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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.

John Odom
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12 Answers12

38

If √ can mean nth root:

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

3 operators. Obviously...

user23013
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22

$$1+23-45-67+89$$

uses four. (I wrote a Python script.)

Lynn
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11

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EDIT: Yes, I know this is no longer valid as of the rule change that doesn't include rounding.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.

SteamCode
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Mwr247
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    Lol nice, this is bit of a gray area so I'm going to wait for a few days. – John Odom Apr 15 '15 at 21:12
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    If we're allowed to round, then allowing 6 zeros would allow just a single operator; ¹²³⁴⁵⁶⁷⁸√9 ≈ 1 after all. – Jon Hanna Apr 16 '15 at 11:23
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    This solution violates rule number 4 "Rounding is not allowed, so it have to equal to 1." – user902383 Apr 17 '15 at 08:59
  • @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. – Mwr247 Apr 17 '15 at 15:26
5

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.

grg
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  • This is not valid, in the binomial coefficient k cannot be greater than n – leoll2 Apr 16 '15 at 15:28
  • Wrong, you can extend the definition from a combinatory point of view. – Francesco De Lisi Apr 16 '15 at 15:37
  • @leoll2 They are nonzero and even very useful when k>n and n is not a nonnegative integer. – user23013 Apr 16 '15 at 18:08
  • How do you define the factorial of a negative number? Check this: http://en.wikipedia.org/wiki/Factorial#Non-extendability_to_negative_integers – leoll2 Apr 16 '15 at 18:25
  • @leoll2 Why don't you think simply at the number of way you can take k objects from n? It's zero if k>n. – Francesco De Lisi Apr 16 '15 at 18:29
  • That's an application of binomial coefficient, not the definition! Check here: http://en.wikipedia.org/wiki/Binomial_coefficient#Definition_and_interpretations – leoll2 Apr 16 '15 at 18:32
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    @leoll2 See here as an example: http://en.wikipedia.org/wiki/Binomial_series It works with any n, negative, real or complex. – user23013 Apr 16 '15 at 21:50
  • If this is allowed, it seems you can get a solution with 0 operators: $${{12\choose34}\choose{56\choose789}}=1$$ – user23013 Aug 12 '15 at 19:06
4

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.

JMP
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Alex
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    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. – John Odom Apr 16 '15 at 21:20
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    @JohnOdom ah so I cannot use it as multiply now, dang! – Alex Apr 16 '15 at 21:21
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      • 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.
  • – Olive Stemforn Jun 20 '15 at 18:57