Check whether an integer is a power of 2 without using +,- operations

Question

Write a program that checks if the integer is a power of 2.

---

Sample input:

8

Sample output:

Yes

---

Sample input:

10

Sample output:

No

---

Rules:

• Don't use +,- operations.

• Use some sort of input stream to get the number. Input is not supposed to be initially stored in a variable.

• The shortest code (in bytes) wins.

You can use any truthy/falsy response (for instance, true/false). You may assume that input number is greater than 0.

Answer 1

GolfScript, 6 chars, no decrements

~.3/&!

Here's a solution that doesn't use the x & (x-1) method in any form. It uses x & (x/3) instead. ;-) Outputs 0 if false, 1 if true.

Explanation:

• ~ evals the input string to turn it into a number,
• . duplicates it (for the subsequent &),
• 3/ divides it by three (truncating down),
• & computes the bitwise AND of the divided value with the original, which will be zero if and only if the input is zero or a power of two (i.e. has at most one bit set), and
• ! logically negates this, mapping zero to one and all other values to zero.

Notes:

• Per the clarified rules, zero is not a valid input, so this code is OK, even though it outputs 1 if the input is zero.

• If the GolfScript decrement operator ( is allowed, then the 5-character solution ~.(&! posted by aditsu is enough. However, it seems to go against the spirit of the rules, if not the letter.

• I came up with the x & (x/3) trick years ago on the Fun With Perl mailing list. (I'm sure I'm not the first to discover it, but I did (re)invent it independently.) Here's a link to the original post, including a proof that it actually works.

Answer 2

APL (7)

Yes, that's 7 bytes. Assume for the moment that I'm using IBM codepage 907 instead of Unicode and then each character is a byte :)

0=1|2⍟⎕

i.e. 0 = mod(log(input(),2),1)

Answer 3

GolfScript, 11 (for 1 (true) and 0 (false))

.,{2\?}%?0>

Put the number on the stack and then run.

GolfScript, 22 (for Yes/No)

.,{2\?}%?0>'Yes''No'if

I love how converting 1/0 to Yes/No takes as much code as the challenge itself :D

Warning: EXTREMELY inefficient ;) It does work fine for numbers up to 10000, but once you get that high you start to notice slight lag.

Explanation:

• .,: turns n into n 0..n (. duplicate, , 0..n range)
• {2\?}: to the power of 2
• %: map "power of 2" over "0..n" so it becomes n [1 2 4 8 16 ...]
• ?0>: checks to see if the array contains the number (0 is greater than index)

Answer 4

Mathematica 28

Numerator[Input[]~Log~2]==1

For integer powers of 2, the numerator of the base 2 log will be 1 (meaning that the log is a unit fraction).

Here we modify the function slightly to display the presumed input. We use # in place of Input[] and add & to define a pure function. It returns the same answer that would be returned if the user input the number in the above function.

    Numerator[#~Log~2] == 1 &[1024]
    Numerator[#~Log~2] == 1 &[17]

True
False

Testing several numbers at a time.

    Numerator[#~Log~2] == 1 &&/@{64,8,7,0}

{True, True, False, False}

Answer 5

Perl 6 (17 characters)

say get.log(2)%%1

This program gets a line from STDIN get function, calculates logarithm with base 2 on it (log(2)), and checks if the result divides by 1 (%%1, where %% is divides by operator). Not as short as GolfScript solution, but I find this acceptable (GolfScript wins everything anyway), but way faster (even considering that Perl 6 is slow right now).

~ $ perl6 -e 'say get.log(2)%%1'
256
True
~ $ perl6 -e 'say get.log(2)%%1'
255
False

Answer 6

Octave (15 23)

EDIT: Updated due to user input requirement;

~mod(log2(input('')),1)

Lets the user input a value and outputs 1 for true, 0 for false.

Tested in Octave, should work in Matlab also.

Answer 7

R, 13 11

Based on the Perl solution. Returns FALSE or TRUE.

!log2(i)%%1

The parameter i represents the input variable.

An alternative version with user input:

!log2(scan())%%1

Answer 8

GolfScript, 5

Outputs 1 for true, 0 for false. Based on user3142747's idea:

~.(&!

Note: ( is decrement, hopefully it doesn't count as - :)
If it does (and the OP's comments suggest that it might), then please refer to Ilmari Karonen's solution instead.
For Y/N output, append 'NY'1/= at the end (7 more bytes).

