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Hey all, you guys are a damn bright bunch, I'm not gonna lie! Those questions were tough buggers and they all got solved so quickly, so well played all!

So here's number 4! Enjoy:

*** When I use root(x), it means the square root of x, root3(x) means cube root of x, rootN(x) is nth root of x etc. ***

1. Suppose a, b, c are the three roots of the equation x^3 - 8x^2 + 5x + 7 = 0. What are the values of:

1a. a + b + c

1b. a^2 + b^2 + c^2

1c. a^3 + b^3 + c^3 ?

2. There are 1988 towns and 4000 roads in a certain country called Amashakashaka! (haha lol) Each road connects two towns, prove that there is a closed path passing through no more than 20 towns.

3. Prove, WITHOUT A CALCULATOR!!! That 7^(root(5)) > 5^(root(7)).

4. The positive integers a and b are such that the numbers 15a + 16 b and 16a - 15b are both squares of positive integers. Find the least possible value that can be taken by the minimum of these two.

(Excited to see how people come up with answers with this one)

5. A problem from 1994, when I was still a wittle 5 year old :)

1 = 19 - 9 * root(4)

2 = 1 * 9 - 9 + root(4)

3 = -root(1 * 9) + root(9 * 4)

4 = 1^99 * 5

5 = -1 - 9 - 9 + 4!

6 = root(1 + 99) - 4

How much higher can you go? Longest list wins! :P

See how many you can solve :)

Weo weo weo, kkkkk GOOOOOOO!!

Edited by Twinhelix
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Sorry there was a typo in 5. I dunno why it won't let me edit my post.

It should obviously read "4 = 1^99 * 4"

Also I forgot to state that you have to use the numbers 1 9 9 4 in order when forming your lists, and you man use + - * / ^ roots factorials and also brackets.

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Easy to calculate by hand that

75>56.

So,

75>56>5root(35)

Thus,

(7root(5))root(5)>(5root(7))root(5)

Which leads immediately to

(7root(5))>(5root(7)).

Yup well done homie ^^

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7 = ((1^9) * root(9)) + 4

8 = (1 + (9/9)) * 4

9 = (1^9) * ((root(9))^(root(4)))

10 = (1^9) + ((root(9))^(root(4)))

11 = ((1^9) * 9) + root(4)

12 = ((1^9) * root(9)) * 4

I'm going to keep going!

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7 = 1^9 * 9 - root(4)

8 = (1 + 9/9) * 4

9 = 1 + root(9) + root(9) + root(4)

10 = 1 + root(9) + root(9 * 4)

11 = (1 * root(9) * root(9)) + root(4)

12 = (-1 + root(9)) * root(9 * 4)

13 = (1 * root(9 * 9)) + 4

14 = 1 + root(9 * 9) + 4

15 = 1 + 9 + 9 - 4

16 = 1 * (9 + 9) - root(4)

17 = 1 + 9 + 9 - root(4)

18 = ((1 - root(9)) * root(9)) + 4!

19 = -1 + 9 + 9 + root(4)

20 = 1 * (9 + 9 + root(4))

21 = 1 + 9 + 9 + root(4)

22 = 1 * (9 + 9 + 4)

23 = 1 + 9 + 9 + 4

24 = (1 / 9 * 9) * 4!

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I pretty sure the answer is 231361 which is 4812.

Here's what I have so far:

We have the two equations

16a - 15b = x2

15a + 16b = y2.

Solving these as linear equations in a and b we get

a = (16x2 + 15y2)/481 and

b = (16y2 - 15x2)/481.

Now if we assume x = y, then we get

a = (31x2)/481 and

b = x2/481

in which case, the smallest x2 can be is 4812.

I have no time to finish as I must go now.

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I pretty sure the answer is 231361 which is 4812.

Here's what I have so far:

We have the two equations

16a - 15b = x2

15a + 16b = y2.

Solving these as linear equations in a and b we get

a = (16x2 + 15y2)/481 and

b = (16y2 - 15x2)/481.

