Making any integer with four 2s (2025)

(eli.thegreenplace.net)

172 points | by germanrabbit 17 hours ago

10 comments

  • hahajk 14 hours ago
    i know everyone plays with their own rules, but from my perspective, a square root spends a 2 - it's not an operator like + or ! . there's an implicit "2" since you could also take the cube root or fourth root
    • jjmarr 11 hours ago
      You can use `sin(arctan())` as the repeatable operator instead of sqrt(). Then "unwrap" by applying ^-2 - (2^-2)

      3=(sin(arctan(sin(arctan(sin(arctan(2))))))^−2 − (2)^−2

      • shiandow 3 hours ago
        There's a 2 in the definition of sin :-)

        Though it really depends which definitions you find more fundamental.

        Still, exp and log seem like good choices.

      • schiffern 4 hours ago
        Perfect answer, I hate it :)
    • JumpCrisscross 13 hours ago
      > it's not an operator like + or !

      Squaring is to multiplication as is the latter to addition. If roots are invalid then so should be multiplications.

      • penteract 12 hours ago
        Exponentiation is to multiplication as multiplication is to addition. Squaring is to multiplication as doubling is to addition.
        • jolmg 8 hours ago
          > Squaring is to multiplication as doubling is to addition.

          as halving is to addition.

          • penteract 8 hours ago
            Square rooting is to multiplication as halving is to addition.
      • SamBam 12 hours ago
        GP didn't say roots are invalid, just that they burn a two.

        If we agree that 2^2 uses two twos (and, indeed, that 2^3 isn't valid, because it uses a 3), then what's the equivalent of a square root?

        Of course to my simple math brain, a square root is ^1/2, which includes the digit 1. But I agree there is someone 2-ish about the square root.

        • direwolf20 9 hours ago
          It can be written as ²√2 and there's also ³√2 and so on
          • schiffern 4 hours ago
            Yep.

            It would be like if we had a stylistic convention that you can drop the second number in an addition and it becomes 1. So "6 +" means "6 + 1."

            Now I can trivially make any number, but I'm just 'exploiting' a convenience feature in the notation:

            0 = 2 + 2 - 2 - 2

            1 = 2 + 2 - 2 - 2 +

            2 = 2 + 2 - 2 - 2 + +

            Etc

      • danparsonson 13 hours ago
        Would you elaborate on that? That doesn't make any sense to me
        • drfloyd51 12 hours ago
          Multiplication is repeated addition.

          2x3= 2+2+2 AxB= A+A B times.

          Raising to a power is repeated multiplication.

          2^3= 2x2x2 A^B = AxA B times

          Roots are the inverse of powers

          Division is the inverse of multiplications.

          Given this… the OC’s comment about giving exceptions to powers in the rules is nonsense because they are still operations.

          • mojomark 12 hours ago
            Multiplication is not the same as repeated addition, which you can observe as soon as you try the two different operations on fractions or negative numbers. For example, you can't add a number to itself 1/3 times, but you can certainly scale a number by 1/3.

            Multiplication is a scaling operation.

            • reichstein 4 hours ago
              Multiplication is the generalization of repeated addition. Repeated addition can be generalized to fractional values.

              In the integers, it is the same thing, with a rule for negative times.

              In the rational numbers, you can generalize to adding a number to itself a rational number of times. Adding a/b to itself p/q times is defined as adding a to itself p times, with a divisor that is adding b to itself q times.

              So adding the rational 5 (5/1) with itself 2_and-1/3 (4/3) times just means (5+5+5+5) / (1 + 1 + 1) = 20/3, a rational that we can also write as 6 -and-2/3. Only addition of integers is used.

              Which is the same as rational multiplication (a/b)(p/q) = (ap)/(b*q).

              But it's true that you can't say that rational multiplication is the same as _integer_ repeated addition.

            • goodmythical 10 hours ago
              Can you not? 9+(9(1/3)) [adding nine to itself a third of a time] is nine plus one third of nine {or nine divided by three} which is one of the three in three times three that equals nine which is three therefore 9+(9(1/3))=11 because 9+3=11
              • mojomark 1 hour ago
                By dividing by 3 you are introducing a scaler into the operation.

                Multiplication also flips signs (direction). That's why a negative * negative = positive, whereas a negative + negative = negative. The addition operator does not flip direction.

                Multiplication is not generally the same as multiple addition operations. However, it has the same result for positive whole numbers.

                I am not writing this to flippant. I'm a 40-something year old engineer and I litterally thought the same thing until a year ago-ish. I was trying to build a function (I can't recall what exactly) for a project and I said, well instead of multiplying these terms (which wasn't efficient for what I was doing or something), I'll just cycle through and add the terms X-number of times. ChatGPT said "uh, no, that's not the same thing," to which I said NFW and proceeded down a deep rabbit hole on the difference. I'm glad I did because I also now have a much better intuition for imaginary numbers, which I was severely lacking before.

              • red369 2 hours ago
                :)

                I like it when people who are clearly smart do things like this! Makes me feel better.

                Reminds me of this story: https://news.ycombinator.com/item?id=5663709

              • srean 2 hours ago
                That's Grothendieck multiplication.
              • alpaca128 4 hours ago
                9+3=12, but otherwise I agree
        • jibalt 11 hours ago
          Because it's just wrong. hahajk said that taking the 2nd root spends a 2, not that "roots are invalid". (FWIW, I'm not necessarily endorsing the view that the square root symbol implicitly spends a 2, I'm just saying that the counterarguments given are logically wrong ... actually more like semantically wrong.)

