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This question inspired me to write the same puzzle but instead replace the "= 100" with "= 1" with similar requirements and restrictions.

What is the expression with the fewest number of operators inserted that evaluates to 1?

Restrictions:

  1. The numbers need to be in the order that's shown in the question.
  2. Only use the operators +,−,×,÷ and √ and ! (Implies that modulus "%", exponent "^", binomial coefficients, and other operators are not allowed).
  3. Parentheses will not be counted, so they can be used to change the order of operations.
  4. Rounding is not allowed, so it have to equal to 1.

Verify your calculations in that calculator application that comes with your PC, if it ever did came with your PC.

This is my first time writing a puzzle here so obviously I should have thought this out a lot more instead of adding rules when situation comes.

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  • $\begingroup$ If you updated this so we could use the modulo operator (%), it could be beaten in one move ^_^: 1%23456789 = 1 $\endgroup$
    – Mwr247
    Commented Apr 15, 2015 at 20:47
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    $\begingroup$ What about the ceiling function? :) $\lceil1234/56789\rceil = 1$ $\endgroup$
    – Ian MacDonald
    Commented Apr 16, 2015 at 1:28
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    $\begingroup$ OP, we need a ruling: Does rounding count or should it be exactly 1? If rounding is OK, how many digits are required? $\endgroup$
    – Engineer Toast
    Commented Apr 16, 2015 at 12:30
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    $\begingroup$ How about $123456789 != 1$, as in not equal? $\endgroup$
    – dmg
    Commented Apr 16, 2015 at 14:05
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    $\begingroup$ get all 69 answers here. $\endgroup$
    – user12496
    Commented May 15, 2015 at 23:04

12 Answers 12

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38
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If √ can mean nth root:

$$\sqrt[1234567]{-8+9}$$

3 operators. Obviously...

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  • $\begingroup$ Too bad you can't re-arrange the numbers or else it would of been 2 operators, but nice answer! $\endgroup$
    – John Odom
    Commented Apr 16, 2015 at 14:52
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    $\begingroup$ If reverse polish notation is allowed, you can pull this off with one operator $$1\sqrt[23456789]{}$$ :-P $\endgroup$
    – Kyle G
    Commented Apr 16, 2015 at 18:03
  • $\begingroup$ Congrats! I was hoping for someone to beat you with an even smaller one :P. $\endgroup$
    – John Odom
    Commented Apr 17, 2015 at 21:52
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$$1+23-45-67+89$$

uses four. (I wrote a Python script.)

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  • $\begingroup$ This will also work 1.0 - 23.0 + 45.0 + 67.0 - 89.0 $\endgroup$
    – sberry
    Commented Apr 15, 2015 at 23:24
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    $\begingroup$ Here is a list of all that I could come up with using only +, -, *, /. gist.github.com/schleppy/4bed5e5a668cf5393a3b $\endgroup$
    – sberry
    Commented Apr 15, 2015 at 23:27
  • $\begingroup$ Can you please share with us your python script? $\endgroup$
    – hytromo
    Commented Apr 17, 2015 at 6:52
  • $\begingroup$ bpaste.net/show/15f01879245f $\endgroup$
    – lynn
    Commented Apr 17, 2015 at 18:43
  • $\begingroup$ Ah @Mauris, nicely done with the space as an operator to deal with the number joining. My way was more complex as I generated the combinations of numbers and operators per. $\endgroup$
    – sberry
    Commented Apr 21, 2015 at 4:56
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How many significant digits matter here for rounding? Because if it's anything less than $3,456,789$ zeroes, I can solve it in three ;)

$1+2/3456789! = ~1$

Many programming languages will evaluate it as "1". Even Wolfram Alpha can't show me enough decimal digits to tell me I'm wrong ;P

EDIT: Yes, I know this is no longer valid as of the rule change that doesn't include rounding. I didn't expect it would be allowed anyways, just figured it would be worth submitting, since it comes so infinitesimally close to 1. Besides, kgull managed to get even closer using a similar method.