Now if we assume x = y, then we get

a = (31x2)/481 and

b = x2/481

in which case, the smallest x2 can be is 4812.

I have no time to finish as I must go now.

Woah! Nice! You got the right answer! However I think the proof and working still lacks a bit. I didn't know how to do this, and I saw the sample solution does something like this to begin it:

Set the equations to equal x squared and y squared as you have done, then square both equations and add. You get:

481(a^2 + b^2) = x^4 + y^4. And 481 = 13 x 37.

Thus (x^4 + y^4) % 13 = 0 or (x^4 + y^4) % 37 = 0.

So consider the first one, if z^4 % 13 = 0, the z^12 % 13

= 0. Then use Fermat's Little Theorem ;) and run from there.

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You state that...

16a - 15b = x2

15a + 16b = y2

and then later that...

x = y

Doesn't that mean that...

16a - 15b = x2

15a + 16b = x2

or...

16a - 15b = 15a + 16b

so...

a = 31b

16(31b) - 15b => 496b - 15b => 481b

Factors of 481 are 13 and 37. Only way to make a squared number with those factors is to multiply by 13 and 37 or 481.

b has to be 481^n where n is an odd number (lowest 481)

a is then 31 * 481^n (lowest 14911)

---edit---

The only problem I see is there is no proof that x = y

Edited by curr3nt
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7 = 1^9*9-root(4)

8 = (1+9/9)*4

9 = (-1*9)+9*root(4)

10 = (-1+root(9)+root(9))*root(4)

11 = 1^9*9+root(4)

12 = 1^9*root(9)*4

13 = 1^9*9+4

14 = (1+root(9)+root(9))*root(4)

15 = 1*root(9)*(root(9)+root(4))

16 = 1+root(9)*(root(9)+root(4))

17 = -1^9+9*root(4)

18 = 1^9*9*root(4)

19 = 1+root(9)*root(9)*root(4)

20 = (-1+root(9)+root(9))*4

21 = 1*root(9)*(root(9)+4)

22 = 1*9+9+4

23 = 1+9+9+4

24 = (1*root(9)+9)*root(4)

25 = 1^(9-9)+4!

26 = (1+root(9)+9)*root(4)

27 = (1*9)+9*root(4)

28 = (1+root(9)+root(9))*4

29 = -1+root(9)+root(9)+4!

30 = 1*root(9)+root(9)+4!

31 = 1+root(9)+root(9)+4!

32 = -1^9+9+4!

33 = 1^9*9+4!

34 = (-1+9+9)*root(4)

35 = -1^9+9*4

36 = 1^9*9*4

37 = 1+root(9)*root(9)*4

38 = (1+9+9)*root(4)

39 = 1*root(9)*(9+4)

40 = 1+root(9)*(9+4)

41 = -1+9+9+4!

42 = 1*9+9+4!

43 = 1+9+9+4!

44 = -1+9+9*4

45 = 1*9+9*4

46 = 1+9+9*4

47 = -1+(root(9)+9)*4

48 = (1*root(9)+9)*4

49 = 1+(root(9)+9)*4

50 = -1+root(9)*9+4!

51 = 1*root(9)*9+4!

52 = (1+root(9)+9)*4

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7 = -1 + root(9) + 9 - 4

8 = 1*root(9) + root(9) + root(4)

9 = 1 + root(9) + root(9) + root (4)

10 = (-1 - root(9) + 9)*root(4)

11 = (1*root(9))*root(9) + root(4)

12 = (1*9)/root(9)*4

13 = 1 + (9/root(9))*4

14 = (1 + root(9) + root(9))*root(4)

15 = 1 + 9 + 9 - 4

16 = 1 - root(9) + 9*root(4)

17 = 1 + 9 + 9 - root(4)

18 = 19 + root(9) - 4

19 = 1*9 + 9 + root(4)

20 = 19 - root(9) + 4

21 = 1*9 + root(9)*4

22 = 1 + 9 + root(9)*4

23 = 1 + 9 + 9 + 4

24 = 1 + root(9)*9 - 4

25 = 1*root(9)*9 - root(4)

