ArnoldC, 299 283 bytes

IT'S SHOWTIME
HEY CHRISTMAS TREE i
YOU SET US UP 0
GET YOUR ASS TO MARS i
DO IT NOW
I WANT TO ASK YOU A BUNCH OF QUESTIONS AND I WANT TO HAVE THEM ANSWERED IMMEDIATELY
GET TO THE CHOPPER i
HERE IS MY INVITATION i
I LET HIM GO 2
ENOUGH TALK
TALK TO THE HAND i
YOU HAVE BEEN TERMINATED

This outputs 1 (which is truthy) for odd input and 0 (which is falsy) for even input.

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Retina, 8 bytes

[02468]$

A Retina answer for decimal input. This is also a plain regex solution that works in almost any regex flavour. Matches (and prints 1) for even inputs and doesn't match (and prints 0) for odd inputs.

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An alternative, also for 8 bytes, uses a transliteration stage to turn all even digits to x first (because transliteration stages have a built-in for even/odd digits):

T`E`x
x$

Of course, the shortest input format (even shorter than unary) would be binary in this case, where a simple regex of 0$ would suffice. But since the challenge is essentially about finding the least-signficant binary digit, binary input seems to circumvent the actual challenge.

Mathematica, 4 bytes

OddQ

Gives True for odd inputs and False for even inputs, who knew?

There's also EvenQ, but who would want to type all of that?

brainfuck, 8 bytes

+[,>,]<.

Input is in unary. Output is the 1 (truthy) for odd numbers and NUL (falsy) for even numbers.

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Python, 11 10 bytes

^{-1 byte thanks to Griffin}

1 .__and__

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Using bitwise and, returns 0 for even and 1 for odd

brainfuck, 14 bytes

Input and output is taken as character codes as per this meta.
Byte value 1 correspond to odd numbers and 0 to even.

+>>,[-[->]<]<.

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Taxi, 1,482 1,290 1,063 bytes

I've never written a program in Taxi before and I'm a novice in programming for general, so there are probably better ways to go about this. I've checked for errors and managed to golf it a bit by trying different routes that have the same result. I welcome any and all revision.

Returns 0 for even and 1 for odd.

Go to Post Office:w 1 l, 1 r, 1 l.Pickup a passenger going to The Babelfishery.Go to The Babelfishery:s 1 l, 1 r.Pickup a passenger going to Divide and Conquer.2 is waiting at Starchild Numerology.Go to Starchild Numerology:n 1 l, 1 l, 1 l, 2 l.Pickup a passenger going to Divide and Conquer.Go to Divide and Conquer:e 1 l, 2 r, 3 r, 2 r, 1 r.Pickup a passenger going to Cyclone.Go to Cyclone:e 1 l, 1 l, 2 l.Pickup a passenger going to Trunkers.Pickup a passenger going to Equal's Corner.Go to Trunkers:s 1 l.Pickup a passenger going to Equal's Corner.Go to Equal's Corner:w 1 l.Switch to plan "b" if no one is waiting.Pickup a passenger going to Knots Landing.Go to Knots Landing:n 4 r, 1 r, 2 r, 1 l.[a]Pickup a passenger going to The Babelfishery.Go to The Babelfishery:w 1 l.Pickup a passenger going to Post Office.Go to Post Office:n 1 l, 1 r.Go to Taxi Garage:n 1 r, 1 l, 1 r.[b]0 is waiting at Starchild Numerology.Go to Starchild Numerology:n 1 r.Pickup a passenger going to Knots Landing.Go to Knots Landing:w 1 r, 2 r, 1 r, 2 l, 5 r.Switch to plan "a".

You're right, that's awful to read without line breaks. Here's a formatted version:

