What is the fewest number of 0s that are needed to make the number 120, using only two of the operators from the set $\{+, -, \times, \div, !\}$ between the 0s?
Details and Assumptions :
Parentheses are allowed, but concatenating digits is not.
You may use each of the two operators selected multiple times.
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Change ker diya mene
Explaining further: The two mathematical operators to be used are factorial(!) and addition/plus(+) Now, Factorial basically is as follows: For example factorial of 5 will be like 5!= 5x4x3x2x1=120 But, over here we have only zeroes, so it is necessary to know that 0!=1. Thus, we will have (0!+0!+0!+0!+0!)!=> (1+1+1+1+1)!=>(5)!=>120(as we saw above) So, N(that is the number of zeroes)=5 P.S. If u like my solution please like Ask for any further queries relating this
The fewest is "4" not 5
Here's why ((0!+0!+0!)!-0!)!=((1+1+1)!-1)!=(3!-1)!=(6-1)!=120
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You can only use 2 operators, u used 3
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Yeah true! See my solution is correct,#thanks victoria
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@Rishabh Sood – We can still use 2 operators and 4 zeros to make 120: ((0!-(-(0!))-(-(0!)))!-(0!))! = ((1-(-1)-(-1))!-(1))! = ((1+1+1)!-(1))! = (3!-1)! = (6-1)! = 5! = 120
@Rishabh Sood – Hey, who downvoted your solution. It is correct
Addition and subtraction are the same, subtraction is adding a negative number.
and you used a negative 0. DNE.
Use only 2 operator but u have used 3 I.e +,-,!
yeah we can use 3 operations that's good..
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So guys, concluding we can use 5 zeroes minimum. Thus me and those who answered 5 win!
yeah that's right....
Did it the same way dude
This is 10 mathematical operators, but only 2 kinds of mathematical operators.
Give me a hi5 bro!(prince)
I think it should be four.. we have the same answer by Mu'amar Musa Nurwigantara .. ((0!-(-(0!))-(-(0!)))!-(0!))!
Thanks. I've updated the answer to 4. Those who previously answered 4 has been marked correct, while those who previously answered 5 has been marked wrong.
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$[cos (0)+ cos (0)+cos (0)+cos (0)+cos (0)]!$
=
$[1+1+1+1+1]!$
=
$5!$
=
$120$
.
$_\square$
(cos(0)+cos(0)+cos(0)+cos(0)+cos(0))!=(1+1+1+1+1)!=5!=120
cos is not in the given range of operators
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$\Large ((0!-(-0!)-(-0!))!-0!)! = 120$
The answer is 4.