[Math Game] The Four Fours

31=!4+!4+!4+4

32=4*4+4*4

Rap

1 Reply

33=4!+!4+4-4

34=4!+!4+4/4

35=4*!4-4/4

36=4*4*Sqrt(4)+4

Rqp

0 Replies

37=4*!4+4/4

38=4*!4+4/Sqrt(4)

39=4/(.4)-4/4

40=4/(.4)+4-4

Rap

1 Reply

@raprap,

4/.4 = 10, not 40

2 Replies

@markr,

Oopsey.

40=4*!4+Sqrt(4*4)

Rap

0 Replies

@markr,

39 = (4 + 4/sqrt(4))!! - !4
40 = !(4 + 4/4) - 4

1 Reply

@markr,

41 = S(4 + sqrt(4), 4) - 4!

Stirling numbers of the second kind (S(n,k)) are described at http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind

0 Replies

42=4!+4!-4-sqrt(4)

Rap

0 Replies

43=4*!4+!4-sqrt(4)

44=4!+4!-Sqrt(4)-Sqrt(4)

Rap

1 Reply

@raprap,

45 = !4 * (4 + 4/4)

1 Reply

@markr,

46=4!*4!-4/Sqrt(4)

47=4!*4!-4/4 (trivial I know)

48=4!*4!-4+4 (more trivia)

48=4*(4+4+4)

Prime series (2,3,5,7,11,13....) 4th prime is 7, P(4)=7

49=P(4)*P(4)+4-4

Rap

1 Reply

@raprap,

I think you meant:
46 = 4! * sqrt(4) - 4/sqrt(4)
47 = 4! * sqrt(4) - 4/4
48 = 4! * sqrt(4) - 4 + 4

1 Reply

@markr,

49 = 4! * sqrt(4) + 4/4

1 Reply

@markr,

The successor function is described at http://en.wikipedia.org/wiki/Successor_function

For N > 4
N = N-4 instances of the successor function, S(x), + 4 - sqrt(4) - sqrt(4).

For example:
5 = S(4) + 4 - sqrt(4) - sqrt(4)
6 = S(S(4)) + 4 - sqrt(4) - sqrt(4)
7 = S(S(S(4))) + 4 - sqrt(4) - sqrt(4)

We can go as high as we wish.

This points out what I think is a problem with this sort of puzzle. I'd rather see exactly what is permitted, with anything not listed being disallowed.

2 Replies

@markr,

Perhaps we could just throw out the functions we know will allow us to generate everything.

50 = 4! + 4! + √4 + !(!√4)
51 = P(√4) * P(P(4)) + 4 - 4

0 Replies

@markr,

markr wrote:

We can go as high as we wish.

This points out what I think is a problem with this sort of puzzle. I'd rather see exactly what is permitted, with anything not listed being disallowed.

People will lose interest if you use too many S's. Also, there's a limited number of characters allowed in one post. Twisted Evil

52 = 4 * P((P(√4))!) + 4 - 4
53 = 4 * P((P(√4))!) + 4 / 4

1 Reply

@Kolyo,

54 = (!4 +!4 + !4) * √4
55 = 44 / 4 * P(P(√4))
56 = (4 + 4) * [4 + P(√4)]
57 = P(√4) * P(4!!) + 4 - 4
58 = P(√4) * P(4!!) + 4 / 4

0 Replies

59 = P(P(P(4))) + 4 - sqrt(4) - sqrt(4)

0 Replies

@raprap,

raprap wrote:

31=!4+!4+!4+4

32=4*4+4*4

Rap

OK, I'm not familiar with the bang in front of the number.... Obviously !4=9, but what is the function?

1 Reply

@DrewDad,

It's called "derangement".

I first learned about it in this thread.

4! is the number of ways you can permute 4 objects.
!4 is the number of ways you can permute 4 objects and have them all end up in new places.

0 Replies

[Math Game] The Four Fours

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