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I made this puzzle myself. As the title says, you must make the numbers 1 to 100 using the numbers 1,7,3,4 - in that order. All numbers must be used. The allowed operators are below:

  • Addition, Subtraction, Multiplication and Division.

  • Square roots and exponents.

  • Factorials.

  • Concatenation.

  • Floor and Ceiling Functions.

  • Parentheses.

Anything that is not on this list must NOT be used.

Those are all the rules, Have fun!

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  • $\begingroup$ Allowed as part of a solution: non integer factorals? (e.g. sqrt(24)! = 4.898! = 101.096) $\endgroup$
    – Retudin
    Commented Oct 18, 2022 at 9:44

5 Answers 5

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It required some cups of coffee, and it became quite messy for some specific numbers, but hopefully the below solution is without errors and typos. Looking forward to seeing better (read "easier") solutions to these numbers where possible.

$1: 1^{734}$
$2: 1^7-3+4$
$3: 1-\text{floor}(7/3)+4$
$4: -1-7+3*4$
$5: -1*7+3*4$
$6: 1*7+3-4$
$7: 1+7+3-4$
$8: 1^7+3+4$
$9: 1+7-3+4$
$10: 1^7+3^{√4}$
$11: -1^7+3*4$
$12: \text{floor}(1/7)+3*4$
$13: 1^7+3*4$
$14: 1*7+3+4$
$15: 1+7+3+4$
$16: -1+7*3-4$
$17: 1*7*3-4$
$18: 1+7*3-4$
$19: 1*7+3*4$
$20: 1+7+3*4$
$21: 1+(7+3)*√4$
$22: -1+7*3+√4$
$23: 1*7*3+√4$
$24: 1+7*3+√4$
$25: 1*7*3+4$
$26: 1+7*3+4$
$27: 1*\text{floor}(\sqrt{734})$
$28: (1+7)*3+4$
$29: 1*(\text{ceil}(√7))^3+√4$
$30: 1+(\text{ceil}(√7))^3+√4$
$31: 1*(\text{ceil}(√7))^3+4$
$32: 1+(\text{ceil}(√7))^3+4$
$33: -1^7+34$
$34: \text{floor}(1/7)+34$
$35: 1^7+34$
$36: 17*\text{ceil}(√3)+√4$
$37: 1*\text{ceil}(√7)+34$
$38: 1+\text{ceil}(√7)+34$
$39: \text{ceil}(√17)+34$
$40: -1+7+34$
$41: 1*7+34$
$42: 1+7+34$
$43: 1+7*\text{ceil}(√34)$
$44: -1+7^{\text{ceil}(√3)} -4$
$45: 1*7^{\text{ceil}(√3)} -4$
$46: 1+7^{\text{ceil}(√3)} -4$
$47: 17*3-4$
$48: 1+7^{\text{ceil}(√3)}-√4$
$49: 17*3-√4$\ $50: -1+7^{\text{ceil}(√3)} +√4$
$51: 17+34\\ 52: 1+7^{\text{ceil}(√3)} +√4$
$53: 1*7^{\text{ceil}(√3)} +4$
$54: (-1+7)^3/4 $
$55: -1+\text{ceil}(\sqrt(7!/√3) )+√4 $
$56: -1+\text{ceil}(\sqrt(7!/√3) )+√4$
$57: -1+\text{floor}(7^{√3}*√4)$
$58: \text{floor}(17*√(3*4))$
$59: \text{ceil}(17*√(3*4))$
$60: (1+\text{ceil}(√7))^3-4$
$61: 1*\text{ceil}(√(7!*3)/√4)$
$62: (1+\text{ceil}(√7))^3-√4$
$63: \text{ceil}(√17 !)+34$
$64: -17+3^4$
$65: \text{floor}(√((1+7)!)/3-√4)$
$66: \text{floor}(√(17^3 ))-4$
$67: -1+\text{floor}(√7)*34$
$68: -1+73-4$
$69: 1*73-4$
$70: -1+73-√4$
$71: 1*73-√4$
$72: 1+73-√4$
$73: -1-7+3^4$
$74: -1*7+3^4$
$75: 1-7+3^4$
$76: -1+73+4$
$77: -1-\text{ceil}(√7)+3^4 $
$78: 1*\text{ceil}(√7)+3^4$
$79: 1-\text{ceil}(√7)+3^4$
$80: -1^7+3^4$
$81: \text{floor}(1/7)+3^4$
$82: 1^7+3^4$
$83: -1+7*3*4$
$84: 1*7*3*4$
$85: 1+7*3*4 $
$86: \text{floor}(173/√4)$
$87: -1+7*3^4$
$88: 1*7+3^4$
$89: 1+7+3^4$
$90: 1*\text{ceil}(√7*34)$
$91: 1+\text{ceil}(√7*34)$
$92: \text{floor}(√((-1+7)!*3*4))$
$93: \text{ceil}(√((-1+7)!*3*4))$
$94: \text{floor}((1+7+√3)^{√4})$
$95: \text{ceil}((1+7+√3)^{√4})$
$96: \text{floor}(√(1+7)*34)$
$97: \text{ceil}(√(1+7)*34)$
$98: 1*\text{ceil}(√(7! *√3) )+4$
$99: -1+(7+3)^{√4}$
$100:1*(7+3)^{√4}$

