Emanuele Ricci

# EVM Puzzle 10 solution

This is Part 10 of the “Let’s play EVM Puzzles” series, where I will explain how to solve each puzzle challenge.

EVM Puzzles is a project developed by Franco Victorio (@fvictorio_nan) that a perfect fit if you are in the process of learning how the Ethereum EVM works and you want to apply some of the knowledge you have just acquired.

## EVM Puzzle 10

```
00 38 CODESIZE
01 34 CALLVALUE
02 90 SWAP1
03 11 GT
04 6008 PUSH1 08
06 57 JUMPI
07 FD REVERT
08 5B JUMPDEST
09 36 CALLDATASIZE
0A 610003 PUSH2 0003
0D 90 SWAP1
0E 06 MOD
0F 15 ISZERO
10 34 CALLVALUE
11 600A PUSH1 0A
13 01 ADD
14 57 JUMPI
15 FD REVERT
16 FD REVERT
17 FD REVERT
18 FD REVERT
19 5B JUMPDEST
1A 00 STOP
```

This puzzle is similar to the Puzzle 9 we have just completed. It's mostly about understanding what opcodes do and solve a system of equations.

Let's see what new opcodes have been introduced:

- GT: pop 2 values from the stack and push the result of
`value0 > value1`

to the stack. If the result is`true`

, it pushes`1`

otherwise`0`

- MOD: pop 2 values from the stack and push back to the stack the result of
`value0 % value1`

. Note that the denominator (`value1`

) is`0`

the result will be`0`

- ISZERO: pop a value from the stack and push the result of
`value0 === 0`

to the stack

## Block 1: check calldata size and call value

```
00 38 CODESIZE
01 34 CALLVALUE
02 90 SWAP1
03 11 GT
04 6008 PUSH1 08
06 57 JUMPI
07 FD REVERT
08 5B JUMPDEST
```

The block adds the size of the code to the stack, add the value sent with the transaction to the stack, swap them in position (you could have achieved the same result with less gas) and then perform `GT(CALLVALUE, CODESIZE)`

.

If the result of that is **0** it will not follow the `JUMPI`

jump and revert.
`CODESIZE`

push to the stack the number of bytes of the contract's code. In this case, it will push to the stack the value `0x1b`

(27 in decimal).

**Note:** The number of code's instructions are 24 (so 24 bytes) but you must add to those also the bytes pushed by the `PUSH*`

opcodes. In this case, we have 2 `PUSH1`

and 1 `PUSH2`

so in total we need to add 3 bytes. That's why the `CODESIZE`

return 27 → 24 bytes for the number of instructions + 3 bytes from the values of the `PUSH`

in the code.

We have found our first equation to not revert: `GT(27, CALLVALUE) = 1`

so we must have `CALLVALUE <= 27`

to not revert.

## Block 2: check the calldata size

```
08 5B JUMPDEST
09 36 CALLDATASIZE
0A 610003 PUSH2 0003
0D 90 SWAP1
0E 06 MOD
0F 15 ISZERO
```

The opcodes push to the stack the `CALLDATASIZE`

, push `0x0003`

, swap them, perform a `MOD(0x0003, CALLDATASIZE)`

and perform `ISZERO`

on the value0 present in the stack. Because we have just performed the MOD operation, it will be `ISZERO(MOD(0x0003, CALLDATASIZE))`

This value will be used by the `JUMPI`

from the instruction in position `14`

. If the result of the `ISZERO`

is not **1** the contract will revert because it will not perform the jump.

The size of our `calldata`

must be a multiple of 3 to make `MODE(3, CALLDATASIZE)`

be equal to **0**.

This is the second part of the system of equations.

## Block 3: find the correct call value to jump to a valid `JUMPDEST`

```
10 34 CALLVALUE
11 600A PUSH1 0A
13 01 ADD
14 57 JUMPI
```

Currently, in our stack we have the result of `ISZERO(MOD(0x0003, CALLDATASIZE))`

and we know that it will be 1 otherwise we are going to revert.

Performing the other operation will make the stack be like

```
PUSH 0A
CALLVALUE
ISZERO(MOD(0x0003, CALLDATASIZE))
```

At this point, we perform the `ADD`

so we have the stack that will be

```
ADD(0A, CALLVALUE)
ISZERO(MOD(0x0003, CALLDATASIZE))
```

`JUMPI`

will perform a jump to the position with value `ADD(0x0A, CALLVALUE)`

. The `JUMPDEST`

that we want to reach is the one in position `19`

(25 in decimal).

This mean that `ADD(0x0A, CALLVALUE) === 19`

. The only possible value for that is that our `CALLVALUE`

is 10 (in hex is 0x0F)

## Solution

The system of equations we have to solve is this:

`CODESIZE = 27`

(`1b`

in hex) is always`CALLVALUE`

must be`<= 27`

to make`GT(CALLVALUE, CODESIZE)`

return`1`

`CALLVALUE = 15`

(`0F`

in hex) to make`ADD(0A, CALLVALUE)`

return`19`

`CALLDATASIZE`

must be a multiple of`3`

to make`ISZERO(MOD(0x0003, CALLDATASIZE))`

return`1`

A possible solution could be:

`CALLVALUE`

=**15**`CALLDATA`

=**0xFFFFFF**

Here's the link to the solution of Puzzle 10 on EVM Codes website to simulate it.