Answer 9

Python, 31

print 3>bin(input()).rfind('1')

Answer 10

C, 48

main(x){scanf("%i",&x);puts(x&~(x*~0)?"F":"T");}

Answer 11

I decided to to use another approach, based on the population count or sideways sum of the number (the number of 1-bits). The idea is that all powers of two have exactly one 1 bit, and no other number does. I added a JavaScript version because I found it amusing, though it certainly won't win any golfing competition.

J, 14 15 chars (outputs 0 or 1)

1=##~#:".1!:1]1

JavaScript, 76 chars (outputs true or false)

alert((~~prompt()).toString(2).split("").map(Number).filter(Boolean).length)

Answer 12

Clip, 9 8 7

!%lnxWO

Reads a number from stdin.

Explanation:

To start with, Z = 0, W = 2 and O = 1. This allows placing of W and O next to each other, whereas using 2 and 1 would be interpreted as the number 21 without a separating space (an unwanted extra character). In Clip, the modulo function (%) works on non-integers, so, to work out if some value v is an integer, you check if v mod 1 = 0. Using Clip syntax, this is written as =0%v1. However, as booleans are stored as 1 (or anything else) and 0, checking if something is equal to 0 is just 'not'ing it. For this, Clip has the ! operator. In my code, v is lnx2. x is an input from stdin, n converts a string to a number and lab is log base b of a. The program therefore translates (more readably) to 0 = ((log base 2 of parseInt(readLine)) mod 1).

Examples:

8

outputs

1

and

10

outputs

0

Edit 1: replaced 0, 1 and 2 with Z, O and W.

Edit 2: replaced =Z with !.

Also:

Pyth, 5

Compresses the Clip version even further, as Pyth has Q for already evaluated input and a log2(a) function instead of just general log(a, b).

!%lQ1

Answer 13

Javascript (37)

a=prompt();while(a>1)a/=2;alert(a==1)

Simple script that just divides by 2 repeatedly and checks the remainder.

Answer 14

Mathematica (21)

IntegerQ@Log2@Input[]

Without input it is a bit shorter

IntegerQ@Log2[8]

True

IntegerQ@Log2[7]

False

Answer 15

JavaScript, 41 40 characters

l=Math.log;alert(l(prompt())/l(2)%1==0);

How this works: you take the logarithm at base 2 using l(prompt()) / l(2), and if that result modulo 1 is equal to zero, then it is a power of 2.

For example: after taking the logarithm of 8 on base 2, you get 3. 3 modulo 1 is equal to 0, so this returns true.

After taking the logarithm of 7 on base 2, you get 2.807354922057604. 2.807354922057604 modulo 1 is equal to 0.807354922057604, so this returns false.

Answer 16

JavaScript, 35

Works for bytes.

alert((+prompt()).toString(2)%9==1)

46 character version, Works for 16 bit numbers.

x=(+prompt()).toString(2)%99;alert(x==1|x==10)

The trick works in most dynamic languages.

Explanation: Convert the number to base 2, interpret that string as base 10, do modulo 9 to get the digit sum, which must be 1.

Answer 17

K/Kona (24 17)

d:{(+/(2_vs x))~1

Returns 1 if true and 0 if false. Any power of 2 has a single bit equal to 1:

2_vs'_(2^'(!10))
(,1
1 0
1 0 0
1 0 0 0
1 0 0 0 0
1 0 0 0 0 0
1 0 0 0 0 0 0
1 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0)

(this prints out all the powers of 2 (from 0 to 9) in binary)

So I sum up all the components of the binary expression of x and see if it's equal to 1; if yes then x=2^n, otherwise nope.

...knew I could make it smaller

Answer 18

python 3, 38

print(1==bin(int(input())).count('1'))

python, 32

However, the code doesn't work in every version.

print 1==bin(input()).count('1')

Notice that the solution works also for 0 (print False).

Answer 19

Perl 5.10+, 13 + 1 = 14 chars

say!($_&$_/3)

Uses the same method from an old FWP thread as my GolfScript entry. Prints 1 if the input is a power of two, and an empty line otherwise.

Needs to be run with perl -nE; the n costs one extra char, for a total of 14 chars. Alternatively, here's an 18-character version that doesn't need the n:

say!(($_=<>)&$_/3)

Answer 20

Ruby — 17 characters (fourth try)

p /.1/!~'%b'%gets

My current best is a fusion of @steenslag's answer with my own. Below are my previous attempts.

