Find the smallest integer multiple of a Decimal

Question

The Challenge

Given a rational number, determine the smallest number which is a positive integer multiple of it. Eg.

Given: 1.25
 x1: 1.25
 x2: 2.5
 x3: 3.75
 x4: 5
 Thus: Answer is 5

Clarifications

Inputs will be a rational number in the range of (0,100] of which the Decimal Representation has no more than 4 digits after the point.

Output must be an integer that is the total after multiplication (5 in the example, rather than 4).

I/O may be in any Standard Format, including as a Fraction type.

Be careful if using repeated addition with floats, as 0.3 + 0.3... results in 2.9999....

Test Cases

1.25 -> 5
4.0 -> 4
0.3 -> 3
4.625 -> 37
3.1428 -> 7857

Rules

• Standard Loopholes Apply
• This is code-golf so fewest bytes wins!
• Have Fun!

• Neil: Surely a fraction type trivialises the question?
• ATaco: @Neil If you've got a trivial solution, I implore you to submit it. Also worth noting many, even modern languages, don't have fractional types.
• lyxal: Essentially, this is just finding the numerator of the fraction form of a decimal
• Jonathan Allan: "determine the smallest number which is a positive integer multiple of it." -> "determine the smallest positive integer which is a positive integer multiple of it."? To me the first sounds like we can just output 1*input.
• ATaco: @DominicvanEssen Inaccuracies due to floating-point errors are disallowed. Answers that fail on this point should clarify explicitly, and attempt to provide an answer that does not. I'll add this as a test case.

Answer 1

Vyxal 2.4.1, 2 bytes

ƒh

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A version downgrade improves my byte count somehow.

Well not somehow. 2.4.1 and earlier versions didn't use exact numbers/rationals under the hood, so there was still a "convert decimal to numerator and denominator" built-in (because I've always wanted some form of rationals).

This is a back port of the Factor answer which half ported my original answer.

Takes input as floats because, well, only float input was supported in 2.4.1.

Converts to numerator, denominator. Then gets the first item of that.

Vyxal >=2.6, 35 bits^v2 , 4.375 bytes

S\//h

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Maths? No thank you!

Takes input as either a fraction object or a float. Both are interpreted as rational implicitly so it doesn't really matter.

Simply casts input to string (which'll be a fraction because vyxal 2 be like that), split on / and get the first item.

• ATaco: Oh! Smart approach!
• lyxal: Funny thing is that I noticed the pattern after having written a math inspired answer and running test cases :p

Answer 2

J, 4 bytes

1&*.

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LCM with 1 to get the numerator of a fractional value.

J has a built-in to extract the numerator/denominator from a number too, but is longer to get just the numerator:

J, 6 bytes

0{2&x:

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• Jonah: TIL about dyadic x:

Answer 3

Excel, 37 bytes

=0+TEXTBEFORE(TEXT(A1,"0/00000"),"/")

Input in cell A1.

Answer 4

Factor, 9 bytes

numerator

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Same idea as lyxal's Vyxal answer, just no need for string manipulation.

Less "cheaty," perhaps? This version takes floats as input instead of ratios:

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Factor + math.floating-point, 26 bytes

[ double>ratio numerator ]

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And my original answer for posterity:

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Factor, 34 bytes

[ 0 [ 1 + 2dup * ratio? ] loop * ]

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• 0 push zero — this is our multiplier
• [ ... ratio? ] loop do ... while top of stack is a ratio (a number like 5/5 promotes itself to integer automatically)
• 1 + increment the multiplier
• 2dup * multiply but keep the input and multiplier on the stack too
• * at this point, the input and correct multiplier are on the stack so multiply them

• lyxal: I managed to backport your answer by downgrading my vyxal version :p

Answer 5

Julia, 11 bytes

~x=lcm(x,1)

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Takes rational number inputs. Julia also has the function rationalize for converting floating point values.

Julia 0.6, 33 bytes

>(x,i=1)=isinteger(i*x)?i*x:x>i+1

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Recursive approach which takes an iterator as its own optional argument. The > operator is overloaded, since it has lower precedence than +, which saves a set of parentheses. Spaces are required around ternary operators beginning in Julia 1.0.The function isinteger is a good fit; last("$(i*x)")<'1' would be another (longer) approach.

