# /= operator

dear experts

i am trying to find out numbers power of 3 between 1 and 100
what is the best way to do this (with constant time performance)
I found this code on internet

``````for( int i=3; i < 100; i++){
if (i % 3 == 0) {
i /= 3;   //what does this statement do?
}
}
``````

any sugessions appreciated
thanks
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Commented:
Your code won't find powers of 3. It will find numbers that are evenly divisible by 3.

the /= operator is the division assignment operator. i /= 3 is the same as saying i = i / 3.

Additionally, your code will be an endless loop. i will start at 3. 3 mod 0 does in fact equal 0 so it will divide i by 3 which makes i = 1. The loop will then run again until you get back to i = 3 then it will divide by 3 and i will be 1 again. This is an endless loop.
0
Author Commented:
I just ran my code and you are correct.
Any idea what the solution should be?
Thanks.
0
Commented:
This problem has a simple enough solution that we can work backwards. We know that all numbers between 3 and 100 that are powers of 3 are easy to find. They are:

3
9
27
81

Now we can figure out how to get only those numbers. Rather than looping through all numbers between 3 and 100 and comparing them you just need to raise each number to a power and compare it to your lower and upper bounds. I'd do it like this:

``````double maxValue = 100;
double power = 1;
while (Math.Pow(3, power++) < maxValue)
{
// do whatever you want in here. Add it to an array, write to a file, whatever. Each result of Math.Pow(3, power) will be a power of 3 that is between 3 and 100.
}
``````
0
Author Commented:
is this what you mean?

``````		double maxValue = 100;
double power = 1;
while (Math.pow(3, power++) < maxValue)
{

System.out.println("Number is power of 3>>>"+ Math.pow(3, power++));

}
``````

but 3 and 27 seem to be missing.
thanks
0
Commented:
Almost, you're double incrementing the power variable there which won't quite work. Probably my mistake in how I wrote the initial code which would lead you the wrong way. Try this instead:

``````		double maxValue = 100;
double power = 0;
while (Math.pow(3, ++power) < maxValue)
{

System.out.println("Number is power of 3>>>"+ Math.pow(3, power));

}
``````

I changed the initial value of power to 0 and used a pre increment operator rather than the post increment operator.

You could also do this and get the same result:

``````		double maxValue = 100;
double power = 1;
while (Math.pow(3, power) < maxValue)
{

System.out.println("Number is power of 3>>>"+ Math.pow(3, power++));

}
``````
0

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Author Commented:
thanks, learnt something new today.
Just incase i dont want to use Math.pow, I was also going through this url
http://stackoverflow.com/questions/1804311/how-to-check-if-an-integer-is-power-of-3

I was trying to use this code from that url
``````while (n % 3 == 0) {
n /= 3;
}
return n == 1;
``````

will this also do the job?

thx
0
Commented:
Nope. You'll keep dividing n by 3 until your number gets to be too small to fit in whatever variable type you are using and throw an exception.

Why would you not want to use Math.Pow?
0
Commented:
If you really don't want to use Math.Pow (although I don't know why), you could use something like this -
int pwr = 3;
int num = 1;
while (num < 100 ) {
num *= pwr;
if (num > 100) {
continue;
}
}
0
Commented:
Or this could tell you if a number is a power of 3
``````int iiiPow =3;
int isthisAPowerOfThree = 129140164;
while (iiiPow<isthisAPowerOfThree){

iiiPow *= 3;
if (iiiPow==isthisAPowerOfThree){System.out.println(isthisAPowerOfThree+" is a power of 3");break;}
else if(iiiPow>isthisAPowerOfThree){System.out.println(isthisAPowerOfThree+" is not a power of 3");}
}
``````
0
Author Commented:
Hi
Math.pow is fine , was just trying to see what the other options are to check which one has best performance (o)
Thanks.
0
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