26 = 1 + root(9)*9 - root(4)

27 = -(1*9) + 9*4

28 = 1 - 9 + 9*4

29 = 1*9*root(9) + root(4)

30 = 1 + 9*root(9) + root(4)

31 = 1*9*root(9) + 4

32 = 1 + 9*root(9) + 4

33 = 1*root(9)*(9 + root(4))

34 = 1 + root(9)*(9 + root(4))

35 = -1 + root(9)*root(9)*4

36 = 1*root(9)*root(9)*4

37 = 1 + root(9)*root(9)*4

38 = -1 + root(9) + 9*4

39 = 1*(root(9) + 9*4)

40 = 1 + root(9) + 9*4

41 = -1 + 9 + 9 + 4!

42 = 1*(9 + 9 + 4!)

43 = 1 + 9 + 9 + 4!

44 = -1 + 9 + 9*4

45 = 1*(9 + 9*4)

46 = 1 + 9 + 9*4

47 = (-1 + (root(9))!)*9 + root(4)

48 = (-(1 + root(9)!) + 9)*4!

49 = (1 - root(9) + 9)^root(4)

50 = (1 + 99)/root(4)

it can still go further.

Edited by hussein
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51 = 1 + (root(9)!)*9 - 4

52 = 1*(root(9)!)*9 - root(4)

53 = 1 + (root(9)!)*9 - root(4)

54 = 1*root(9)*9*root(4)

55 = -1 + (root(9)!)*9 + root(4)

56 = 1*(root(9)!)*9 + root(4)

57 = 1 + (root(9)!)*9 + root(4)

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7 = 1^9*9-root(4)

8 = (1+9/9)*4

9 = (-1*9)+9*root(4)

10 = (-1+root(9)+root(9))*root(4)

11 = 1^9*9+root(4)

12 = 1^9*root(9)*4

13 = 1^9*9+4

14 = (1+root(9)+root(9))*root(4)

15 = 1*root(9)*(root(9)+root(4))

16 = 1+root(9)*(root(9)+root(4))

17 = -1^9+9*root(4)

18 = 1^9*9*root(4)

19 = 1+root(9)*root(9)*root(4)

20 = (-1+root(9)+root(9))*4

21 = 1*root(9)*(root(9)+4)

22 = 1*9+9+4

23 = 1+9+9+4

24 = (1*root(9)+9)*root(4)

25 = 1^(9-9)+4!

26 = (1+root(9)+9)*root(4)

27 = (1*9)+9*root(4)

28 = (1+root(9)+root(9))*4

29 = -1+root(9)+root(9)+4!

30 = 1*root(9)+root(9)+4!

31 = 1+root(9)+root(9)+4!

32 = -1^9+9+4!

33 = 1^9*9+4!

34 = (-1+9+9)*root(4)

35 = -1^9+9*4

36 = 1^9*9*4

37 = 1+root(9)*root(9)*4

38 = (1+9+9)*root(4)

39 = 1*root(9)*(9+4)

40 = 1+root(9)*(9+4)

41 = -1+9+9+4!

42 = 1*9+9+4!

43 = 1+9+9+4!

44 = -1+9+9*4

45 = 1*9+9*4

46 = 1+9+9*4

47 = -1+(root(9)+9)*4

48 = (1*root(9)+9)*4

49 = 1+(root(9)+9)*4

50 = -1+root(9)*9+4!

51 = 1*root(9)*9+4!

52 = (1+root(9)+9)*4

lol. i was really stuck on 47, i cant believe its that simple. great work!

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You have done nicely too. I wonder if there is any number that has no solution for this?

56 = 1*root(9)!*9+root(4)

57 = 1+root(9)!*9+root(4)

58 = 1*root(9)!*9+4

59 = 19*root(9)+root(4)

60 = (1+root(9))*9+4!

61 = 19*root(9)+4

62 = -1-9+root(9)*4!

63 = -1*9+root(9)*4!

64 = 1-9+root(9)*4!

65 = -1-root(9)!+root(9)*4!