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  • $\begingroup$ As far as I understand the OP, the rules do not allow negative numbers and -1 would not be allowed as this is not a subtraction. $\endgroup$
    – lukas.j
    Commented Oct 18, 2022 at 9:12
  • $\begingroup$ @lukas.j Negative numbers are allowed. $\endgroup$
    – one23456789
    Commented Oct 21, 2022 at 22:46
  • $\begingroup$ Then I think your question should have mentioned negation resp. the unary minus. Basically you have a contradiction in your question because now with allowing the unary minus, you have 8 digits to chose of, whereby 1 or -1 must be used first, then 7 or -7, etc. $\endgroup$
    – lukas.j
    Commented Oct 22, 2022 at 8:55
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I doubt if a binary operator is needed at all, some solutions

41 = ⌊√1734⌋
42 = ⌈√1734⌉

35 = ⌊√√√√√41!⌋ i.e. ⌊√√√√√⌊√1734⌋!⌋
36 = ⌈√√√√√41!⌉
39 = ⌊√√√√√42!⌋
40 = ⌈√√√√√42!⌉

28 = ⌊√√√√√39!⌋
29 = ⌈√√√√√39!⌉
31 = ⌊√√√√√40!⌋
32 = ⌊√√√√√40!⌋
17 = ⌊√√√√√35!⌋
18 = ⌈√√√√√35!⌉
19 = ⌊√√√√√36!⌋
20 = ⌈√√√√√36!⌉

69 = ⌊√√√√28!⌋
70 = ⌈√√√√28!⌉
8 = ⌊√√√√√28!⌋
9 = ⌈√√√√√28!⌉
85 = ⌊√√√√29!⌋
86 = ⌈√√√√29!⌉
10 = ⌈√√√√√29!⌉
11 = ⌊√√√√√31!⌋
12 = ⌈√√√√√31!⌉
13 = ⌈√√√√√32!⌉
65 = ⌊√√√17!⌋
66 = ⌈√√√17!⌉
94 = ⌊√√√18!⌋
95 = ⌈√√√18!⌉
14 = ⌊√√√√20!⌋
15 = ⌈√√√√20!⌉

33 = ⌈√√√15!⌉
23 = ⌊√√√14!⌋
24 = ⌈√√√14!⌉
16 = ⌊√√√13!⌋
17 = ⌈√√√13!⌉
79 = ⌊√√11!⌋
80 = ⌈√√11!⌉
43 = ⌊√√10!⌋
44 = ⌈√√10!⌉
25 = ⌊√√√√23!⌋
26 = ⌈√√√√23!⌉
30 = ⌊√√√√24!⌋

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As noted in comments, each number must be used exactly once, in the order listed.

Here are solutions for a couple dozen numbers:

1 = 1 + 7 - 3 - 4

2 = 1 + (7 / (3 + 4))

7 = 1 + 7 + 3 - 4

8 = (1 + 7) ^ (-3 + 4)

9 = 1 + 7 - 3 + 4

14 = (1 * 7) + 3 + 4

15 = 1 + 7 + 3 + 4

16 = -1 + (7 * 3) - 4

18 = 1 + (7 * 3) - 4

19 = (1 * 7) + (3 * 4)

20 = 1 + 7 + (3 * 4)

26 = 1 + (7 * 3) + 4

42 = (-1 + 7) * (3 + 4)

49 = 1 * 7 * (3 + 4)

56 = (1 + 7) * (3 + 4)

64 = -17 + (3 ^ 4)

73 = -1 - 7 + (3 ^ 4)

75 = 1 - 7 + (3 ^ 4)

80 = -(1 ^ 7) + (3 ^ 4)

81 = (1 ^ 7) * (3 ^ 4)

82 = (1 ^ 7) + (3 ^ 4)

87 = -1 + 7 + (3 ^ 4)

88 = (1 * 7) + (3 ^ 4)

89 = 1 + 7 + (3 ^ 4)

96 = (1 + 7) * (3 * 4)

98 = 17 + (3 ^ 4)

Improve this answer
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2
  • $\begingroup$ Good job finding quite a few important numbers! $\endgroup$
    – Stevo
    Commented Oct 16, 2022 at 0:28
  • $\begingroup$ As far as I understand the OP, the rules do not allow negative numbers and -1 would not be allowed as this is not a subtraction. $\endgroup$
    – lukas.j
    Commented Oct 18, 2022 at 9:12
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PARTIAL ANSWER, NOT FINISHED, however I have made quite significant progress.