Go to Post Office:w 1 l, 1 r, 1 l.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery:s 1 l, 1 r.
Pickup a passenger going to Divide and Conquer.
2 is waiting at Starchild Numerology.
Go to Starchild Numerology:n 1 l, 1 l, 1 l, 2 l.
Pickup a passenger going to Divide and Conquer.
Go to Divide and Conquer:e 1 l, 2 r, 3 r, 2 r, 1 r.
Pickup a passenger going to Cyclone.
Go to Cyclone:e 1 l, 1 l, 2 l.
Pickup a passenger going to Trunkers.
Pickup a passenger going to Equal's Corner.
Go to Trunkers:s 1 l.
Pickup a passenger going to Equal's Corner.
Go to Equal's Corner:w 1 l.
Switch to plan "b" if no one is waiting.
Pickup a passenger going to Knots Landing.
Go to Knots Landing:n 4 r, 1 r, 2 r, 1 l.
[a]
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery:w 1 l.
Pickup a passenger going to Post Office.
Go to Post Office:n 1 l, 1 r.
Go to Taxi Garage:n 1 r, 1 l, 1 r.
[b]
0 is waiting at Starchild Numerology.
Go to Starchild Numerology:n 1 r.
Pickup a passenger going to Knots Landing.
Go to Knots Landing:w 1 r, 2 r, 1 r, 2 l, 5 r.
Switch to plan "a".

Here's my best attempt to explain the logic:

Go to Post Office to pick up the stdin value in a string format.
Go to The Babelfishery to convert the string to a number.
Go to Starchild Numerology to pickup the numerical input 2.
Go to Divide and Conquer to divide the two passengers (stdin & 2).
Go to Cyclone to create a copy of the result.
Go to Trunkers to truncate the original to an integer.
Go to Equal's Corner to see if the two passengers are the same.
Equal's Corner returns the first passenger if they're the same (no .5 removal so the stdin was even) or nothing if they're not.
If nothing was returned, it was odd, so go pick up a 0 from Starchild Numerology.
Go to Knots Landing to convert any 0s to 1s and all other numbers to 0s.
Go to The Babelfishery to convert the passenger (either a 1 or 0 at this point) to a string.
Go to Post Office to print that string.
Go to Taxi Garage to terminate the program.

Fireball, 1 byte

◘

Explanation:

◘    Implicit input mod 2

Fireball is a heavy WIP programming language created by me.

Prints 0 for even values, and 1 for odd.

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LOLCODE, 67 bytes

HOW DUZ I C YR N
  VISIBLE BOTH SAEM MOD OF N AN 2 AN 0
IF U SAY SO

Function that returns WIN (true) if number is not prime, else (prime) it will return FAIL (false).

Call with C"123".

MATL, 5 3 bytes

Because builtins are boring

:He

This outputs a matrix of nonzero values (which is truthy) for even input, and a matrix containing one zero (which is falsy) for odd input.

Try it online! The footer code is an if-else branch to illustrate the truthiness or falsihood of the result. Removing that footer will implicitly display the matrix.

Explanation

Consider input 5 as an example

:     % Implicitly input n. Push row vector [1 2 ... n]
      % STACK: [1 2 3 4 5]
He    % Reshape into a 2-row matrix, padding with zeros if needed
      % STACK: [1 3 5;
                2 4 0]

Japt, 1 byte

v

Returns 1 for even numbers, 0 for odd.

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Explanation

One of Japt's defining features is that unlike most golfing languages, functions do not have fixed arity; that is, any function can accept any number of arguments. This means that you can sometimes leave out arguments and Japt will guess what you want. v on numbers is a function that accepts one argument and returns 1 if the number is divisible by the argument, else 0. It just so happens that the default argument is 2, thereby solving this challenge in a single byte.

2-byte solution (1 byte +1 byte for the -h flag):

¢

¢ converts the input into a base-2 string. The -h flag returns the last char from the string.

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brainfuck, 12 bytes

,++[>++]>++.

This requires an interpreter with a circular tape and cells that wrap around. The one on TIO has 65,536 8-bit cells and satisfies the requirements.

I/O is in bytes. Odd inputs map to 0x00 (falsy), even inputs to a non-zero byte (truthy).

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How it works

We start by reading a byte of input with , and adding 2 to its value with ++. We'll see later why incrementing is necessary.

Next, we enter a loop that advances to the cell at the right, add 2 to it, and repeats the process unless this set the value of the cell to 0.

Initially, all cells except for the input cell hold 0. If the input is odd, adding 2 to it will never zero it out. However, after looping around the tape 127 times, the next iteration of the loop will set the cell to the right of the input cell to 128 × 2 = 0 (mod 256), causing the loop to end. >++ repeats the loop body one more time, so next cell is also zeroed out and then printed with ..