Ruby — 19 characters (third try)

p /10*1/!~'%b'%gets

Ruby — 22 characters (second try)

p !('%b'%gets)[/10*1/]

Ruby — 24 characters (first try)

p !!('%b'%gets=~/^10*$/)

Answer 21

C# (54 characters)

 Math.Log(int.Parse(Console.ReadLine()),2)%1==0?"Y":"N"

Answer 22

Rebmu (9 chars)

z?MOl2A 1

Test

>> rebmu/args [z?MOl2A 1] 7
== false

>> rebmu/args [z?MOl2A 1] 8 
== true

>> rebmu/args [z?MOl2A 1] 9 
== false

Rebmu is a constricted dialect of Rebol. The code is essentially:

z? mo l2 a 1  ; zero? mod log-2 input 1

Alternative

14 chars—Rebmu does not have a 'mushed' bitwise AND~

z?AND~aTIddA 3

In Rebol:

zero? a & to-integer a / 3

Answer 23

SmileBASIC, 29 bytes

This is a fairly uncreative and basic (no pun intended) answer. 0 is false, 1 is true.

INPUT N
A=LOG(N,2)?(0OR A)==A

We know that if n is a power of 2, then log2(n) will be an integer. So, this calculates the base-2 log of N, our input number, and checks if it is an integer using the following method.

In SB there are actually two number types, doubles and integers. It also has automatic coercion between doubles and ints depending on the context. When a float is coerced to an int, it simply truncates the fraction and produces an integer which represents the whole part (assuming that number is in the valid integer size–signed 32 bit.) The bitwise OR operator OR expects both of its terms to be integers, so it will automatically cast doubles to ints. Thus, 0 OR double simply produces that double's whole part, which we then check against the original value. If these two are equal, it is a power of two.

Answer 24

Haskell, 47 bytes

(or 21 bytes if input/output can be function argument/return)

f 0=0
f 1=1
f n=f$n/2
main=interact$show.f.read

Returns 1 for powers of 2, and 0 otherwise.

Answer 25

Python 2, 29,25 bytes

x=input();print x&x*~0==x

Try it online!

using x&x*~~x/~x is a modified version of x&(~x+1)! Since the later uses + had to use ~x which is equivalent to -x-1!

• saved 4 bytes by using x*~0 instead of x*~~x/~x!

Answer 26

GTB, 46 bytes

2→P`N[@N=Pg;1P^2→P@P>1E90g;2]l;2~"NO"&l;1"YES"

Answer 27

Python 2.7 (30 29 39 37)

EDIT: Updated due to user input requirement;

a=input()
while a>1:a/=2.
print a==1

Brute force, try to divide until =1 (success) or <1 (fail)

Answer 28

APL (12 for 0/1, 27 for yes/no)

≠/A=2*0,ιA←⍞ 

or, if we must output text:

3↑(3x≠/A=2*0,ιA←⍞)↓'YESNO '

Read in A. Form a vector 0..A, then a vector 2⁰..2^A (yes, that's way more than necessary), then a vector comparing A with each of those (resulting in a vector of 0's and at most one 1), then xor that (there's no xor operator in APL, but ≠ applied to booleans will act as one.) We now have 0 or 1.

To get YES or NO: multiply the 0 or 1 by 3, drop this number of characters from 'YESNO ', then take the first 3 characters of this.

Answer 29

Ruby, 33, 28, 25

p /.1/!~gets.to_i.to_s(2)

Answer 30

Python, 35

print bin(input()).strip('0')=='b1'

Doesn't use not only +/- operations, but any math operations aside from converting to binary form.

Other stuff (interesting, but not for competition):

I have also a regexp version (61):

import re;print re.match(r'^0b10+$',bin(input())) is not None

(Love the idea, but import and match function make it too long)

And nice, but boring bitwise operations version (31):

x=input();print x and not x&~-x

(yes, it's shorter, but it uses ~-x for decrement which comtains - operation)

Answer 31

C, 65 bytes

main(k){scanf("%i",&k);while(k&&!(k%2))k/=2;puts(k==1?"T":"F");}

Answer 32

Haskell (52 50)

k n=elem n$map(2^)[1..n]
main=interact$show.k.read

Answer 33

Python (33)

print int(bin(input())[3:]or 0)<1

Answer 34

OCaml - 43 or 45

Since no-one has submitted an OCaml solution, I submit one inspired by nightcrackers C solution.

