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Apr 4th, 2009

c++ I did not understand

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I have a question I did not understand if anyone of you understood it please explain to me

suppose that m and n are integers and m is nonzero. Recall that m is called a divisor of n if n=mt for some integer t; that is, when m divides n, the remainder is 0. moreover .m is called a proper divisor of n if m<n and m divides n. a positive integer is called perfect if it is the sum of its positive proper divisors. for example, the positive proper divisors of 28 are 1,2,4,7 and 14 and 1+2+4+7=28. therefore 28 is perfect. write a program that does the following;
a.output the first four perfect integers.
b.takes as input a positive integer and then outputs whether the integer is perfect.

and thank you a lot

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MoOou

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Apr 4th, 2009

Re: c++ I did not understand

Do you have a problem with the math involved or the C++ involved? The easiest way to do the math you need using C++ is to use the modulo operator. Assuming m and n are both postive integers, if the result of m % n is zero then, n is a positive proper divisor of m according to the description you provided.

Lerner

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Re: c++ I did not understand

Click to Expand / Collapse Quote originally posted by MoOou ...

I have a question I did not understand if anyone of you understood it please explain to me

suppose that m and n are integers and m is nonzero. Recall that m is called a divisor of n if n=mt for some integer t; that is, when m divides n, the remainder is 0. moreover .m is called a proper divisor of n if m<n and m divides n. a positive integer is called perfect if it is the sum of its positive proper divisors. for example, the positive proper divisors of 28 are 1,2,4,7 and 14 and 1+2+4+7=28. therefore 28 is perfect. write a program that does the following;
a.output the first four perfect integers.
b.takes as input a positive integer and then outputs whether the integer is perfect.

and thank you a lot

well ur inputted with a number for ex lets take the no 28 ( which is in ur ex)
1. (this one is for proper divisor )
u will have to find such nos that divide 28 completely i.e
28 % number = = 0
the remainder must be zero
thus you will land up in 1 , 2, 4, 7 , 14

2. then when u find these numbers , u will have to find the summation of these in such a manner that
their sum is equal to 28
i.e in this case 1+2+4+7+14 = 28

thus the number 28 becomes perfect positive number .

rahul8590

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Re: c++ I did not understand

there is something I did not understand we prompt the user to put the numbers or one number only??
what's n=mt ??
we have to use if in these statement (m!=0)(m%n==0)(m<n) right??
and we don't need to use loop, right??

Last edited by MoOou; Apr 5th, 2009 at 1:55 pm.

MoOou

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Re: c++ I did not understand

$n=mt$ is the other way of writing that
$\frac{n}{m}=t$ where $t$ is a integer

it means that if you divide n by m, the division should leave no remainder.

>>we have to use if in these statement (m!=0)(m%n==0)(m<n) right??
yes

>>and we don't need to use loop, right??
No

You will have to use the loops for the first part of the question

siddhant3s

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Re: c++ I did not understand

well let me simply for you ..

1. accept number(`s in form of for loop or so ) and perfrom step 1 of my previous post .
you will land up with alll the divisors of that particular number (which will be the current number of for loop ) .
store it in some kinda array .

2. now check the summation of these numbers (which are divisors of the number inputted). if the add up to the number which is inputted , then the number which is inputted is perfect .

now re run the for loop and find such first 4 perfect numbers

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Re: c++ I did not understand

can u please give us the full program for this question

"suppose that m and n are integers and m is nonzero. Recall that m is called a divisor of n=mt for some integers t; that is when m divides n, the remainders is 0,moreover,m is called a proper divisor of n if m<n and m divides n .a positive integers is called perfect if it is the sum of its positive proper divisors .for example the positive divisor of 28 are 1,2,4,7 and 14 and 1+2+4+7+14=28,thrfore 28 is perfect write a program that does the following :a)output the first four perfect integers b)takes as input a positive integers and then outputs whether the integers is perfect"

Click to Expand / Collapse Quote originally posted by rahul8590 ...

well let me simply for you ..

1. accept number(`s in form of for loop or so ) and perfrom step 1 of my previous post .
you will land up with alll the divisors of that particular number (which will be the current number of for loop ) .
store it in some kinda array .

2. now check the summation of these numbers (which are divisors of the number inputted). if the add up to the number which is inputted , then the number which is inputted is perfect .

now re run the for loop and find such first 4 perfect numbers

fefe

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Re: c++ I did not understand

please help me to get the full program for writing this question
"suppose that m and n are integers and m is nonzero. Recall that m is called a divisor of n=mt for some integers t; that is when m divides n, the remainders is 0,moreover,m is called a proper divisor of n if m<n and m divides n .a positive integers is called perfect if it is the sum of its positive proper divisors .for example the positive divisor of 28 are 1,2,4,7 and 14 and 1+2+4+7+14=28,thrfore 28 is perfect write a program that does the following :a)output the first four perfect integers b)takes as input a positive integers and then outputs whether the integers is perfect"

fefe

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Re: c++ I did not understand

C++ Syntax (Toggle Plain Text)
input n
vector<int> factors
for(int i = 2; i <= n/2; ++n)
   //check if i is factor of n
   //if it is add it to the vector
 
//declare variable to act as running total and initalize to zero
 
//loop through vector adding elements together
for(int i = 0; i < size of vector; ++i)
    //add each element to previous total
 
//if n equals total from above loop, then n is a perfect number.
input n
vector<int> factors
for(int i = 2; i <= n/2; ++n)
   //check if i is factor of n
   //if it is add it to the vector

//declare variable to act as running total and initalize to zero

//loop through vector adding elements together
for(int i = 0; i < size of vector; ++i)
    //add each element to previous total

//if n equals total from above loop, then n is a perfect number.

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