On the other hand, if the input is n and n is even, the code before the loop sets the input cell to n + 2. After looping around the tape (256 - (n - 2)) / 2 = (254 - n) / 2 times, the input cell will reach 0, and the cell to its right will hold the value (254 - n) / 2 × 2 = 254 - n. After adding 2 with >++, . will print 256 - n = -n (mod 256), which is non-zero since n is non-zero.

Finally, note that the second case would print 258 - n = 2 - n (mod n) if we didn't increment the input before the loop, since one more loop around the tape would be required to zero out the input cell. The program would thus fail for input 2.

Java 8, 8 bytes

n->n%2<1

Java 7, 30 bytes

Object c(int n){return n%2<1;}

Outputs true for even numbers and false for odd numbers

Try it here.

If 1/0 would be allowed instead of true/false (it doesn't considering the numbers of votes here):

• Java 8 (6 bytes): n->n%2
• Java 7 (25 bytes): int c(int n){return n%2;}

C++, 25 bytes

template<int v>int o=v&1;

This defines a variable template (a function-like construct) with value equal to the bitwise operation input&1. 0 for even values, 1 for odd values. The value is calculated on compile-time.

Requires C++14.

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C#, 8 bytes

n=>n%2<1

Compiles to a Func<int, bool>.

Or if an anonymous function is disallowed, this method for 21 bytes:

bool p(int n)=>n%2<1;

JavaScript, 6 bytes

An anonymous function:

n=>n&1

Alternatively with the same length:

n=>n%2

Both will return 0|1 which should fulfill the requirement for truthy|falsey values.

Try both versions online

Sinclair ZX81 BASIC 124 bytes (listing)

 1 INPUT A
 2 IF A<1 THEN STOP
 3 LET A=INT A
 4 LET B=INT (A/2)
 5 LET M=A-B*2
 6 PRINT A;":";
 7 GOSUB (M+1)*10
 8 RUN
10 PRINT "TRUE"
11 RETURN
20 PRINT "FALSE"
21 RETURN

Line 3 can be removed to save at least 10 bytes, although it is there to ensure that only integer numbers are excepted (i.e., anything after the decimal point is ignored by the program listing). It's basically doing the closest you'll get to a modulo on ZX81 (or most other 8-bit) BASIC interpreters, so A % 2 is passed to M - then we call the sub-routine at (M+1)*10 which will either print TRUE and return, or FALSE.

There are probably a few ways to save bytes, but it's functioning. I may refactor later. Entering a number less that 1 will exit the program.

Pyth, 3

I was expecting pyth to have a 1 or 2 byte builtin for this. Instead here are the best solutions I could find:

%Q2

or

.&1

or

e.B

05AB1E, 1 byte

È

Fairly self-explantory. Returns a % 2 == 0

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Jelly, 1 byte

Ḃ

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Just another builtin.

For people who don't know Jelly: it has quite a bit of ability to infer missing bits of code, thus there isn't much syntactic difference between a snippet, a function, and a full program; the interpreter will automatically add code to input appropriate arguments and output the result. That's pretty handy when dealing with PPCG rules, which allow functions and programs but disallow snippets. In the TIO link, I'm treating this as a function and running it on each integer from 1 to 20 inclusive, but it works as a full program too.

Jelly, 2 bytes

&1

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It's pretty short without the builtin, too. (This is bitwise-AND with 1.)

Retina, 3 bytes

11

The trailing newline is significant. Takes input in unary. Outputs 1 for odd numbers, nothing for even numbers. Try it online!

Labyrinth, 5 bytes

?_2%!

Prints 0 for even and 1 for odd inputs.

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?   Read input.
_2  Push 2.
%   Modulo.
!   Print.

Now the instruction pointer hits a dead end and turns around. Upon attempting the % on an empty stack, the program exits due to division by zero.

alternatively, also 5 bytes

?#&!@

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?   Read input.
#   Push stack depth (1).
&   Bitwise AND (extract least-significant bit).
!   Print.
@   Terminate.

7, 18 characters, 7 bytes

177407770236713353

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7 doesn't have anything resembling a normal if statement, and has more than one idiomatic way to represent a boolean. As such, it's hard to know what counts as truthy and falsey, but this program uses 1 for odd and the null string for even (the truthy and falsey values for Perl, in which the 7 interpreter is written). (It's easy enough to change this; the odd output is specified before the first 7, the even output is specified between the first two 7s. It might potentially need an output format change to handle other types of output, though; I used the two shortest distinct outputs here.)