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  • 1
    $\begingroup$ Lol nice, this is bit of a gray area so I'm going to wait for a few days. $\endgroup$
    – John Odom
    Commented Apr 15, 2015 at 21:12
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    $\begingroup$ If we're allowed to round, then allowing 6 zeros would allow just a single operator; ¹²³⁴⁵⁶⁷⁸√9 ≈ 1 after all. $\endgroup$
    – Jon Hanna
    Commented Apr 16, 2015 at 11:23
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    $\begingroup$ This solution violates rule number 4 "Rounding is not allowed, so it have to equal to 1." $\endgroup$
    – user902383
    Commented Apr 17, 2015 at 8:59
  • $\begingroup$ @user902383 A rule which was not made until this answer was already posted, and which became the reason for the rule ;P But yes, it's no longer valid. $\endgroup$
    – Mwr247
    Commented Apr 17, 2015 at 15:26
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Assuming the binomial coefficient is not an operator itself and parentheses are allowed and not counted, this requires only 1 operator.

$$1+{2345\choose6789}=1$$

Check the Pochhammer symbol too:

$$1+(-2)_{3456789}$$

Some useful information on Wolfram Alpha.

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8
  • $\begingroup$ This is not valid, in the binomial coefficient k cannot be greater than n $\endgroup$
    – leoll2
    Commented Apr 16, 2015 at 15:28
  • $\begingroup$ Wrong, you can extend the definition from a combinatory point of view. $\endgroup$
    – Francesco De Lisi
    Commented Apr 16, 2015 at 15:37
  • $\begingroup$ @leoll2 They are nonzero and even very useful when k>n and n is not a nonnegative integer. $\endgroup$
    – user23013
    Commented Apr 16, 2015 at 18:08
  • $\begingroup$ How do you define the factorial of a negative number? Check this: en.wikipedia.org/wiki/… $\endgroup$
    – leoll2
    Commented Apr 16, 2015 at 18:25
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    $\begingroup$ @leoll2 See here as an example: en.wikipedia.org/wiki/Binomial_series It works with any n, negative, real or complex. $\endgroup$
    – user23013
    Commented Apr 16, 2015 at 21:50
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If parentheses will not be counted and if we could use it as multiply:

$12(34)-5(67)-8(9) = 1$

I used only 2 operators.

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  • 1
    $\begingroup$ Clever way of finding a loophole lol. But I meant that parentheses can be used to change the order of operations. I should of mentioned that sooner but now the question is updated. $\endgroup$
    – John Odom
    Commented Apr 16, 2015 at 21:20
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    $\begingroup$ @JohnOdom ah so I cannot use it as multiply now, dang! $\endgroup$
    – Alex
    Commented Apr 16, 2015 at 21:21
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    $\begingroup$ * * * Puzzle poser (John Odom), if you do not have that explicitly changed in the instructions at present about the limitations of the parentheses, then Alex's solution stands as correct with only two operators. $\endgroup$
    – Olive Stemforn
    Commented Jun 20, 2015 at 18:57
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$$(1 + 23 + 45 + 6! - 789)! = 1$$

$$((1+2-3)\times456789)!=1$$

Everyone is trying with minimum operators.
I guess, with @user23013's solution, we can try with various possibilities :)

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  • $\begingroup$ Any answer that cooperates with the rules is a good answer :). But I am hoping someone will be able to beat the top answer right now that has 3 operators. $\endgroup$
    – John Odom
    Commented Apr 17, 2015 at 14:56
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A simple expression that is exact, and only uses four operations without bending the rules (if exponentiation isn't permitted, I assume that neither is using the numbers to create n-th roots) while using at least one non-basic operation, is $$ ((1+2-3)\times456789)! $$ That's one addition, one subtraction, one multiplication, and one factorial (actually, zero factorial, but you know what I mean). Another similar option is $$ ((12/3-4)\times56789)! $$ A slightly more bendy solution using the fact that negative integer factorials can be considered to be infinite is $$ 1+2/(3-456789)! $$

