Numbers....hehe
53 59 61 67 71 73 79 83 89 97 are the prime numbers between 51 & 100.
40 of them including 51 and 100 38 without.
They are 100 minus 49 = 51
There will be (200-100)/2+1 = 51 numbers divisible by 2.There will be Floor[(200 - 100)/3] = 33 numbers divisible by 3.Multiples of 6 from the 17th through the 33rd will be counted twice by the sum of these numbers (51+33). There are 17 of them.That means the count of numbers you are interested in is 51+33-17 = 67.
There are ten prime numbers between 51-100.
Numbers....hehe
53 59 61 67 71 73 79 83 89 97 are the prime numbers between 51 & 100.
49 if you don't include 0 and 100. 51 if you include 0 and 100.
40 of them including 51 and 100 38 without.
The prime numbers between 51 and 60 are 53 and 59. A prime number is a natural number greater than 1 that is only divisible by 1 and itself. In this range, 53 and 59 are the only prime numbers as they cannot be divided evenly by any other number except 1 and themselves.
They are 100 minus 49 = 51
There will be (200-100)/2+1 = 51 numbers divisible by 2.There will be Floor[(200 - 100)/3] = 33 numbers divisible by 3.Multiples of 6 from the 17th through the 33rd will be counted twice by the sum of these numbers (51+33). There are 17 of them.That means the count of numbers you are interested in is 51+33-17 = 67.
#include <iostream> int main() { for(int i=51; i <= 100; i+=2) { cout << i << endl; } return 0; }
#include<stdio.h> #include<conio.h> void main() { int i,j; for(i=51;i<100;i=i+2) { printf("%d", i); } getch(); }
I believe here are 51 such numbers.
The prime numers between 50 and 100 are 51, 53, 59, 61, 67, 71, 73, 79, 83, 87, 89, 91, and 97.