Code problem

Question

public static String ben (int x){

if (x /5 <=0)

return x %5;

else

return (x % 5) + ben (x/5);

}

When I pass 51 into ben() what will be the output?
Can anyone please explain.
Options are
202
201
102
101

Answer 1

Well first you have to fix the ben() method so that it actually returns String or change the return type. So, I'll answer as if the return type is String since that is what is needed to get the answer from the available choices.

The if statements condition will be false. This is integer division so the fractional part will be truncated.
51/5 = 10 not 10.2 in this case
So the else statement is ran.
x % 5 = 1
Using the modulo operator % will return the remainder from division.
51 / 5 has a remainder of 1. That is 5 goes into 51, 10 times leaving 1 left over.
10 * 5 = 50 51 - 50 = 1
So, "1" + ben(51/5) is returned
The recursive call to ben that is returned is equivalent to
ben(10) due to integer division
The if statement is false 10/5 = 2
So, here (10 % 5) + ben(10/5) is returned
Which is
"0" + ben(2)
now we have
"1" + "0" + ben(2)
The if statement is true due to integer division.
2/5 = 0
Returns 2 % 5 = 2
2 divided by 5 is 0 with a remainder of 2
"1" + "0" + "2"
"102"

Answer 2

In your code sample, it return type is string, so you are forgetting to form string in return statement.. So it is error.
It shloud be
return ""+(x%5); now this return converted string. In next return also should change like this.

If it return string, if you pass 51 then
In if 51/5<=0 => 10<=0 false
then it return "" +(51%5) +ben(51/5) => 1+ben(10)

Next ben(10)=> 10/5<=0 false so it return 10%5 +ben(10/5) => 0+ben(2)

Next ben(2)=> 2%5<=0 is true so it return ""+(2%5) =>2

So finally "" +1+(0+ (2))
"102" a string form.

In your code if you change return type to int, then answer is 1+0+2=3

Answer 3

with correction
return String.valueOf(x % 5);

x=51
if 51/5 = 10; 10 <= 0 false
else return 51 % 5 = 1
+
x= 51/5 = 10
if 10/5 = 2; 2 <= 0 false
else return 10 % 5 = 0
+
x= 10/5 = 2
if 2/5 = 0; 0 <= 0 true
return 2 % 5 = 2

String 1+"0"+"2"