There is some debate as to whether the operator - includes unary negation. Other entries have managed to sidestep the issue by finding a solution that is equally short without unary negation. I instead give two answers:

let x=read_int()in 0=x land lnot(x*(-1));;

and

let x=read_int()in 0=x land lnot(x*lnot 0);;

the ;; ends a group of statements in the Ocaml interpreter and causes the result to be printed as - : bool = false or - : bool = true, which are clear negative/positive answers as required by the question.

Answer 35

Python 2 - 23

x=input();print x&x/3<1

This is a direct port of the golf-script solution above, that beats the current best python solution by 8 characters. The order of operations works out amazingly well here.

Answer 36

Bash + dc + grep, 24

dc -e2o?p|grep -c ^10\*$

Reads from stdin. Outputs "1" if a power of 2 and "0" otherwise:

$ dc -e2o?p|grep -c ^10\*$
8
1
$ dc -e2o?p|grep -c ^10\*$
10
0
$ 

---

Pure Bash (only builtins), 40

[[ `read a;printf %o $a` =~ ^[124]0*$ ]]

Reads from stdin. Returns success (0) if a power of 2 and failure (1) otherwise:

$ [[ `read a;printf %o $a` =~ ^[124]0*$ ]]
8
$ echo $?
0
$ [[ `read a;printf %o $a` =~ ^[124]0*$ ]]
10
$ echo $?
1
$

Answer 37

Java 7, 124

This is pretty short in Java. Most of the program is just boilerplate code and to read the integer.

class H{public static void main(String[]b){int a=new java.util.Scanner(System.in).nextInt();System.out.println((a&a/3)<1);}}

Answer 38

AutoHotkey 33 bytes

a=%1%
While a>1
a:=a/2
Send % a=1

Input parameters are assigned to variables 1, then 2, etc. It means you have to assign it to another variable before using it in math because AHK will think you mean the number 1 and not the variable 1. It would be slightly shorter still but AHK treats a/=2 as a a floor divide by 2 instead of just a divide by 2 if and only if it is the left-most assignment operation on a line. I'm sure it has its reasons.

---

This is, surprisingly, slightly shorter than the method that uses log():

a=%1%
Send % Mod(Log(a)/Log(2),2)=0

Answer 39

Excel 19 bytes

=MOD(LOG(A1,2),2)=0

Nothing magical or tricky. Input is from cell A1. Returns TRUE or FALSE.

Answer 40

Haskell, 41 bytes

f n=mod(2^n)n<1
main=interact$show.f.read

Alternative:

main=interact$show.(<1).(mod=<<(2^)).read

Try it online!

---

43 bytes

main=interact$show.(1==).until(<2)(/2).read

Try it online. Halves x until it's less than 2, and checks if the result is 1.

Answer 41

PHP, 86 chars

function P($a,$c){if($c==$a)return 1;if($c>$a)return 0;return P($a,$c*2);}echo P(x,2);

Replace x with the number you want to test.

Answer 42

PHP (58)

<?php $a=fgets(STDIN);while($a%2==0)$a/=2;echo $a==1?1:0;

Simple divide-by-2 loop and remainder check.

Answer 43

ActionScript 3 (63 characters*)

var r = Math.log(prompt())/Math.log(2);
return ( (r > 0) && (int(r) == r) );

* I'm going to ignore the fact that AS3 doesn't have a built-in prompt() method, (and hope AS3's lack of implementation doesn't get counted against me).

Extra white-space left in for readability and clarity, but not counted, since they can be removed before execution.

Answer 44

GTB, 17

With 1/0

`Ar?A>1:A/2→A~A=1

With YES/NO (31)

`Ar?A>1:A/2→A@A=1$~"YES"#~"NO"&

Answer 45

JavaScript (43 characters when compacted, 35 for some browsers)

Readable version:

var r = Math.log(prompt()) / Math.LN2;
alert( (r > 0) && (~~r == r) );

Compacted version (credit goes to FireFly for helping me shave off even more characters):

r=Math.log(prompt())/Math.LN2;alert(~~r==r)

JS Fiddle: http://jsfiddle.net/IQAndreas/LFYg7/2/

EDIT: On the off chance that you use FireFox 25+ or any other browser with Math.log2(), only 35 characters are required (thanks to xfix for this suggestion):

r=Math.log2(prompt());alert(~~r==r)

Answer 46

Scheme 40

(display(=(mod(/(log(read))(log 2))1)0))

It displays #t for true and #f for false (the booleans in Scheme).

