Operator that combines sum, product and difference between two numbers

Question

Challenge:

There's a silly puzzle circulating on social networks that reads:

8 + 2 = 16106
5 + 4 = 2091
9 + 6 = ?

Implement a function or operator that, when given two positive integer numbers x and y such that x > y > 0, yields the correct answer as an integer, where the answer's digits are the digits of x * y followed by the digits of x + y followed by the digits of x - y. Very simple.

Rules:

• Standard loopholes are disallowed.
• This is code-golf so shortest code in bytes wins.
• Input data validation is not required. This program may crash or return garbage when given invalid input.
• You're allowed to use numeric functions and operators (including integer and floating point, math library functions, and other functions that accept and return numbers).
• You're allowed to use a function that returns the number of digits of a number, if applicable.
• You're not allowed to use strings or any kind of concatenation anywhere in your code.
• The result may be returned or pushed to the stack, whichever applies in the language. The result must be an integer number, not a string.

Sample code:

Dyalog APL:

The following code creates a dyadic operator named X.

X←{(⍺-⍵)+((⍺+⍵)×10*1+⌊10⍟⍺-⍵)+⍺×⍵×10*(2+⌊10⍟⍺+⍵)+⌊10⍟⍺-⍵}

Explanation:

• In APL, you evaluate from right to left.

• ⍺ and ⍵ are the left and right operand, respectively

• ⌊10⍟⍺-⍵ reads: floor of log10(⍺-⍵). First performs substraction then logarithm then floor. From right to left. log10 is done in order to count the digits of ⍺-⍵ (you must sum 1 afterwards).

• ⍺×⍵×10*(...) reads: 10 to the (...)th power, multiplied by ⍵, multiplied by ⍺

• Hence, ⍺×⍵×10*(2+⌊10⍟⍺+⍵)+⌊10⍟⍺-⍵ is the product, shifted to the left by the sum of the number of digits of the sum and the difference. Multiplying by a power of 10 will shift an integer to the left.

• ((⍺+⍵)×10*1+⌊10⍟⍺-⍵) is the sum, shifted to the left by the number of digits of the difference.

• (⍺-⍵) is the difference. No shifting is necessary here.

• X←{...} is how you define an operator in APL.

Examples:

      8 X 2
16106
      5 X 4
2091
      9 X 6
54153

GNU dc:

The following code creates a macro named a:

[sysx10lxly-dseZdsclxly+dsd+Z1+^lxly**10lc^ld*+le+]sa

Explanation:

• sx and sy pop an element from the stack and save it on the registers x and y, respectively.

• lx and ly load an element from registers x and y respectively and push it to the stack.

• d duplicates the last element in the stack.

• ^ computes the power of two numbers.

• Z pops a number and returns its number of digits. This is done because dc has no logarithm function.

• [...]sa stores a macro in register a. la loads it. x executes the macro at the top of the stack.

Examples:

8 2 laxn
16106
5 4 laxn
2091
9 6 laxn
54153

Answer 1

JavaScript (ES7), 63 61 59 bytes

Saved 4 bytes thanks to Neil.

(a,b)=>[a*b,a+b,a-b].reduce((p,c)=>p*10**-~Math.log10(c)+c)

<input id=a type=number oninput="c.innerText=((a,b)=>[a*b,a+b,a-b].reduce((p,c)=>p*10**-~Math.log10(c)+c))(+a.value,+b.value)">
<input id=b type=number oninput="c.innerText=((a,b)=>[a*b,a+b,a-b].reduce((p,c)=>p*10**-!Math.log10(c)+c))(+a.value,+b.value)">
<p id=c>

Answer 2

C, 79 75 bytes

Thanks to @G B for saving 4 bytes!

#define X for(c=1;(c*=10)<=a
c,d;f(a,b){X+b;);d=c*a*b+a+b;X-b;);a=d*c+a-b;}

Try it online!

Answer 3

Bash, 66

• 2 bytes saved thanks to @chepner.

f()(s=$[$1+$2]
d=$[$1-$2]
echo $[($1*$2*10**${#s}+s)*10**${#d}+d])

Try it online.