66 = -1*root(9)!+root(9)*4!

67 = 1-root(9)!+root(9)*4!

68 = (-1+9+9)*4

69 = -1*root(9)+root(9)*4!

70 = 1-root(9)+root(9)*4!

71 = -1^9+root(9)*4!

72 = 1^9*root(9)*4!

73 = 1+(9+9)*4

74 = (1+root(9)!*root(9)!)*root(4)

75 = -19+94

76 = (1+9+9)*4

77 = 1*9*9-4

78 = 1+9*9-4

79 = 1*9*9-root(4)

80 = 1+9*9-root(4)

81 = 19*root(9)+4!

82 = -1+9*9+root(4)

83 = 1*9*9+root(4)

84 = -1+9*9+4

85 = 1*9*9+4

86 = 1+9*9+4

87 = (1+root(9)!)*9+4!

88 = -1*root(9)!+94

89 = +1-root(9)!+94

90 = 19*root(9)!-4!

91 = -1*root(9)+94

92 = +1-root(9)+94

93 = -1^9+94

94 = 1^9*94

95 = 1^9+94

96 = -1+root(9)+94

97 = 1*root(9)+94

98 = 1+root(9)+94

99 = -1+root(9)!+94

100 = 1*root(9)!+94

101 = +1+root(9)!+94

102 = -1+9+94

103 = +1*9+94

104 = +1+9+94

105 = 1*9*9+4!

106 = 1+9*9+4!

107 = (-1+root(9)!)!-9-4

108 = 1*root(9)*9*4

109 = (-1+root(9)!)!-9-root(4)

110 = (-1+root(9)!)!-root(9)!-4

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Adding to Nana7's list

52= (1+9+root(9))*4

53 = -1+(9*root(9))*root(4)

54 = (1*9*root(9))*root(4)

55 = 1+((9*root(9)*root(4)

56 = -1 +9*9-4!

57 = 1*9*9-4!

58 = 1+9*9-4!

59 =1+(root(9)!*9+4

60 = (1+9)*root(9)*root(4)

61 = 1+(root(9)!+9)*4

62 = -1-9+root(9)*4!

From this 63 and 64 are easy variations evolving from 62, just tired of typing.

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I see that while I was calculating and posting that Nana added even more. Nana sure is energetic! I have not been able to figure a way to prove it yet, but I believe that any number is possible. With the ability to use factorials of factorials, there is no max value possible to acquire.

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Are you sure, 4 = 1^99*5, or do you mean 4 = 1^99*4 ?

Yes, Twinhelix posted an apology and correction. Here is a copy of the post from Twinhelix:

Posted Yesterday, 08:27 PM

Sorry there was a typo in 5. I dunno why it won't let me edit my post.

It should obviously read "4 = 1^99 * 4"

Also I forgot to state that you have to use the numbers 1 9 9 4 in order when forming your lists, and you man use + - * / ^ roots factorials and also brackets.

Edited by thoughtfulfellow
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Yes, Twinhelix posted an apology and correction. Here is a copy of the post from Twinhelix:

Posted Yesterday, 08:27 PM

Sorry there was a typo in 5. I dunno why it won't let me edit my post.

It should obviously read "4 = 1^99 * 4"

Also I forgot to state that you have to use the numbers 1 9 9 4 in order when forming your lists, and you man use + - * / ^ roots factorials and also brackets.

Thanks.

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I still think at some point there will be a number that can not be expressed. Eventually.

111 = (-1+root(9)!)!-root(9)^root(4)

112 = (-1+root(9)!)!-root(9)!-root(4)

113 = 19+94

114 = (-1+root(9)!)!-root(9)*root(4)

115 = (-1+root(9)!)!-root(9)-root(4)

116 = (-1+root(9)!)!-root(9)!+root(4)

117 = (-1+root(9)!)!-root(9)!/root(4)

118 = (-1+root(9)!)!-root(9)!+4

119 = (-1+root(9)!)!-root(9)+root(4)

120 = (-1+root(9)!)!*(root(9)-root(4))