For true compatibility you need to add (import(rnrs)) or (import(scheme)) (for r7rs) but it works without that in ikarus and in most REPLs.

Answer 47

Perl 53

perl -pe '$_=unpack"b*",pack"L",$_;s!0!!g;s!111*!no!;s!1!yes!'

I can shave off 1 character if i use + in the re, not used for adding.

Answer 48

Golfscript (17 14)

Works from input, using the base builtin for binary conversion and counting ones;

~2base{1=},,1=

Old version, 17 chars, without builtin conversions and without using the decrement operator as a substitute for -;

~2*:a 1{2*.a<}do=

• ~ lifts the input to an integer on the stack
• 2* multiplies by 2 to solve 2^0 problem
• :a sets the variable a to the number to search for and leaves it on stack
• 1{2*.a<}do sets start value 1 and doubles as long as number<a
• = compares the number to search for with the number generated

Answer 49

J - 17 15 bytes - not . ceiling (0==) . mod 1 . log2 . to_i . read keyboard

Result is 1 for True and 0 for False

0=1|2^.".1!:1,1

Yes/No version

(0=1|2^.".1!:1,1){'No',:'Yes'

Answer 50

D (83 chars with boilerplate, 17 chars to get the solution)

import std.stdio;
void main(){
    int i;
    readf("%d", &i);
    while(!(i%2))i/2;
    writeln(i==1);
}

while the number is even divide by 2, when the result is 2 then it was a power of 2

Answer 51

Perl 69

$_=<>;chomp;$s=1;while($s<$_){$s*=2;$r="Yes" if $s==$_}print $r||"No"

Answer 52

Perl 5 (36 or 29 bytes)

say sprintf('%b',<>)=~/^10*$/?Yes:No

If we don't want to strictly output "Yes" or "No", then we can go shorter:

say sprintf('%b',<>)=~/^10*$/

This second version outputs "1" for true, and "" for false.

Answer 53

C99 94 92 71

int main(){int a,i=1;scanf("%d",&a);while(i){if(a==i)return 1;i<<=1;}}

readable

#include <stdio.h>

int main() {
    int a, i=1;
    scanf("%d",&a);
    while(i){
        if(a==i) return 1;
        i<<=1;
    }
}

What it does

It XORs the input value with a power of 2, until the bit of variable i is shifted out and the value becomes 0. Only if the input value is also a power of 2 (with other words, only one bit in the input was set) the XOR resolves to zero.

The input will be compared with a power of 2 value, that is left shifted in every iteration (in other words multiplied by 2) and returns if both are equal, thus the input is also a power of 2.

On success 1 is returned, 0 otherwise.

Why C99?

Because the return value of a C program is implicitly defined as 0, if not specified, since C99. Compile with: gcc -std=c99 ...

Answer 54

autohotkey -45bytes in txt file and 48 bytes with editor and lang.

MsgBox % mod(log(k)/log(2), 1) ? "no" : "yes"

Answer 55

PHP, 55

<? if(fmod(log($argv[1],2),1)==0) echo 1; else echo 0;

Eg:

# php pow2.php 8
1
# php pow2.php 9
0

PHP w/ interactive I/O, 107

<? $h=fopen('php://stdin','r');$n=trim(fgets($h));fclose($h);if(fmod(log($n,2),1)==0) echo 1; else echo 0;

Answer 56

Python (40 char)

p=lambda n:n>0and(n<2or n%2<1and p(n/2))

Works for negative number also (returning False)

C (39 chars)

p(n){return n>0&&(n<2||n%2<1&&p(n/2));}

or with trick, (work on old compilers without optimizing, withoput return operator returns last evaluated expression value) (32 chars)

p(n){n>0&&(n<2||n%2<1&&p(n/2));}

Answer 57

Java 143

Thinking binary ( a power of 2 has always only one bit set so the compare with the integer with ionly highest bit set can only return true on powers of 2 (including the ominous 0 :P)

Update: shaved 2 bytes off due to a derp

enum P{public static void main(String[]a){int i=new java.util.Scanner(System.in).nextInt();System.out.print(i==(Integer.highestOneBit(i)*1));}}

Ungolfed

enum P;
{
    public static void main(String[]a)
    {
        int i = new java.util.Scanner(System.in).nextInt();
        System.out.print(i==(Integer.highestOneBit(i)*1));
    }
}

Answer 58

BASH, 48

x=`echo "obase=2;$1"|bc|rev|bc`;echo "$x==1"|bc

Takes input as a command line argument and returns 1 for true and 0 for false. You can easily modify it to echo Yes/No, but the bash if-construct makes it much longer.

