and where you are stuck.
  – Ethan Bolker
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  +TheSimpliFire I need an approach that solves this using permutation and combination
  – Ryu
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  +Ethan The question is too easy to solve by listing all divisors but I want to know how this can be related with permutation and combination
  – Ryu
  15 hours ago
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This does not require permut/combinations. You only need to list the proper divisors of $72=2^3cdot3^2$ and add them up.
  – TheSimpliFire
  15 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The approach that does not depend on hand counting is given by the divisor function. It depends on the prime factorization of the number
  – Ross Millikan
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Welcome to stackexchange. That said, I'm voting to close this question because it's simple enough for you to have worked out the answer by listing all the divisors. When you ask a question here it's good practice to include what you've done so far and where you are stuck.
  – Ethan Bolker
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  +TheSimpliFire I need an approach that solves this using permutation and combination
  – Ryu
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  +Ethan The question is too easy to solve by listing all divisors but I want to know how this can be related with permutation and combination
  – Ryu
  15 hours ago
  
  

  
  
  
  

1

1

This does not require permut/combinations. You only need to list the proper divisors of $72=2^3cdot3^2$ and add them up.
– TheSimpliFire
15 hours ago

This does not require permut/combinations. You only need to list the proper divisors of $72=2^3cdot3^2$ and add them up.
– TheSimpliFire
15 hours ago

The approach that does not depend on hand counting is given by the divisor function. It depends on the prime factorization of the number
– Ross Millikan
15 hours ago

The approach that does not depend on hand counting is given by the divisor function. It depends on the prime factorization of the number
– Ross Millikan
15 hours ago

2

2

Welcome to stackexchange. That said, I'm voting to close this question because it's simple enough for you to have worked out the answer by listing all the divisors. When you ask a question here it's good practice to include what you've done so far and where you are stuck.
– Ethan Bolker
15 hours ago

Welcome to stackexchange. That said, I'm voting to close this question because it's simple enough for you to have worked out the answer by listing all the divisors. When you ask a question here it's good practice to include what you've done so far and where you are stuck.
– Ethan Bolker
15 hours ago

+TheSimpliFire I need an approach that solves this using permutation and combination
– Ryu
15 hours ago

+TheSimpliFire I need an approach that solves this using permutation and combination
– Ryu
15 hours ago

+Ethan The question is too easy to solve by listing all divisors but I want to know how this can be related with permutation and combination
– Ryu
15 hours ago

+Ethan The question is too easy to solve by listing all divisors but I want to know how this can be related with permutation and combination
– Ryu
15 hours ago

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

$72=2^3cdot 3^2$, so find the number of ways you can pick from three $2$'s and two $3$'s ... though at least one of them, but not all.

That is, we pick any from zero to three $2$'s (4 options), and anywhere from zero to two $3$'s (3 options), so that's twelve different numbers, but again we can't pick none of them or all of them, so there should be ten such proper divisors.

... but that's all we can do with this problem in terms of combinations ... if you want to know the sum of all those proper divisors, you'll actually have to find them and add them.

share|cite|improve this answer

answered 15 hours ago

Bram28

54.5k33777

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

$72=2^3cdot 3^2$, so find the number of ways you can pick from three $2$'s and two $3$'s ... though at least one of them, but not all.

That is, we pick any from zero to three $2$'s (4 options), and anywhere from zero to two $3$'s (3 options), so that's twelve different numbers, but again we can't pick none of them or all of them, so there should be ten such proper divisors.

... but that's all we can do with this problem in terms of combinations ... if you want to know the sum of all those proper divisors, you'll actually have to find them and add them.

share|cite|improve this answer

answered 15 hours ago

Bram28

54.5k33777

add a comment | 

up vote
1
down vote



accepted

$72=2^3cdot 3^2$, so find the number of ways you can pick from three $2$'s and two $3$'s ... though at least one of them, but not all.

That is, we pick any from zero to three $2$'s (4 options), and anywhere from zero to two $3$'s (3 options), so that's twelve different numbers, but again we can't pick none of them or all of them, so there should be ten such proper divisors.

... but that's all we can do with this problem in terms of combinations ... if you want to know the sum of all those proper divisors, you'll actually have to find them and add them.

share|cite|improve this answer

answered 15 hours ago

Bram28

54.5k33777

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

$72=2^3cdot 3^2$, so find the number of ways you can pick from three $2$'s and two $3$'s ... though at least one of them, but not all.

That is, we pick any from zero to three $2$'s (4 options), and anywhere from zero to two $3$'s (3 options), so that's twelve different numbers, but again we can't pick none of them or all of them, so there should be ten such proper divisors.

... but that's all we can do with this problem in terms of combinations ... if you want to know the sum of all those proper divisors, you'll actually have to find them and add them.

share|cite|improve this answer

answered 15 hours ago

Bram28

54.5k33777

$72=2^3cdot 3^2$, so find the number of ways you can pick from three $2$'s and two $3$'s ... though at least one of them, but not all.

That is, we pick any from zero to three $2$'s (4 options), and anywhere from zero to two $3$'s (3 options), so that's twelve different numbers, but again we can't pick none of them or all of them, so there should be ten such proper divisors.

... but that's all we can do with this problem in terms of combinations ... if you want to know the sum of all those proper divisors, you'll actually have to find them and add them.

share|cite|improve this answer

answered 15 hours ago

Bram28

54.5k33777

share|cite|improve this answer

share|cite|improve this answer

answered 15 hours ago

Bram28

54.5k33777

answered 15 hours ago

Bram28

54.5k33777

answered 15 hours ago

Bram28

54.5k33777

54.5k33777

add a comment | 

add a comment |