Solving with modulus

[Omesh Harrylall]

what number can be divided by 7 without a remainder but when divided by 2 3 4 5 or 6 leaves 1?

I've found the riddle answer for the above question to be 301
but I can't figure out how to write a c++ program to find me the answer using %. Some hints on how to go about this would be highly appreciated,

Thank You.

[ResidentBiscuit]

How did you figure it out without a program? Just translate that thought pattern into something a computer understands

[Pickle Gunner]

Try Something like this:

#include iostream

int main(){
int NumberEntered = 0;

while(NumberEntered % 7 != 0 && NumberEntered % 2 != 1  && NumberEntered % 3 != 1 
 && NumberEntered % 4 != 1  && NumberEntered % 5 != 1  && NumberEntered % 6 != 1)
NumberEntered++;

//You could also do  && NumberEntered != 301 if you do not want 301 as an answer

std::cout >> NumberEntered;
return 0;
}

Check if the number is valid, if not start again with the next number.

[Omesh Harrylall]

I basically did LCM of 2,3,4,5,6 which is 60...

then I added 1 to make it 61 and tested it using % and it wasn't funny divisible by 7.

so I kept adding the LCM 61 121 181 241 301

the first one that left no remainder was 301.

I tried something similar to your code Pickle but it seems very tedious because I need to go through every single number until I get to 301.


this is what I currently have:

void main()
{
  int a[7],n,i,j,c,max,min;
  unsigned long prod;
  
  printf("Enter the no. of entries: ");
  scanf("%d",&n);
  printf("Enter the entries:");
  for(i=0;i<n;i++)
  {

    if(c>0)
      a[i]=c;
    else
    {
      printf("Invalid Entry");
      return;
    }
  }

  max=a[0];
  for(i=0;i<n;i++)
    if(a[i]>=max)
      max=a[i];
  min=a[0];
  for(i=0;i<n;i++)
    if(a[i]<min)
      min=a[i];

  for(i=0,prod=1;i<n;i++)
   ;

  for(i=max;i<=prod;i+=max)
  {

    c=0;
    for(j=0;j<n;j++)
 
		 c+=1;
    if(c==n)
    {
      printf("The LCM of the nos: %d",i);
      break;
    }
  }

 
  getch();
}

I got up to the LCM but I need a test for % and a for loop to add 60 but I haven't been able to find a way to test the equality.


Hope all of that makes sense... I'm pretty new to c++.

[NwvkoG1T]

Pretty simple man!

I am writing only the required code you can Add the user part and main() later on!

int flag;
for(int i=1;i<1000;i++)
{

flag=1;

if(i%7==0)
{

for(int j=2;j<7;j++)
if(i%j!=1)
{
flag=0;
break;
}

if(flag==1)
cout<<"match found:"<<i;
continue;

}
}

[NwvkoG1T]

and its kinda amazing that you figured it out without writing a program! how did you do it?

[Omesh Harrylall]

Apparently by hurting my head on paper ;-; . I guess I needed some sleep because I was heading to the right solution lol...

I'm usually good at programming from PSEUDOCODES but my professor apparently loves giving riddles :'( to dumbfound me haha

Thank you IcyFlame and others the cplusplus community is very helpful and very fast in responding :)

[NwvkoG1T]

Thanks a lot! Omesh!

[NwvkoG1T]

Well, do tell me what is pseudo code?

[Omesh Harrylall]

Well its a description or information in a certain manner to help you make a program.

Kind of like a basic algorithm to get you started on how to start and finish the program.

This riddle took me a while because I wasn't sure how to start it.

I also had to re-edit to be more informative to help a user understand the riddle.

[NwvkoG1T]

oh.... well did the code that i had posted work?