121 = (-1+root(9)!)!+root(9)-root(4)

122 = (-1+root(9)!)!+root(9)!-4

123 = (-1+root(9)!)!+root(9)!/root(4)

124 = (-1+root(9)!)!+root(9)!-root(4)

125 = (-1+root(9)!)!+root(9)+root(4)

126 = (-1+root(9)!)!+root(9)*root(4)

127 = (-1+root(9)!)!+9-root(4)

128 = (-1+root(9)!)!+root(9)!+root(4)

129 = (-1+root(9)!)!+root(9)^root(4)

130 = (-1+root(9)!)!+root(9)!+4

131 = (-1+root(9)!)!+9+root(4)

132 = (-1+root(9)!)!+root(9)!*root(4)

133 = (-1+root(9)!)!+9+4

134 = -1-9+root(9)!*4!

135 = -1*9+root(9)!*4!

136 = +1-9+root(9)!*4!

137 = -1-root(9)!+root(9)!*4!

138 = -1*root(9)!+root(9)!*4!

139 = +1-root(9)!+root(9)!*4!

140 = -1-root(9)+root(9)!*4!

141 = -1*root(9)+root(9)!*4!

142 = +1-root(9)+root(9)!*4!

143 = -1^9+root(9)!*4!

144 = 1^9*root(9)!*4!

145 = 1^9+root(9)!*4!

146 = -1+root(9)+root(9)!*4!

147 = 1*root(9)+root(9)!*4!

148 = 1+root(9)+root(9)!*4!

149 = -1+root(9)!+root(9)!*4!

150 = 1*root(9)!+root(9)!*4!

151 = 1+root(9)!+root(9)!*4!

152 = -1+9+root(9)!*4!

153 = 1*9+root(9)!*4!

154 = 1+9+root(9)!*4!

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#1:

If the roots are a, b, c

x^3-8x^2+5x+7 == (x-a)(x-b)(x-c) = x^3-(a+b+c)x^2+(ab+bc+ac)x-abc

so, comparing coefficients,

a+b+c = 8

(ab+bc+ac) = 5,

& abc = -7

now, (a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ac) evaluates to 64 = a^2+b^2+c^2+10

so a^2+b^2+c^2=54

now, a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)+3abc evaluates to a^3+b^3+c^3=8(54-5)-21

so a^3+b^3+c^3 = 371

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I still think at some point there will be a number that can not be expressed. Eventually.

111 = (-1+root(9)!)!-root(9)^root(4)

112 = (-1+root(9)!)!-root(9)!-root(4)

113 = 19+94

114 = (-1+root(9)!)!-root(9)*root(4)

115 = (-1+root(9)!)!-root(9)-root(4)

116 = (-1+root(9)!)!-root(9)!+root(4)

117 = (-1+root(9)!)!-root(9)!/root(4)

118 = (-1+root(9)!)!-root(9)!+4

119 = (-1+root(9)!)!-root(9)+root(4)

120 = (-1+root(9)!)!*(root(9)-root(4))

121 = (-1+root(9)!)!+root(9)-root(4)

122 = (-1+root(9)!)!+root(9)!-4

123 = (-1+root(9)!)!+root(9)!/root(4)

124 = (-1+root(9)!)!+root(9)!-root(4)

125 = (-1+root(9)!)!+root(9)+root(4)

126 = (-1+root(9)!)!+root(9)*root(4)

127 = (-1+root(9)!)!+9-root(4)

128 = (-1+root(9)!)!+root(9)!+root(4)

129 = (-1+root(9)!)!+root(9)^root(4)

130 = (-1+root(9)!)!+root(9)!+4

131 = (-1+root(9)!)!+9+root(4)

132 = (-1+root(9)!)!+root(9)!*root(4)

133 = (-1+root(9)!)!+9+4

134 = -1-9+root(9)!*4!

135 = -1*9+root(9)!*4!

136 = +1-9+root(9)!*4!

137 = -1-root(9)!+root(9)!*4!

138 = -1*root(9)!+root(9)!*4!

139 = +1-root(9)!+root(9)!*4!