          The comment below by drfloyd51 is similarly wrong ... hahajk didn't say anything about "giving exceptions to powers in the rules".

          And the comment by speglt is also wrong:

          > If taking a square root is an inherent use of a two, then dividing a number by another number should also be a use of a two, because there are two operands.

          The number of operands isn't relevant; the degree of the root is.

          And from SideQuark we get

          > One could even claim division is raising to the -1 power, thereby disqualifying it in exactly the same manner you disqualify sqrt.

          But hahajk said nothing about anything being disqualified. (Not to mention that division is not raising to the -1 power ... raising what? n/d = n * d^-1, but this is completely irrelevant to hahajk's argument.)

        • spieglt 12 hours ago
          Multiplication is adding numbers to themselves. Exponentiation is multiplying numbers by themselves. Ditto division/taking roots. Squaring a number is just the instance of exponentiation that happens to have two operands. If taking a square root is an inherent use of a two, then dividing a number by another number should also be a use of a two, because there are two operands.
      • Biganon 3 hours ago
        Yes, taking the nth root is an operator. Taking the SQUARE root is this operator with the operand 2, which is usually graphically omited. I totally agree with the top comment: it's burning one of the 2's.
    • ta2345fa59be 6 hours ago
      Or use the successor operator.
    • SideQuark 13 hours ago
      Alternatively, one can claim sqrt is a single symbol, and there are extensions requiring a digit. Log, ln, also have hidden bases, but they’re also valid symbols. Cos/sin may be in radians or degrees, both in common use, and no real place to denote it.

      One could even claim division is raising to the -1 power, thereby disqualifying it in exactly the same manner you disqualify sqrt. I’ve never seen sqrt with a 2 index; power -1 is seen quite often for division.

      Go figure….

  • Someone 2 hours ago
    It’s true cheating, but if you assume the Peano axioms, you can write any integer with one zero without having to resort to complex functions such as √. For example (if I didn’t miscount the S-es or parentheses),

      S(S(S(S(S(S(S(0)))))))) = 7
    • rav 2 hours ago
      You don't even need zero in that case, you can use the fact that 2 - 2 = 0 to use the four 2s:

        S(S(S(S(S(S(S(2+2-2+2)))))))) = 7
      
      ...which is essentially the same as Dirac's general solution (which also derives a "successor operator" akin to S that doesn't spend any numeric literals).

      EDIT: I realize that this was already posted in a comment by shoo several hours ago.

      • vessenes 1 hour ago
        You’ve got S(4) in the center there
  • adas0693 15 hours ago
    there is a related problem (inspired by an interview questions from ITA software, a part of Google now) that asks to find the smallest positive integer that cannot be expressed using a digit K (in [2,9]) N times in an expression that involves the four basic arithmetic operations and the parentheses: +, -, *, /, (, and ).

    to make the problem more interesting, the list of operations can be extended to include exponentiation and concatenation.

    for example, using the four basic operations for N=4, and K in [2,9], we get:

    7 cannot be computed in 4 2s; 11 cannot be computed in 4 3s; 10 cannot be computed in 4 4s; 8 cannot be computed in 4 5s; 9 cannot be computed in 4 6s; 4 cannot be computed in 4 7s; 5 cannot be computed in 4 8s; 4 cannot be computed in 4 9s;

    • catlifeonmars 14 hours ago
      This feels very suspiciously like a finite field problem.
    • genter 15 hours ago
      Are you allowed to brute force it?
      • argee 14 hours ago
        That's how you do it right? It's a BFS with 4 being the max depth, and you print whatever numbers in the problem set which you didn't find.
        • twoodfin 2 hours ago
          Given the problems ITA solved, thinking about the question that way makes perfect sense.
      • dooglius 10 hours ago
        Figuring that out is presumably part of the interview (and would depend on N)
  • gnabgib 16 hours ago
    Popular last year (368 points, 165 comments) https://news.ycombinator.com/item?id=43145753
  • bajard 7 hours ago
    22 is cheating because 22 = 2*10+2 is arbitrary in base 10. If you write 22 in some base, you have to write the base, and that's an extra number.
  • teo_zero 16 hours ago
    Never understood why 7 is considered difficult...

    7 = ⌈ 2.2² + 2 ⌉

    • blueblock 15 hours ago
      Rounding spoils the fun of the puzzle.
      • nostrademons 14 hours ago
        It seems very much in the spirit of allowing concatenation and square roots...
      • jtari3333 7 hours ago
        Rounding is a completely valid mathematical operation.
    • anthk 12 hours ago
      (2 << 2) - ( 2 / 2)
    • AnimalMuppet 15 hours ago
      It is difficult when the rules do not allow exponentiation.
      • Jtsummers 15 hours ago

          7 = ⌈ 2.2 + 2 + 2⌉
        
        Would still work.
  • allywilson 7 hours ago
    I'll just drop this here (yes, it's 4 instead of 2 - they're twice as good): https://fourmula.awsum.info/
  • Arnt 16 hours ago
    Spoilsport.

    And JFYI since December is near: https://rant.gulbrandsen.priv.no/hack/advent-calendar-for-ne...

  • shoo 7 hours ago
    There's a direct trivial solution for any natural number using the successor function S(a) = a + 1 from Peano arithmetic.

    Encode 0 as 2 + 2 - 2 - 2 then apply S a times to that 0.

    Only depends upon natural numbers & Peano stuff, no need for construction of rationals, irrationals, logarithms, etc.