• MarcMush: Another fun one: >(x,i=1)=try Int(i*x)catch;x>i+1end

Answer 6

R, (削除) 39 (削除ここまで) 31 bytes

\(x,y=x*1:1e6)y[!(y+1e7)%%1][1]

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Correctly handles difficult inputs that otherwise suffer from floating-point inaccuracies.

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R, 25 bytes (non-competing)

\(x,y=x*1:1e6)y[!y%%1][1]

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Original answer: fails due to floating-point inaccuracies for 3.1428, and probably other difficult inputs.

Answer 7

Google Sheets, 31 bytes

=+split(text(A1,"#/#####"),"/")

Put the input in cell A1 and the formula in B1.

Uses number formatting like the Excel answer to get a fraction, then takes advantage of an undocumented feature of the unary plus operator +, i.e., uplus(), to get the numerator.

Answer 8

Ruby, 26 bytes

->d{(1..).find{_1/d%1==0}}

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Original: Ruby, 31 bytes

f=->d,x=1{d*x%1>0?f[d,x+1]:d*x}

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The more obvious f=->d{d.to_r.numerator} does not seem to work for 0.300.

Thanks Value Ink for the tips, but I think that it should be a separate answer.

• Value Ink: ->d{d.to_s.to_r.numerator} does work and is shorter by 5 bytes, and taking input as a string will lead to the obvious solution you posited. In fact, since the rules explicitly say a fractional input is allowed, proc &:numerator also works for the truly "cheaty" version.

Answer 9

Python, 20 bytes

lambda n:n.numerator

An unnamed function that accepts a Fraction object and returns the integer.

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• Jonathan Allan: Floating point inaccuracy I think - it accepts a float and does not perform repeated addition.
• Jonathan Allan: @NickKennedy removed.

Answer 10

Nekomata, 1 byte

ṇ

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ṇ = 'numerator' = 'Get the numerator of a number'

Answer 11

PARI/GP, 25 bytes

n->numerator(bestappr(n))

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Answer 12

Perl 5 -p, 26 bytes

($\=$_*++$i)=~/\./&&redo}{

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Answer 13

Maxima, 17 bytes

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f(n):=num(rat(n))

Answer 14

APL(Dyalog Unicode), ^{(削除) (削除ここまで)}3 bytes ^SBCS

∧∘1

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LCM between input and 1. This is actually really similar to a problem in the APL Competition, 1:7 Let's Be Rational.

Answer 15

Pyth, 6 bytes

*fsI*Q

Takes input as a float.

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*fsI*Q
 f Find the first integer, starting at 1 and counting upward, such that
 *Q The integer multiplied by the input
 I Is unchanged when
 s Cast to an integer
* Multiply the result (the denominator) by the input

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Same length:

fsIcTQ

Finds the numerator directly using division.

• Nick Kennedy: Fails for 3.1428
• izzyg: @NickKennedy My other answer, fsIcTQ, works for that input - do you know if it also has a counterexample?

Answer 16

Rust, (削除) 39 (削除ここまで) 38 bytes

-1 thanks to @corvus_192

|mut x|{let n=x;while x%1.>0.{x+=n};x}

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Closure naively loops and increments x by its intial value until it becomes a whole number.

• corvus_192: I think using > instead of != works.
• code: @corvus_192 you're right, thanks.

Answer 17

Python 3, 53 bytes

Trivial solution (thanks to lyxal for the comment).

lambda n:Fraction(n).numerator
from fractions import*

Originally (65 bytes, returns a float which can be converted to an int while being the same value):

def x(n,i=1):
 while not str(i*n).endswith('.0'):i+=1
 return i*n

20 bytes from SuperStormer:

lambda n:n.numerator

• SuperStormer: 20 bytes: lambda n:n.numerator meta
• Rhaixer: The number is not explicitly passed as a Fraction (I assume it is an int), however, I guess I'll add that as an extra answer.

Answer 18

AWK, (削除) 35 (削除ここまで) 24 bytes

{while(1ドル*++a%1);1ドル*=a}1

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(Several improvements by @Dominic van Essen to my original answer: -11 bytes)

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Original answer: AWK, 35 bytes

{while(1ドル*++a>int(1ドル*a));print1ドル*a}

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• Dominic van Essen: 24 bytes...
• Nick Kennedy: Fails for 3.1428

Answer 19

Vyxal, 39 bits^v2 , 4.875 bytes

{D⌊≠|+

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Posting as a separate answer because it doesn't use any fraction string cheese :p

Takes input as either a fraction or a float, as both are implicitly converted to rationals.

