Re: Logic Quiz
Okay to me, the question is a pure brute force i.e., you pick a prime number and figure your way up ... I could have done it on papers but i didn't have anything available so i wrote a Java program
public class PsyahQuestion {
static int *a*;
static int *b*;
static int *c*;
public static void main (String] args){
int] primeNumbers = {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97};
for (int pk : primeNumbers) {
*calculate*(pk);
System.*out*.printf("a = %d, b = %d, c = %d, pk = %d
", *a*, *b*, *c*, pk);
}
}
private static void calculate(int pk) {
for (int i = 1; i <= 100; i++){
for (int j = 1; j <= 100; j++){
*a *= i;
*b *= j;
if ((i != j) && (i/j == pk)){
if (i * j < 100) {
*c *= i * j;
if (*c ** pk > 100){
return;
}
}
}
}
}
}
}
and the output is
a = 11, b = 5, c = 55, pk = 2
a = 12, b = 4, c = 48, pk = 3
a = 11, b = 2, c = 22, pk = 5
a = 14, b = 2, c = 28, pk = 7
a = 11, b = 1, c = 11, pk = 11
a = 13, b = 1, c = 13, pk = 13
a = 17, b = 1, c = 17, pk = 17
a = 19, b = 1, c = 19, pk = 19
a = 23, b = 1, c = 23, pk = 23
a = 29, b = 1, c = 29, pk = 29
a = 31, b = 1, c = 31, pk = 31
a = 37, b = 1, c = 37, pk = 37
a = 41, b = 1, c = 41, pk = 41
a = 43, b = 1, c = 43, pk = 43
a = 47, b = 1, c = 47, pk = 47
a = 53, b = 1, c = 53, pk = 53
a = 59, b = 1, c = 59, pk = 59
a = 61, b = 1, c = 61, pk = 61
a = 67, b = 1, c = 67, pk = 67
a = 71, b = 1, c = 71, pk = 71
a = 73, b = 1, c = 73, pk = 73
a = 79, b = 1, c = 79, pk = 79
a = 83, b = 1, c = 83, pk = 83
a = 89, b = 1, c = 89, pk = 89
a = 97, b = 1, c = 97, pk = 97
Bhaijan thats supposed to be a “logic quiz” not a programming contest :halo:
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