Use exactly four 4's to form every integer from 0 to 50

Steven G

Steven G

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Use exactly four 4's to form every integer from 0 to 50, using the usual operators used on similar type problems in the past. This was a challenge at a recent NYMATYC conference (NYS Mathematical Association of Two Year Colleges)
I'll start off with 0=44-44=4/4! - 4/4! ....
 
44/44 = 1

4/4 + 4/4 = 2

(4 + 4 + 4)/4 = 3

4 + 4*(4 - 4) = 4
 
\(\displaystyle 5 = \frac{\sqrt{4}*\sqrt{4}}{4} + 4\)

Is that allowed?

If not, then

5 = 4!/4 - 4/4

6 = 4!/4 * 4/4

7 = 44/4 - 4

8 = 4 * 4/4 + 4
 
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9 = 4 + 4 + 4/4

\(\displaystyle 10 = 4 + 4 + 4/\sqrt{4}\)

\(\displaystyle 11 = \frac{44}{\sqrt{4}*\sqrt{4}}\)

12 = 4! - 4 - 4 - 4
 
27 = 4! - (4/4) + 4

28 = 4! + (4 - 4) + 4

36 = 4! + 4 + 4 + 4

44 = 4! + (4∙4) + 4
 
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13: 4! / √4 + 4/4

14: 4 * √4 + 4 + √4

15: 4 * 4 - 4/4

16: 4 * 4 + 4 - 4

17: 4 * 4 + 4/4

18: 4 * √4 + 4 / .4

19: 4! - 4 - 4/4

20: 4 * (4 + 4 / 4)

21: √4 - 4 + 4 / 4 : EDIT; sh. be 4! - 4 + 4/4

22: 4 * 4 + 4 + √4

Ok Mark: go fill in the terrible gaps you left (before Harry gets a heart attack!)
 
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Harry_the_cat said:
You're playing havoc with my OCD!
Click to expand...
Sorry 'bout that. We're supposed to post 'em in order? (I had already seen at least one duplicate posted.)

Note: I edited post #7 (math error).

'Nuther note: I don't have to go to the corner.
 
mmm4444bot said:
Sorry 'bout that. We're supposed to post 'em in order? (I had already seen at least one duplicate posted.)

Note: I edited post #7 (math error).

'Nuther note: I don't have to go to the corner.
Click to expand...
But Denis does, for 21 minutes.
 
mmm4444bot said:
Sorry 'bout that. We're supposed to post 'em in order? (I had already seen at least one duplicate posted.)

Note: I edited post #7 (math error).

'Nuther note: I don't have to go to the corner.
Click to expand...
Ummmm, yes you do! Your 23 and 24 only use three 4s. Rules say you have to use four 4s.
 
31: ASIN(√4 / 4) + 4/4
32: 4 * √4 * √4 * √4
33: 4!! * 4 + 4/4 (see note below)
34: 44 - 4/.4
35: 4! + 44/4
36: 44 - 4 - 4
37: √4 baker's dozen + 44/4 ; ahem!!
38: 44 - 4 - √4
39: (√4 + √4 - 4/4) baker's dozen
40: 44 - √4 - √4

note below!
4!! = 8
https://en.wikipedia.org/wiki/Double_factorial
 
Denis said:
31: ASIN(√4 / 4) + 4/4
32: 4 * √4 * √4 * √4
33: 4!! * 4 + 4/4 (see note below)
34: 44 - 4/.4
35: 4! + 44/4
36: 44 - 4 - 4
37: √4 baker's dozen + 44/4 ; ahem!!
38: 44 - 4 - √4
39: (√4 + √4 - 4/4) baker's dozen
40: 44 - √4 - √4

note below!
4!! = 8
https://en.wikipedia.org/wiki/Double_factorial
Click to expand...

Double factorial could come in handy!! Must admit I've never heard of that.

I don't know about 31 (although pretty smart). Here's an alternative:

\(\displaystyle 31 = 4! + \frac {4! + 4}{4}\)

37 and 39 are dodgy, but not sure if these are any better:

37 = 44 - 4 - F4 where F4 =3 , the fourth Fibonacci number

39 = F4 * F4 * 4 + F4

Moving on …

\(\displaystyle 41 = (4 + F4)^{\sqrt4} - 4!!\) …. dodgy? (Edit: even LaTex doesn't like it)

\(\displaystyle 42 = 44 - 4 + \sqrt{4}\)

\(\displaystyle 43=44 - \frac{4}{4}\)

\(\displaystyle 44 = 44 + 4 - 4\)

\(\displaystyle 45 = 44 + \frac{4}{4}\)

\(\displaystyle 46 = 44 +4 - \sqrt{4}\)
 
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41: T(4) + 4 + 4/4 : T(4) = 4th triangular number = 10

47: 4! * √4 - 4/4
48: 44 + √4 + √4
49: 4! + 4! + 4/4
50: 44 + 4 + √4
 
51: F(4/.4) - √4 - √4

52: 44 + 4 + 4

53: F(4/.4) + √4 - 4

54: 44 + 4/.4

55: F(4/.4) + 4 - 4

56: F(4/.4) - 4/4

57: F(4/.4) - √4 + 4

58: Acos(√4/4) - √4

59: F(4/.4) + √4 + √4

60: (4 + 4/4)! / √4
 
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