Explained

{D⌊≠|+­⁡​‎‎⁡⁠⁡‏⁠‎⁡⁠⁢⁡‏‏​⁡⁠⁡‌⁢​‎⁠‎⁡⁠⁣‏⁠‎⁡⁠⁤‏‏​⁡⁠⁡‌⁣​‎‎⁡⁠⁢‏‏​⁡⁠⁡‌⁤​‎‎⁡⁠⁢⁢‏‏​⁡⁠⁡‌­
{ | # ‎⁡While
 ⌊≠ # ‎⁢ The floor of the top of the stack doesn't equal the top of the stack
 D # ‎⁣ (triplicate before the check so there's a copy left over)
 + # ‎⁤Add the input to the top of the stack
💎

Created with the help of Luminespire.

Answer 20

Charcoal, 27 bytes

≔I−θ.ηF⊟⪪θ.F25F¬%ηIκ≧÷IκηIη

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≔I−θ.η

Remove the decimal point from the input and cast it to integer.

F⊟⪪θ.

Loop over each decimal place.

F25F¬%ηIκ≧÷Iκη

Divide the integer by 2 and/or 5 if possible.

Iη

Output the resulting integer.

Answer 21

Retina, (削除) 95 (削除ここまで) 71 bytes

\.(\d+)$
1,ドル$.1*0
.*5,0
$.(*2*),0
.*[2468],0
$.(*5*),0
}`0,0
,
^0+|,0*

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\.(\d+)$
1,ドル$.1*0

On the first pass, delete the decimal point but keep track of how many digits there were after it. (It's golfier to repeat this stage but it has no effect on subsequent passes.)

(.*5),0
$.(*2*),0

Try doubling the number to create a trailing zero to match.

(.*[2468]),0
$.(*5*),0

Try multiplying the number by 5 to create a trailing zero to match.

0,0
,

Remove a pair of matching zeros.

}`

Repeat until there are no more trailing zeros to remove or the number ends in 1, 3, 7 or 9.

^0+|,0*

Delete any leading zeros and any remaining working digits.

Answer 22

T-SQL, 51 bytes

DECLARE @2 real=4.625
DECLARE @ REAL=@2
WHILE @>floor(@)SET @+=@2
PRINT @

Answer 23

Swift 2.2, floating-point, 65 bytes

func f(x:Float,y:Float=1)->Float{return x*y%1==0 ?x*y:f(x,y:y+1)}

Back in Swift 2, % was supported for floating-point types. I also can't figure out what Decimal was called back then.

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Swift 5.8 + Foundation, decimal, 109 bytes

func f(x:Decimal,y:Decimal=1)->Decimal{var z=x,w=x*y
NSDecimalRound(&z,&w,0,.up)
return w==z ?w:f(x:x,y:y+1)}

Rounding mode doesn't matter; I chose .up because it's the shortest. The function signature of NSDecimalRound makes more sense in Objective-C where you only work with NSDecimal* and not the struct directly.

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Answer 24

05AB1E, 8 bytes

∞.ΔI/DïQ

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Explanation:

∞ # Push an infinite positive list: [1,2,3,...]
 .Δ # Pop and find the first integer that's truthy for:
 I/ # Divide the current integer by the input-decimal
 D # Duplicate the decimal
 ï # Cast/floor its copy to an integer
 Q # Check if both are the same value
 # (after which the found integer is output implicitly as result)

• Kevin Cruijssen: @NickKennedy Should be fixed, although it's plausible another test case might still fail due to floating point accuracy tbh.

Answer 25

JavaScript (V8), 31 bytes

f=(n,m=n)=>(m+1e9)%1?f(n,m+n):m

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Take float

JavaScript (V8), 4 bytes

x=>x

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• Dominic van Essen: Is the 4-byte answer a joke? It obviously fails if input is given as a non-reduced fraction.
• l4m2: @DominicvanEssen I'd say requiring reduced fraction input is usually reasonable, but yeah it's not serious here

Answer 26

Ruby, 16 bytes

Builtin solution posted as a separate answer from G B's answer per their suggestion. Takes a Rational as input.

proc &:numerator

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Ruby -n, 18 bytes

The same idea but takes input from STDIN via -n and prints the answer.

p$_.to_r.numerator

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