MATHS riddle

[Neoxter]

Thanks Google

EDIT: Show us how did u solve it

[Tawa]

Originally Posted by Neoxter

Thanks Google

EDIT: Show us how did u solve it

#include <iostream>
using namespace std;

bool f(float);

int main(){
    long i;
    float a;
    for(i = 0; i < 10000; i++){
        a = (float)i;
        if(f(a/6) && f(a/7) && f(a/8) && f(a/9) && f(a/10))
            cout << "I = " << i << endl;
    }
}

bool f(float a){
    int b = a;
    return (a==b);
}

This is how I figured it out

[Kingroudy]

don't you dare to question my IQ and claim that i used google.
to form 2,4,8 you need (2,2,2)
to form a 9 u need (3,3)
to form a 7 you need a (7)
to form a 5 you need a (5)
to form a 6 u need nothing because you already have 3 and 2
to form a 10 u need nothing because you already have 5 and 2
you don't need a 1 do you
so the answer is 2*2*2*3*3*5*7

[Tawa]

#include <iostream>
using namespace std;

bool f(float);

int main(){
    long i;
    int max;
    bool display;

    cout << "The number should be divisible by all digits from 1 to ";
    cin >> max;
    for(i = 1; i < 100000000; i++){
        display = true;
        for(int j = 1; j <= max; j++){
            if(!f((float)i/j))
                display = false;
        }
        if(display){
            cout << "I = " << i << endl;
            break;
        }
        else
            cout << i << "\x8\x8\x8\x8\x8\x8";
    }
    if(display)
        cout << "Done!" << endl;
    else
        cout << "Failed!" << endl;

    return 0;
}

bool f(float a){
    int b = a;
    return (a==b);
}

Compile this one
Note: the execution might be very slow if you use a large number.

[Google]

@Tawa, your algorithm can be enhanced.
Instead of calling the stupid "f" function I don't know how much times, you could just use the modulus operator like this: if(i%j != 0)
instead of: if(!f((float)i/j))
As for the "Decore", no need for "\x8\x8\x8\x8\x8\x8", \r will do the job.

[Google]

Originally Posted by Kingroudy

don't you dare to question my IQ and claim that i used google.
to form 2,4,8 you need (2,2,2)
to form a 9 u need (3,3)
to form a 7 you need a (7)
to form a 5 you need a (5)
to form a 6 u need nothing because you already have 3 and 2
to form a 10 u need nothing because you already have 5 and 2
you don't need a 1 do you
so the answer is 2*2*2*3*3*5*7

An AI bot can be taught to think this way. Unlike the brute force method of Tawa that is expensive in terms of resources.

[Kingroudy]

Originally Posted by Google

An AI bot can be taught to think this way. Unlike the brute force method of Tawa that is expensive in terms of resources.

i'm an android LOL