misc
No flag for you
Solved By : Taz
 We found out that
ls
,cat
,echo
commands were available.  So started looking around for these.
 Found a way to list dir using echo:
echo /*
 started looking around for the flag and found it in the
/home/user/run/opt
dir
echo /home/user/run/opt/*
 Did some research on ways to read a file using echo, and got something.
 Reference link: https://stackoverflow.com/questions/22377792/howtouseechocommandtoprintoutcontentofatextfile
echo "$(</home/user/run/opt/flagb01d7291b94feefa35e6.txt)"
Alternative Arithmetic
Solved By : Ava and choco
 After connection we are given a java questions which we have to research for and solve
Question 1 :
1. Find a nonzero long x` such that `x == x`, enter numbers only, no semicolons
 answer for 1st question is
9223372036854775808
 Explanation :
The first question given is to find long x such that
x == x
Here we will need to exploit the use of 2’s compliment of binary representation of long. example: in a 4 bit representation of 1, the binary is0001
but to represent 1, the binary will take 2’s compliment so it will be1111
The MSB here represents the sign of the binary the limitation of 2’s compliment is that, the number that has reached the minimum negative limit (i.e. the maximum 2’s compliment value), it’s positive counter part will be the same binary value as it’s two’s compliment. Let’s take the one for 4 bit representation again, the minimum value in this will be 8, i.e,1000
if you take +8 here, the binary will also be1000
So we got the concept for the answer. But now, the data type is long (8 bytes) The answer must be9223372036854775808
Question 2 :
Find 2 different `long` variables `x` and `y`, differing by at most 10, such that `Long.hashCode(x) == Long.hashCode(y)`
 answer for 2nd question is
x = 1 & y = 0
 Explanation :
For the second one the given algorithm for Long.hashCode() is
public static int hashCode(long value) { return (int)(value ^ (value >>> 32)); }
As we see here, the hash code for long compresses the value to an int from 64 bits to 32 bits. The hash values of positive long values are itself (hashcode of 8 is 8) The exploitation here is that negative numbers behaves different. let’s take 12. it’s binary representation is
1111111111111111111111111111111111111111111111111111111111110100
after it got shifted to the right by 32 bits it will be
0000000000000000000000000000000011111111111111111111111111111111
this value is xor with the original value giving
1111111111111111111111111111111100000000000000000000000000001011
when we convert it to int only the first 32 from LSBs are taken i.e.,
00000000000000000000000000001011
Hence, the Hash value of 12 will be 11 So
Long.hashCode(i) = i for i >=0 and (i)1 for i < 0
but the question is asking for x and y where
x  y <=10
Hence using the advance stages of arithmetic mathematics, we can take answers as
{0,1},{1,2},{2,3},{3,4} and {4,5}
Courtesy of https://www.devdiaries.net/blog/JavaRecruitmentQuestionsHashCode/for Hashcode algorithm.
Question 3 
Enter a `float` value `f` that makes the following function return true
// code given along with Q3 :
boolean isLucky(float magic) {
int iter = 0;
for (float start = magic; start < (magic + 256); start++) {
if ((iter++) > 2048) {
return true;
}
}
return false;
 answer for 3rd Question is
2.14E9
 Explanation:
The third question is a little tricky We need to find float value magic such that we need to break the loop The other catch is that it have to be less than 7 characters
The trick here is that we need to exploit the precision error in java
in java, the floating points can go upto
3.4 * 10^34
but the secret is that it will take a precision of 8 digits. i.e., if the value in decimal points is more than 8 digits. It will round them off to 8 digits itself. so for example ‘345678’ in float will be 345678.0 but if digits has reached 8 like ‘12345678
’, float will save the value as1.2345678E7
The maximum digits in float is actually 10 (1999999999) if you want to represent a number higher than that, you must make it in2.45E10
like formatnow, for the exploit, if you add any number that is negligibly small to a float value, this will not affect the float value at all. In fact, float values can only consider changes made in the first 8 most significant digits. In other words, if converted in decimal, only the first 7 digits will be considered, rest all are static Rather, it is interesting to note that the remaining digits will face binary accuracy error (when you convert some numbers to IEEE754 and back to the number, it won’t be the same. This will be addressed in the next question)
example for
123456789 + 1
it will be written as(1.23456792 + 0.00000001)*10^8
you see that one is negligent when converted to this form hence the answer will be1.23456792*10^8
(binary rep error in last digits)now if we take
123456789 + 100
it will be written as(1.23456792 + 0.00000100)*10^8
you see that 100 falls under the 8 digit range hence the answer will be1.23456892*10^8
We got the idea about the exploit. but keep in mind, we need to select a number such that incrementing it will not affect the value but adding 256 to it should So we can discard numbers that are 7 digits or less since incrementing it affects the value Also we can discard numbers that are 10 digits or above since adding 256 won’t affect the valueHence, we can only take numbers are that 8 and 9 digits long.
The question says, we can only use 7 characters or less.
We can simply represent the float value as
1.0E8
toohence, the answer is within the range of
1E8 to 9.999E9
Alternative Arithmetic (Final Flag)
Solved By : nigamelastic and choco

I will use the following to run this java compiler to run my code online.

on connecting we see the first question which is:
4th question :
Enter 3 `String` values `s1`, `s2`, and `s3` such that:
new BigDecimal(s1).add(new BigDecimal(s2)).compareTo(new BigDecimal(s3)) == 0
but
Double.parseDouble(s1) + Double.parseDouble(s2) != Double.parseDouble(s3)
Do not enter quotation marks.
 answer 4 worked but very stupidly: i will paste my inputs and outputs and you just see :
First attempt
Do not enter quotation marks.
s1 = 0.004
s2 = 0.003
s3 = 0.001
Sorry, your answer is not correct. Try something else next time.
Second attempt
s1 = 20000000000000000000000000000000000000000000000.0000000000000000000000000000000000000000004
s2 = 10000000000000000000000000000000000000000000000.0000000000000000000000000000000000000000003
s3 = 10000000000000000000000000000000000000000000000.0000000000000000000000000000000000000000001
Sorry, your answer is not correct. Try something else next time.
Third attempt
s1 = 0.04
s2 = 0.03
s3 = 0.01
Perfect! You are correct.

credits to choco, who guided me, and after our discussion we realised for some reason
0.04
is working but0.004
isn’t choco also pointed out this amazing link https://docs.oracle.com/cd/E1995701/8063568/ncg_goldberg.html 
finally we discussed and realized the following :

according to double : ```bash 0.003 will be represented as 3.00000000000000006245004513517E3 0.004 will be represented as 4.00000000000000008326672684689E3 0.02 will be represented as 2.00000000000000004163336342344E2
but
0.03 will be represented 2.99999999999999988897769753748E2
* after this stupidity we finally get the 5th and final option:
 Final question! [🎶BOSS MUSIC🎶]
Fill in
```java
var i = (<type>) <num1>; var j = (<type>) <num2>;
such that after running the code above, the following expression:
i < j  i == j  i > j
evaluates to `false`.
<num1> and <num2> are Java code that satisfies this regex: [09]*\.?[09]*
 which is very easy , I remember it from my college days
 its
Integer 128
 this is due to the integer cache limits
 the following link would be a better explanation
 https://www.programmersought.com/article/91462031238/