#include
void main()
{
long int n;
clrscr();
printf("Enter a number:");
scanf("%ld",&n);
printf("%d",n%10);
getch();
}
#include<stdio.h> int main (void) { float x,y; int z,w; printf("enter a number:"); scanf("%6.2f",&x); y=x/10.0; z=x-y; w=z%10; printf("the rightmost digit is %d\n",w); return 0; }//main
One place to find information on visual basic coding is a local library. The website msdn microsoft has extensive tutorials and guides on visual basic coding.
The rightmost digit represents how many 1s (in this example 1) 1 The next digit left represents how many 2s (in this example 1) 2 The next digit left represents how many 4s (in this example 0) 0 The next digit left represents how many 8s (in this example 1) 8 The next digit left represents how many 16s (in this example 1) 16 The next digit left represents how many 32s (in this example 1) 32 The next digit left represents how many 64s (in this example 1) 64 Total 123
int n; // the number you want to reverse int rev_n = 0; while (n > 0) { // shift rev_n digits left rev_n *= 10; // put rightmost digit of n onto right of rev_n rev_n += n % 10; // remove rightmost digit of n n /= 10; }
160 and 192.
The units digit of a whole number is always the rightmost digit.
25 and 36 are each equal to their rightmost digit squared. 25 = 5 X 5 36 = 6 X 6
4 Every digit between the leftmost non-zero digit and EITHER the right most non-zero digit OR the rightmost digit after a decimal.
It is the integer that you get by dividing the three digit number by 10. Alternatively, the number of the multiples of ten is that you get when you truncate the rightmost number. for example assume a three digit number 517 Divide 517 by ten, you get 51.7 The integer number is 51 Alternatively, if you trnucate the rightmost number (7) you get 51 also then the answer the three digit number 517 has 51 multiples of 10.
Yes - any further zeroes to the right of the rightmost digit after a decimal point has no effect on the value of the number.
int getReverse(int n) { int reversedN = 0; // Divide n by 10 on each pass to remove the rightmost digit for (; n > 0; n /= 10) { // Multiple reversedN to shift all digits to the left // (to make room for the next digit) reversedN *= 10; // Add on (n%10), which is the rightmost digit of n reversedN += (n % 10); } return reversedN; }
What does this mean? The rightmost digit of {eq}n^j{/eq} is the remainder when {eq}n^j{/eq} is divided by {eq}10{/eq}. yep totaly nor random :))
Rightmost column is called the ones.
#include<stdio.h> int main (void) { float x,y; int z,w; printf("enter a number:"); scanf("%6.2f",&x); y=x/10.0; z=x-y; w=z%10; printf("the rightmost digit is %d\n",w); return 0; }//main
You discard all digits after the first four digits, in this case. Then, as is usual in rounding, you look at the first digit discarded, and if it's 5 or more, you add one to the rightmost digit you keep.
This is a weekly question posted by Columbus State University. See http://www.colstate.edu/mathcontest/problem.php?CategoryID=3&LinkID=CurrentPlease do not answer this question as this person is too lazy to figure it out themselves.
"rightmost " is not a proper word, however it's meaning is 'furthest to the right'