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To consecutive numbers cannot have any common prime factors. Thus the HCF of p and (p+1) will be 1. To find the LCM, you multiply two numbers together and divide by the HCF. In this case, you'd do p(p+1)/1

This simplifies to p2+p

So the HCF is 1 and the LCM is p2+p

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โˆ™ 2012-02-25 22:46:06
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Q: HCF and LCM of p and p plus 1?
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C program that accepts 15 different numbers and find the HCF and LCM?

#include <iostream.h> #include <conio.h> main() { int num[3],i,j,p,q,tmp,n1,n2,LCM,rem,flag; clrscr(); for(i=0;i<3;i=i+1) { cout<<"Enter No - "<<i+1<<"="; cin>>num[i]; } clrscr(); n1=num[0]; p=n1; for(i=1;i<3;i=i+1) { n2=num[i]; /* Finding HCF */ q=n2; if(p<q) { tmp=q; q=p; p=tmp; } while(p%q!=0) { rem=p%q; q=p; p=rem; if(p<q) { tmp=q; q=p; p=tmp; } } /*finding LCM */ LCM=1; for(j=1;n1%j==0n2%j==0;j=j+1) { if(n1%j==0) { n1=n1/j; flag=1; } if(n2%j==0) { n2=n2/j; flag=1; } if(flag==1) { LCM=LCM*j; } } LCM=LCM*n1*n2; n1=LCM; } cout<<"the LCM ="<<LCM; cout<<"the hcf ="<<q; getch(); } ----- int gcd (int a, int b) { . int tmp; . if (a<0) a= -a; . if (b<0) b= -b; . if (a<b) tmp= a, a= b, b= tmp; . while (b) { . . tmp= a%b; . . a= b; . . b= tmp; . } . return a; } int gcd_n (int n, const int *vect) { . int i, gcdtmp; . for (i=0, gcdtmp=0; i<n && gcdtmp!=1; ++i) . . gcdtmp= gcd (gcdtmp, vect[i]); . return gcdtmp; } int LCM (int a, int b) { . int d= gcd (a, b); . if (d==0) return d; . else return a/d*b; } int lcm_n (int n, const int *vect) { . int i, lcmtmp; . for (i=0, lcmtmp=1; i<n; ++i) . . lcmtmp= LCM (lcmtmp, vect[i]); . return lcmtmp; }


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Shell program to find LCM and hcf of two no.s tput clear echo "Enter first no" read a echo "Enter 2nd no" read b p= 'expr $a \* $b' while [$b -ne 0] do r= 'expr $a % $b' a=$b b=$r done LCM = 'expr $p / $a' echo "LCM = $LCM" echo "Hcf = $a"


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/* Program to find LCM/HCF of 15 Nos in C++/ C (replace cin with scanf & cout with printf for c) */ /* Developed by - Kishore Kr. Banerjee - papillon_kish@yahoo.com */ #include <iostream.h> #include <conio.h> main() { int num[3],i,j,p,q,tmp,n1,n2,LCM,rem,flag; clrscr(); for(i=0;i<3;i=i+1) { cout<<"Enter No - "<<i+1<<"="; cin>>num[i]; } clrscr(); n1=num[0]; p=n1; for(i=1;i<3;i=i+1) { n2=num[i]; /* Finding HCF */ q=n2; if(p<q) { tmp=q; q=p; p=tmp; } while(p%q!=0) { rem=p%q; q=p; p=rem; if(p<q) { tmp=q; q=p; p=tmp; } } /*finding LCM */ LCM=1; for(j=1;n1%j==0n2%j==0;j=j+1) { if(n1%j==0) { n1=n1/j; flag=1; } if(n2%j==0) { n2=n2/j; flag=1; } if(flag==1) { LCM=LCM*j; } } LCM=LCM*n1*n2; n1=LCM; } cout<<"the LCM ="<<LCM; cout<<"the hcf ="<<q; getch(); } ----- int gcd (int a, int b) { . int tmp; . if (a<0) a= -a; . if (b<0) b= -b; . if (a<b) tmp= a, a= b, b= tmp; . while (b) { . . tmp= a%b; . . a= b; . . b= tmp; . } . return a; } int gcd_n (int n, const int *vect) { . int i, gcdtmp; . for (i=0, gcdtmp=0; i<n && gcdtmp!=1; ++i) . . gcdtmp= gcd (gcdtmp, vect[i]); . return gcdtmp; } int LCM (int a, int b) { . int d= gcd (a, b); . if (d==0) return d; . else return a/d*b; } int lcm_n (int n, const int *vect) { . int i, lcmtmp; . for (i=0, lcmtmp=1; i<n; ++i) . . lcmtmp= LCM (lcmtmp, vect[i]); . return lcmtmp; }


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