Answer 4

EXCEL, 61 Bytes

=A1-B1+(A1+B1)*10^LEN(A1-B1)+A1*B1*10^(LEN(A1-B1)+LEN(A1+B1))

Excel, 18 Bytes not valid

=A1*B1&A1+B1&A1-B1

Answer 5

Stacked, 36 bytes

,@A$(*+-){!A...n!}"!{%y#'10\^x*y+}#\

Try it online!

Previously: ,@A$(-+*){!A...n!}"!:inits$#'"!$summap:[email protected]\^1\,\*sum

I'm going to try to squeeze out a byte or two before writing an explanation. (#' = size of, and " is "do on each", no strings attached here.)

Noncompeting at 26 bytes: $(*+-)#!!:{%y#'10\^x*y+}#\.

Answer 6

TI-Basic, 34 33 bytes

Prompt A,B
A-B+(A+B)10^(1+int(log(A-B
Ans+AB10^(1+int(log(Ans

Answer 7

GNU dc, 36

Defines a macro m that takes the top two members of the stack, applies the macro and leaves the result on the stack (as per the example in the question):

[sadsbla-dZAr^lalb+*+dZAr^lalb**+]sm

Try it online.

Answer 8

Perl 6,  81 61  58 bytes

->\x,\y{($/=($/=x- y)+(x+y)*({10**$++}...*>$/).tail)+x*y*({10**$++}...*>$/).tail}

Try it

->\x,\y{(x*y,x+y,x- y).reduce:{$^a*10**Int(1+log10($^b))+$b}}

Try it

->\x,\y{[[&({$^a*10**Int(1+$^b.log10)+$b})]] x*y,x+y,x- y}

Try it

Answer 9

Jelly, 27 bytes

+,ạ,.1Fl⁵Ḟ‘€Ṛ+\⁵*
ạ,+,×Fæ.ç

Defines a dyadic link / function, callable with ç. Takes two integers as input and returns an integer. It has the added bonus of being able to take x<y or x>y by using absolute difference.

Try it online!

Explanation:

+,ạ,.1Fl⁵Ḟ‘€Ṛ+\⁵* -- Create link which computes what order of magnitude
                        to multiply the difference, sum, and product by
ạ,+,×Fæ.ç         -- Main link, applies this using dot product

Details:

+,ạ,.1Fl⁵Ḟ‘€Ṛ+\⁵* -- Create dyadic like which does the following:
       l⁵Ḟ‘       -- Create operation which computes number of digits
                       (log base 10 (⁵ is the literal 10), floored, incremented)
           €      -- Apply this to each element in
+,ạ,.1F           -- ... the list [sum,difference,.1]
            R     -- Reverse the list
             +\   -- Add up first n elements to get list.
               ⁵* -- Raise 10 (⁵ is literal 10) to the power of each element

ạ,+,×Fæ.ç         -- Main link, applies above link
ạ,+,×F            -- The list [difference, sum, product]
      æ.          -- Dot product (multiply corresponding elements) with
        ç         -- The above link.

Answer 10

PHP, 79 75 bytes

two versions:

[,$a,$b]=$argv;echo(10**strlen($s=$a+$b)*$a*$b+$s)*10**strlen($d=$a-$b)+$d;
[,$a,$b]=$argv;echo(10**strlen($a+$b)*$a*$b+$a+$b)*10**strlen($a-$b)+$a-$b;

takes input from command line arguments; run with -r.

I guess strlen qualifies as "function that returns the number of digits",
although it uses the number as a string. Let me know if not.

Answer 11

C (gcc), 70 bytes

#define _ for(c=1;a+b>=(c*=10););d=c*d+a-(b=-b);
c,d;f(a,b){d=a*b;_ _}

Try it online!

based on Steadybox answer, putting everything in a macro to golf it a little more.

(Note: assigning the result to d instead of a works, unexpectedly. I had a look at the generated assembly code and it seems to be ok.)

Answer 12

Haskell, 54 bytes

a%0=a
a%b=10*a%div b 10+mod b 10
a#b=(a*b)%(a+b)%(a-b)

The puzzle is implemented via an infix function #, e.g. 8#2 = 16106. The other function, %, defines base-10 concatenation (assuming the RHS is greater than 0).