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1
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Three Operators

Similar idea to Mwr247's solution, but even more significant figures:

$$\left(\frac1{23456789!}\right)! = 1$$

Wolfram Alpha seems to think it is exactly 1. Good enough for me >_>

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6
  • $\begingroup$ Very clever... =P $\endgroup$
    – Mwr247
    Commented Apr 16, 2015 at 13:38
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    $\begingroup$ That is very clever, but I'm going to have to say that rounding is not allowed. Nice try though :) $\endgroup$
    – John Odom
    Commented Apr 16, 2015 at 14:47
  • $\begingroup$ How does this work? $\endgroup$
    – QuyNguyen2013
    Commented Apr 16, 2015 at 22:50
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    $\begingroup$ The factorial of a fractional number is defined by the Gamma function. Since 1/23456789!=~10^-162695685 is really close to zero and 0!=1 then (1/23456789!)! is so ridiculously close to one, even WolframAlpha says 'yep, that's a one'. I'm honestly curious what it actually looks like in scientific notation. $\endgroup$
    – Kyle G
    Commented Apr 16, 2015 at 23:39
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    $\begingroup$ @kgull You can use the series expansion: $x!\approx 1 - \gamma x + O(x^2)$ ($\gamma$ is the Euler-Mascheroni constant). So, the expression is around $1-10^{-10^{8.211}}$. $\endgroup$
    – 2012rcampion
    Commented Feb 13, 2016 at 8:19
0
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$$1-23+45+67-89$$ uses only 4 operators.

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1
  • $\begingroup$ This is the same as Maurius Van Hauwe with transposed operators. $\endgroup$
    – Engineer Toast
    Commented Apr 16, 2015 at 12:40
0
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Edit: I realized Alex did it using 2 operations using this loophole so this is nothing special. I'd delete this normally but I think this solution is still kinda cool.

I got three operations without using an nth root:

$(12)(3)(4)(.5)(-6+7)/((8)(9))=1$

Taking advantage of parentheses not counting. (Yes I know the rule wasn't meant for them to be used this way but I spent a long time thinking of how to exploit this loophole so cut me some slack.)

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-3
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In languages like C# and Java dividing two integers will always result in an integer (decimals will be omitted). Therefore only one operation is required to solve this problem:

12345/6789

Which will result in 1.

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  • 4
    $\begingroup$ Welcome to Puzzling SE! This puzzle isn't about the behaviour of programming languages/computers, it's just a matter of math. 12345/6789 is a rational number in math! $\endgroup$
    – leoll2
    Commented Apr 16, 2015 at 12:34
  • $\begingroup$ And yet this solution evaluates to 1 while complying with the rules. $\endgroup$
    – Mike.
    Commented Apr 16, 2015 at 12:43
  • $\begingroup$ It did comply with the rules but since I updated it to not allow rounding this answer will no longer be valid, unless you store the answer as a float and it does return 1. Sucking up with my coding languages will not work lol :P. $\endgroup$
    – John Odom
    Commented Apr 16, 2015 at 14:50
  • $\begingroup$ @leoll2 "in math" What does that mean? When the "/" operator is defined as integer division, it's perfectly logical. $\endgroup$
    – bjb568
    Commented Apr 16, 2015 at 23:56
  • $\begingroup$ @bjb568 It's logical but the integer division is considered rounding the answer which will break rule #4. $\endgroup$
    – John Odom
    Commented Apr 17, 2015 at 15:17
-4
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One operator.

Taking advantage of user23013 's loop hole.

If √ can mean nth root:

$$\sqrt[23456789]{1}$$

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    $\begingroup$ Rule 1: The numbers need to be in the order that's shown in the question. $\endgroup$
    – Engineer Toast
    Commented Apr 16, 2015 at 12:30

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