140 = -1-root(9)+root(9)!*4!

141 = -1*root(9)+root(9)!*4!

142 = +1-root(9)+root(9)!*4!

143 = -1^9+root(9)!*4!

144 = 1^9*root(9)!*4!

145 = 1^9+root(9)!*4!

146 = -1+root(9)+root(9)!*4!

147 = 1*root(9)+root(9)!*4!

148 = 1+root(9)+root(9)!*4!

149 = -1+root(9)!+root(9)!*4!

150 = 1*root(9)!+root(9)!*4!

151 = 1+root(9)!+root(9)!*4!

152 = -1+9+root(9)!*4!

153 = 1*9+root(9)!*4!

154 = 1+9+root(9)!*4!

Ahh nana you beast!

I do think that 155 is one of the problems numbers to make though...... If anyone comes up with the solution to that, then you are a legend! I think 156 is possible

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#1:

If the roots are a, b, c

x^3-8x^2+5x+7 == (x-a)(x-b)(x-c) = x^3-(a+b+c)x^2+(ab+bc+ac)x-abc

so, comparing coefficients,

a+b+c = 8

(ab+bc+ac) = 5,

& abc = -7

now, (a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ac) evaluates to 64 = a^2+b^2+c^2+10

so a^2+b^2+c^2=54

now, a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)+3abc evaluates to a^3+b^3+c^3=8(54-5)-21

so a^3+b^3+c^3 = 371

Yup that is the right method, well done. I believe it's called Veita's formula or method or something... In case you wanna look it up :)

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I still think at some point there will be a number that can not be expressed. Eventually.

111 = (-1+root(9)!)!-root(9)^root(4)

112 = (-1+root(9)!)!-root(9)!-root(4)

113 = 19+94

114 = (-1+root(9)!)!-root(9)*root(4)

115 = (-1+root(9)!)!-root(9)-root(4)

116 = (-1+root(9)!)!-root(9)!+root(4)

117 = (-1+root(9)!)!-root(9)!/root(4)

118 = (-1+root(9)!)!-root(9)!+4

119 = (-1+root(9)!)!-root(9)+root(4)

120 = (-1+root(9)!)!*(root(9)-root(4))

121 = (-1+root(9)!)!+root(9)-root(4)

122 = (-1+root(9)!)!+root(9)!-4

123 = (-1+root(9)!)!+root(9)!/root(4)

124 = (-1+root(9)!)!+root(9)!-root(4)

125 = (-1+root(9)!)!+root(9)+root(4)

126 = (-1+root(9)!)!+root(9)*root(4)

127 = (-1+root(9)!)!+9-root(4)

128 = (-1+root(9)!)!+root(9)!+root(4)

129 = (-1+root(9)!)!+root(9)^root(4)

130 = (-1+root(9)!)!+root(9)!+4

131 = (-1+root(9)!)!+9+root(4)

132 = (-1+root(9)!)!+root(9)!*root(4)

133 = (-1+root(9)!)!+9+4

134 = -1-9+root(9)!*4!

135 = -1*9+root(9)!*4!

136 = +1-9+root(9)!*4!

137 = -1-root(9)!+root(9)!*4!

138 = -1*root(9)!+root(9)!*4!

139 = +1-root(9)!+root(9)!*4!

140 = -1-root(9)+root(9)!*4!

141 = -1*root(9)+root(9)!*4!

142 = +1-root(9)+root(9)!*4!

143 = -1^9+root(9)!*4!

144 = 1^9*root(9)!*4!

145 = 1^9+root(9)!*4!

146 = -1+root(9)+root(9)!*4!

147 = 1*root(9)+root(9)!*4!

148 = 1+root(9)+root(9)!*4!

149 = -1+root(9)!+root(9)!*4!

150 = 1*root(9)!+root(9)!*4!

151 = 1+root(9)!+root(9)!*4!

152 = -1+9+root(9)!*4!

153 = 1*9+root(9)!*4!

154 = 1+9+root(9)!*4!

:blink:

that's some great work!

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