# 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 is 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          CODESIZE01      34          CALLVALUE02      90          SWAP103      11          GT04      6008        PUSH1 0806      57          JUMPI07      FD          REVERT08      5B          JUMPDEST09      36          CALLDATASIZE0A      610003      PUSH2 00030D      90          SWAP10E      06          MOD0F      15          ISZERO10      34          CALLVALUE11      600A        PUSH1 0A13      01          ADD14      57          JUMPI15      FD          REVERT16      FD          REVERT17      FD          REVERT18      FD          REVERT19      5B          JUMPDEST1A      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 1 to the stack. If the result is `true` it push `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 CODESIZE01 34 CALLVALUE02 90 SWAP103 11 GT04 6008 PUSH1 0806 57 JUMPI07 FD REVERT08 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 amount 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 JUMPDEST09 36 CALLDATASIZE0A 610003 PUSH2 00030D 90 SWAP10E 06 MOD0F 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 CALLVALUE11 600A PUSH1 0A13 01 ADD14 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 0ACALLVALUEISZERO(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 = 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.

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