Answer 59

~-~! - 37

'=|*;~~==%['&*/~~]*|:'&^==~[@|1|]@|0|

I had to improvise with the number input...it's the Unicode value of the character you enter. Teehee.

This uses a recursive /2 solution.

Answer 60

JavaScript (EcmaScript 6) - 13 Characters

f=x=>!(x&x/3)

Creates a function f that takes a value and returns true (or false) if it is a power of 2 (or not) and will output the answer to the console.

Based on Ilmari Karonen's answer (so I included a few others of my own, below).

EcmaScript 6 can be run on FireFox or SpiderMonkey - no other browser/JavaScript engine yet supports EcmaScript 6 arrow functions. If you want to prompt for input then you can call it as an anonymous function (23 characters):

(x=>!(x&x/3))(prompt())

But it's less characters skip the function bit and just do:

JavaScript - 19 Characters

!((a=prompt())&a/3)

Will do the same as above but should run in all browsers and will prompt for input and output to the console (or in SpiderMonkey, you can replace prompt() [which doesn't exist as SpiderMonkey is command-line based] with readline()).

JavaScript (EcmaScript 6) - 20 Characters

f=x=>x>1?f(x/2):x==1

Creates a function f and, when called, will output the answer to the console.

JavaScript - 29 Characters

for(a=prompt();a>1;)a/=2;a==1

Same as above but will prompt the user for input and output to the console.

JavaScript - 32 Characters

for(a=prompt();!(a%2);)a/=2;a==1

JavaScript - 34 Characters

for(a=prompt();a==(a|0);)a/=2;a==1

Answer 61

dc, 13 bytes

?[2~0=m]dsmxp

Reads input from STDIN. Input is not explicitly stored in a variable, though it is stored on the stack - not sure if this too much rule-bending.

Successively divides by 2 until a 1 remainder is found. Since dc doesn't have any native concept of truthy/falsey, I'm taking the liberty of defining my own here. 0 means true, any +ve integer means false:

$ for i in 1 2 3 15 16 17 65535 65536; do dc pow2.dc <<< $i; done
0
0
1
7
0
8
32767
0
$ 

Answer 62

Batch - 24 bytes

Stealing Ilmari's winning GolfScript answer...

@cmd/c"set/a!(%1/3^&%1)"

See explanation here.

Answer 63

Note: This question is older than Pyth, so this answer is not eligible to win.

Pyth, 5 bytes

!%lQ1

This is essentially identical to @marinus's APL answer. We take the log base 2 of the input, take it mod 1, and take the logical not.

---

Pyth, 6 bytes

!stjQ2

This answer converts the input to base 2, removes the first digit, sums the remainder, and takes the logical not.

Answer 64

Python (23)

int(bin(input())[3:])<1

Answer 65

Python 3.6, 30 bytes

from enum import _power_of_two

... there's a builtin lurking in the stdlib ...

Answer 66

Python:

print 0==int(bin(input())[3:] or 1)

Answer 67

Easiest way: An unsigned number is a power of 2 if only 1 BIT is set: if (!(n&(n-1))). Subtracting 1 inverts all BITs. if n=10000000b (80h/128), n&01111111b=0

if (!(n&(n-1)))
  printf("%d is a power of 2", n);
else
 printf("%d is NOT a power of 2", n);

// to get specific power of 2, search from
// right to left for 1st 0 BIT

int pow2(unsigned n) { // is power of 2?
  if (n&(n-1) || n<2) // if not, return 0
  return 0;
  n--;
  for (int i=1; n&1<<i; i++);
  return i; // 2^i
}

Answer 68

Basic BASH command

# assume integer is stored in variable QUERY
[ "$[(${QUERY}/2)*2]" = "$QUERY" ] && echo "Is even" || echo "Is odd"

Takes advantage of how the shell does integer division, and drops any remainder.

weylin

Answer 69

#include <stdio.h>

int main(){
    int i=16;
    if(i&(i-1))
        printf("No\n");
    else
        printf("Yes\n");
    return 0;
}