product consecutive numbers

by John Doe

JavaScript

function p(x){
    for(n=2;n<13;n++){/*number of factors*/
        r=Math.pow(x,1/n)|0,
        q=(a,b)=>b>0?a*q(a+1,b-1):1;
        for(s=-2*n;s<n;s++)if(q(r+s,n)==x) return r+s;   
    }
    return false;
}
alert(p(10*11*12*13*14*15));