7 uses a compressed octal encoding in which three bytes of source represent eight bytes of program, so 18 characters of source are represented in 7 bytes on disk.

Explanation

177407770236713353
 77  77     7       Separate the initial stack into six pieces (between the 7s)

        023         Output format string for "output integers; input one integer"
       7   6        Escape the format string, so that it's interpreted as is
             13     Suppress implicit looping
               3    Output the format string (produces input)
                5   Run the following code a number of times equal to the input:
   40                 Swap the top two stack elements, escaping the top one
                 3  Output the top stack element

Like many output formats, "output integers" undoes any number of levels of escaping before outputting; thus 40, which combined make a swap-and-escape operation, can be used in place of 405, a swap operation (which is a swap-and-escape followed by an unescape). If you were using an output format that isn't stable with respect to escaping, you'd need the full 405 there. (Incidentally, the reason why we needed to escape the format string originally is that if the first output contains unrepresentable characters, it automatically forces output format 7. Escaping it removes the unrepresentable characters and allows format 0 to be selected.)

Of the six initial stack elements, the topmost is the main program (and is consumed by the 13 that's the first thing to run); the second is the 023 that selects the output format and requests input, and is consumed by that operation; the third is consumed as a side effect of the 3 operation (it's used to discard stack elements in addition to producing output); the fourth, 40, is the body of the loop (and consumed by the 5 that executes the loop); and the fifth and sixth are swapped a number of times equal to the input (thus end up in their original positions if the input is even, or in each others' positions a

You could golf off a character by changing the leading 177 to 17 (and relying on an implicit empty sixth stack element), but that would change the parity of the outputs to a less idiomatic method than odd-is-true, and it doesn't save a whole byte (the source is still seven bytes long). As such, I decided to use the more natural form of output, as it doesn't score any worse.

Brain-Flak, 22 20 bytes

Here is annother cool answer in Brain-Flak you should also check out

(({})){({}[()]<>)}<>

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Explanation

To start we will make a copy of our input with (({})).

The bottom copy will serve as a truthy value while the top one will be used for the actual processing. This is done because we need the input to be on the top and it is rather cumbersome to put a 1 underneath the input.

Then we begin a loop {({}[()]<>)}. This is a simple modification on the standard countdown loop that switches stacks each time it decrements.

Since there are two stacks an even number will end up on the top of the stack it started on while an odd number will end on the opposite stack. The copied value will remain in place and thus will act as a marker of where we started.

Once we are done with the loop we have a 0 (originally the input) sitting on top of either a truthy (the copy of the input) or falsy (empty stack) value. We also have the opposite value on the other stack.

We need to get rid of the 0 which can be removed either by {} or <>. Both seem to work and give opposite results, however {} causes a falsy value for zero, when it should return truthy. This is because our "truthy" value is a copy of the input and zero is the only input that can be falsy.

This problem is resolved by ending the program with <> instead.

(Of course according to the specification I do not technically have to support zero but give two options I would prefer to support it)

Piet, 15 codels / 16 bytes

5njaampjhompppam

Online interpreter available here.

This program returns 0 if the input is even and 1 if the input is odd.

The text above represents the image. You can generate the image by pasting it into the text box on the interpreter page. For convenience I have provided the image below where the codel size is 31 pixels. The grid is there for readability and is not a part of the program.

Explanation

This program uses the modulo builtin to determine if the input is even or odd.

Instruction    Δ Hue   Δ Lightness   Stack
------------   -----   -----------   -------
In (Number)    4       2             n
Push [2]       0       1             2, n
Modulo         2       1             n % 2
Out (Number)   5       1             [Empty]
[Exit]         [N/A]   [N/A]         [Empty]

The dark blue codels in the bottom-left are never visited and can be changed to any color other than a color of a neighboring codel. I chose dark blue as I think it looks nice with the rest of the program. The top-left black codel could also be white, but not any other color. I have chosen black as I think it looks nicer.

I have provided the program in both image form and text form as there is no clear consensus on how to score Piet programs. Feel free to weigh in on the meta discussion.

TIS-100, 39 bytes

Of course, this is, more precisely, a program for the T21 Basic Execution Node architecture, as emulated by the TIS-100 emulator.

