Answer:
91
Explanation:
You know that the number must end in 6 or 1 to get a left over of 1 when divided by 5. Six won't work because 2 divides evenly into a number ending into 6.
So the number ends in 1.
21 doesn't work because 3 divides into 21 evenly.
31 doesn't work because 7 does not divide into it.
41 doesn't work. 41 is prime. 7 won't divide into it.
51 is divisible by 3
61 seven does not evenly divide into 61.
71 seven leaves a remainder of 1
81 7 leaves a remainder.
91 Answer
Question↓
You are purchasing a new printer. Which of the following is the most important requirement?
Answer↓
A. Is the printer compatible with your computer's operating system?
If The Printer is Not Compatible With your Computer you will not Be able to Use Therefore, you Will have to buy a New One.
xXxAnimexXx
Hope this Helps! ^ω^
Answer:
#include<stdio.h>
#include<conio.h>
int m=0,n=4;
int cal(int temp[10][10],int t[10][10])
{
int i,j,m=0;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
{
if(temp[i][j]!=t[i][j])
m++;
}
return m;
}
int check(int a[10][10],int t[10][10])
{
int i,j,f=1;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
if(a[i][j]!=t[i][j])
f=0;
return f;
}
void main()
{
int p,i,j,n=4,a[10][10],t[10][10],temp[10][10],r[10][10];
int m=0,x=0,y=0,d=1000,dmin=0,l=0;
clrscr();
printf("\nEnter the matrix to be solved,space with zero :\n");
for(i=0;i < n;i++)
for(j=0;j < n;j++)
scanf("%d",&a[i][j]);
printf("\nEnter the target matrix,space with zero :\n");
for(i=0;i < n;i++)
for(j=0;j < n;j++)
scanf("%d",&t[i][j]);
printf("\nEntered Matrix is :\n");
for(i=0;i < n;i++)
{
for(j=0;j < n;j++)
printf("%d\t",a[i][j]);
printf("\n");
}
printf("\nTarget Matrix is :\n");
for(i=0;i < n;i++)
{
for(j=0;j < n;j++)
printf("%d\t",t[i][j]);
printf("\n");
}
while(!(check(a,t)))
{
l++;
d=1000;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
{
if(a[i][j]==0)
{
x=i;
y=j;
}
}
//To move upwards
for(i=0;i < n;i++)
for(j=0;j < n;j++)
temp[i][j]=a[i][j];
if(x!=0)
{
p=temp[x][y];
temp[x][y]=temp[x-1][y];
temp[x-1][y]=p;
}
m=cal(temp,t);
dmin=l+m;
if(dmin < d)
{
d=dmin;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
r[i][j]=temp[i][j];
}
//To move downwards
for(i=0;i < n;i++)
for(j=0;j < n;j++)
temp[i][j]=a[i][j];
if(x!=n-1)
{
p=temp[x][y];
temp[x][y]=temp[x+1][y];
temp[x+1][y]=p;
}
m=cal(temp,t);
dmin=l+m;
if(dmin < d)
{
d=dmin;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
r[i][j]=temp[i][j];
}
//To move right side
for(i=0;i < n;i++)
for(j=0;j < n;j++)
temp[i][j]=a[i][j];
if(y!=n-1)
{
p=temp[x][y];
temp[x][y]=temp[x][y+1];
temp[x][y+1]=p;
}
m=cal(temp,t);
dmin=l+m;
if(dmin < d)
{
d=dmin;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
r[i][j]=temp[i][j];
}
//To move left
for(i=0;i < n;i++)
for(j=0;j < n;j++)
temp[i][j]=a[i][j];
if(y!=0)
{
p=temp[x][y];
temp[x][y]=temp[x][y-1];
temp[x][y-1]=p;
}
m=cal(temp,t);
dmin=l+m;
if(dmin < d)
{
d=dmin;
for(i=0;i < n;i++)
for(j=0;j < n;j++)
r[i][j]=temp[i][j];
}
printf("\nCalculated Intermediate Matrix Value :\n");
for(i=0;i < n;i++)
{
for(j=0;j < n;j++)
printf("%d\t",r[i][j]);
printf("\n");
}
for(i=0;i < n;i++)
for(j=0;j < n;j++)
{
a[i][j]=r[i][j];
temp[i][j]=0;
}
printf("Minimum cost : %d\n",d);
}
getch();
}
Explanation:
Answer:
#include <iostream>
using namespace std;
void printmultiples(int n) //function to print first five multiples between 3168 and 376020
{
int a =3168,c=1;
cout<<"First five multiples of "<<n<<" are : ";
while(a%n!=0 && a<=376020) //finding first mutiple of n after 3168.
{
a++;
}
while(c<6)//printing multiples.
{
cout<<a<<" ";
a+=n;
c++;
}
cout<<endl;
}
int main() {
int t,n;
cin>>t;//How many times you want to check.
while(t--)
{
cin>>n;
printmultiples(n);//function call..
}
return 0;
}
Input:-
3
15
43
273
Output:-
First five multiples of 15 are : 3180 3195 3210 3225 3240
First five multiples of 43 are : 3182 3225 3268 3311 3354
First five multiples of 273 are : 3276 3549 3822 4095 4368
Explanation:
I have used a function to find the first five multiples of the of the numbers.The program can find the first five multiples of any integer between 3168 and 376020.In the function I have used while loop.First while loop is to find the first multiple of the integer n passed as an argument in the function.Then the next loop prints the first five multiples by just adding n to the first multiple.
In the main function t is for the number of times you want to print the multiples.In our case it is 3 Then input the integers whose multiples you want to find.