Answer 13

Dyalog APL, 31 bytes

{a⊥⍨10*1+⌊10⍟a←(⍺×⍵)(⍺+⍵)(⍺-⍵)}

based on the sample APL code from the problem statement

Answer 14

PHP, 87 Bytes

[,$a,$b]=$argv;echo($s=$a-$b)+($t=$a+$b)*10**($l=strlen($s))+$a*$b*10**($l+strlen($t));

and a not valid solution for 37 Bytes

[,$a,$b]=$argv;echo$a*$b,$a+$b,$a-$b;

Answer 15

Ruby, 61 bytes

->a,b{[a*b,a+b,a-b].reduce{|x,y|z=y;x*=10while(z>z/=10);x+y}}

Which suspiciously looks a lot like this Javascript answer, but without using a logarithm.

Answer 16

Python, 92 91 Chars

def g(x,y):
    l=lambda x,i=0:l(x/10,i+1)if x else 10**i
    a=x-y
    a+=(x+y)*l(a)
    return x*y*l(a)+a

Thanks to Wizards suggestion;)

Answer 17

R (3.3.1), 104 bytes

function(x,y)Reduce(function(p,q)p*10^(floor(log10(q)+1))+q,lapply(c(`*`,`+`,`-`),function(z)z(x,y)),0)

returns an anonymous function.

This is my first golfing attempt, so any feedback is appreciated.

Answer 18

REXX, 70 bytes

f:arg a b
c=a-b
c=(a+b)*10**length(c)+c
c=a*b*10**length(c)+c
return c

Of course, the native way would be much shorter:

f:arg a b
return a*b||a+b||a-b

Answer 19

PowerShell, 88 Bytes

param($x,$y)$l=2;if(($a=$x+$y)-gt9){$l++};($x*$y)*[math]::Pow(10,$l)+$a*10+$x-$y

PowerShell does not have a to the power operator which does not help. Also can't count the length of an integer unless you count it as a string, which we can not do, so I check to see if it is -gt 9 to get know the length. Likely could be more terse but I have to get back to work.

Answer 20

Python 2.7, 109 96 bytes

import math
a=lambda n:10**int(math.log10(10*n))
b,c=input()
d=b-c+(b+c)*a(b-c)
print d+b*c*a(d)

Corrected after following rules of contest. Credits to mbomb007 for bringing down the code from 109 bytes to 96 bytes

Answer 21

J, 25 bytes

X=.10#.[:;10#.inv&.>*;+;-

1. *;+;- Box the results of each operation.
2. 10#.inv&.> Convert each result into an array of base-10 digits. (inv is ^:_1)
3. [:; Unbox and join the arrays.
4. 10#. Convert array of base-10 digits into an integer.
5. X=. define the above as the operator X.

Results:

   8 X 2
16106
   5 X 4
2091
   9 X 6
54153

Answer 22

Mathematica, 67 bytes

c=Ceiling;l=Log10;#-#2+(#+#2)10^(c@l[#-#2]/. 0->1)+10^c@l[2#]10#2#&

Takes x-y, then takes the log10 of x-y, rounds it up, calculates 10 to the power of that and then multiplies it by x+y. But we also need to consider log10(x-y) being 0, so we replace 0 with 1. Then we take the log10 of 2x, rounded up, plus 1, and find 10 to the power of that. Multiply that by xy, and add that on.

Answer 23

05AB1E, 23 22 16 bytes

-Dg°¹²+*Dg°¹²**O

Try it online!

We could have saved a few bytes if we'd been allowed to use strings in the program (but not in calculations) by looping over a string containing the operations "-+*", as the code performed for each operation is the same.
Of course, if we'd been allowed to use concatenation we'd saved a lot more.

Answer 24

R, 64 bytes

x=scan();(b=(a=diff(-x))+10^nchar(a)*sum(x))+10^nchar(b)*prod(x)

Usage:

> x=scan();(b=(a=diff(-x))+10^nchar(a)*sum(x))+10^nchar(b)*prod(x)
1: 8 2
3: 
Read 2 items
[1] 16106