I'll refer you to this answer for a fantastically in-depth explanation of the scoring for TIS-100 programs, as well as their structure.

@0
ADD UP
G:SUB 2
JGZ G
MOV ACC ANY

Explanation:

@0          # Indicates that this is node 0
ADD UP      # Get input and add it to ACC, the only addressable register in a T-21
G:          # Defines a label called "G"
SUB 2       # Subtracts 2 from ACC
JGZ G       # If ACC is greater than 0, jumps to G
MOV ACC ANY # Sends the value of ACC to the first available neighbor; in this case, output.
            # Implicitly jump back to the first line

In pseudocode, it'd look something like:

while (true) {
    acc = getIntInput()
    do {
        acc -= 2
    } while (acc > 0)
    print(acc)
}

The T21 doesn't have boolean types or truthy/falsy values, so the program returns -1 for odd numbers and 0 for even numbers, unless the previous input was odd, in which case it returns -1 for even numbers and 0 for odd numbers - if that fact disturbs you, this is a full-program answer, so you can just restart your T21 between uses.

Batch, 16 bytes

@cmd/cset/a%1%%2

Outputs 1 for odd, 0 for even. Alternative 16-byte version also works on negative numbers:

@cmd/cset/a"%1&1

17 bytes to output 1 for even, 0 for odd:

@cmd/cset/a"~%1&1

Octave, 12 bytes

@(x)mod(x,2)

Takes x modulus 2. Returns 0 for even numbers and 1 for odd numbers. These evaluates to false and true respectively in Octave. This works in MATLAB too.

Sesos, 2 bytes

SASM

set numin
get
jmp, sub 2

SBIN

00000000: 1228                                              .(

Output is via exit code, zero for even and non-zero for odd. Odd inputs may take a long time.

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How it works

This is a bit cheaty, but it complies with our rules.

The program reads a decimal integer from STDIN and keeps subtracting 2 until it reaches 0. For even inputs, this finishes in linear time and does nothing, successfully.

Programs that do not contain the set mask directive use arbitrary precision integers, so odd inputs will slowly continue to allocate memory. Once the available memory is exhausted, the program will get killed, resulting in a non-zero exit code. Don't expect this to happen anytime soon...

Alternate version, 3 bytes

SASM

set numin
get
jmp, sub 2

SBIN

00000000: 124601                                            .F.

Output is via exit code, zero for even and non-zero for odd. Stray output to STDOUT should be ignored.

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How it works

Marginally less cheaty. The program works as before, but also prints the tape cell as a character in each iteration. This will exit with an error once the cell becomes negative.

Hexagony, 7 bytes

?{2\%'!

Prints 0 for even numbers (falsy) and 1 for odd numbers (truthy).

Try it online!

Explanation

Here is the folded code:

 ? {
2 \ %
 ' !

I believe this is optimal although not unique. As usual I've run a brute force search (not an exhaustive one, but the characters I've excluded are very unlikely to be useful for this program). It did find a whole bunch of other solutions, which I haven't investigated in detail yet:

^?"2^%!
^?"2}%!
{?"2^%!
{?"2}%!
2^?\%"!
2}?\%"!
\{?2%'!
2^?")%!
2^?"1%!
2^?"2%!
2^?=^%!
2{?"}%!
2{?'=%!
2{?={%!
2}?")%!
2}?"1%!
2}?"2%!
2}?=^%!
|"?^2%!
)>"?}!%
\}2?%"!
2^=?^%!
2{'?^%!
2{=?{%!
2}=?^%!

As for the solution I've picked above:

?   Read input.
{   Move to the left memory edge.
    The IP wraps around to the left corner.
2   Set the memory edge to 2.
\   Deflect the IP southwest.
'   Move to the memory edge that points at the input and at the 2.
    The IP wraps around to the right corner.
%   Take the input modulo 2.
!   Print the result.
    The IP wraps to the top right corner.
{   Move to the input edge, which points at two empty edges.
%   Attempt to take the modulo of those, which terminates the program 
    due to the attempted division by zero.

I did look at the next program the above list, which is quite fun because it loops through the first two lines 6 times, filling an entire hexagonal ring with 2s before wrapping back around to the first and taking the modulo. At that point it actually uses the second 2 for the computation. I'm sure there are some gems in the others as well, but I don't think I'll have the time to go through them in detail.