Ok... Let's have a look.

1 : 1 + 7 - 3 - 4
2 : 1 x 7 - 3 + 4
3 : 1 + 7 - 3 - √4
4 : 1 x ⌊(7 / 3)⌋ x √4
5 : ⌊(1/7)⌋ + 3 + √4
6 : ⌊(1/7)⌋ + 3 x √4
7 : ⌊(1/7)⌋ + 3 + 4
8 : 1 + 7 - ⌊(3/4)⌋
9 : 1 + 7 + ⌈(3/4)⌉
10 : ⌊(1 x 7 x 3/√4)⌋
11 : 1 + ⌊(7x3/√4)⌋
12 : 1 + ⌈(7x3/√4)⌉
13 : ⌈(1/7)⌉ + 3 x 4
14 : 1 x 7 + 3 + 4
15 : 1 + 7 + 3 + 4
16 : -(1 - 7 x 3 + 4)
17 : 1 + (7 - 3) x 4
18 : -(1 - (7 + 3 x 4))
19 : 1 x (7 + 3 x 4)
20 : 1 + 7 + 3 x 4
21 : 1 - 7 + 3 + 4!
22 : -1 - 7 + 3! + 4!
23 : 1 + 7 + 3! + 4!
24 : -1 + 7 x 3 + 4
25 : 1 x 7 x 3 + 4
26 : 1 + 7 - 3! + 4!
27 : ⌊1/7⌋ + 3 + 4!
28 : ⌈1/7⌉ + 3 + 4!
29 : 17 + 3 x 4
30 : 1^7 x 3! + 4!
31 : ⌊(⌈√17⌉^3/4)⌋
32 : ⌊√17⌋^3/√4
33 : -1 + 7 + 3 + 4! (thank you wimi)
34 : -1 + (7!/3!)/4!
35 : 1 x (7!/3!)/4!
36 : 1 + (7!/3!)/4!
37 : ⌈1/7⌉ + √(3!^4)
38 : 1 + 7 + 3! + 4!
39 : -1 + (7 + 3) x 4 (thank you wimi)
40 : (1 x 7 + 3) × 4 (thank you wimi)
41 : ⌊√1734⌋
42 : (-1 + 7) x (3 + 4)
43 : ⌊173/4⌋
44 : ⌈173/4⌉
45 : -1 + 7 x (3!) + 4
46 : 1 x 7 x (3!) + 4
47 : 17 x 3 - 4
48 : -1 + 7 ^ ⌊(√(3+4))⌊
49 : 1 x 7 x (3 + 4)
50 : 1 + 7 ^ ⌊(√(3+4))⌋
51 : 17 + 34
52 : ⌊√173⌋ x 4
53 : 17 + √(3!^4)
54 : (1 + ⌊√√7!⌋) x 3 x √4
55 : 17 x 3 + 4
56 : (1 + 7) x (3 + 4) 57 :
58 :
59 : -1 + ⌊√(7)⌋ x (3!+4!)
60 : ⌊√(1+7)⌋ x (3!+4!)
61 : 1+ ⌊√(7)⌋ x (3!+4!)
62 : ⌊(⌈√17⌉^3/√4)⌋
63 : ⌈(⌈√17⌉^3/√4)⌉
64 : -17 + (3 ^ 4)
65 :
66 :
67 :
68 : -1 + 73 - 4
69 : 1 x 73 - 4
70 : 1 + 73 - 4
71 :
72 :
73 : -1 + (-7 + 3 ^4)
74 : 1 x (-7 + 3 ^ 4)
75 : 1 + (-7 + 3 ^4)
76 : -1 + 73 + 4
77 : 1 x (73 + 4)
78 : 1 + 73 + 4
79 : 1 x 7 + 3! x 4! 80 : -(1 ^ 7) + (3 ^ 4)
81 : (1 ^ 7) x (3 ^ 4)
82 : (1 ^ 7) + (3 ^ 4)
83 : 1 + ⌊√7!⌋ + 3 x 4
84 : -1 + ⌊7^3/4⌋
85 : 1 x ⌊7^3/4⌋
86 : 1 + ⌊7^3/4⌋
87 : -1 + 7 + (3 ^ 4)
88 : (1 * 7) + (3 ^ 4)
89 : 1 + 7 + (3 ^ 4)
90 : ⌈√(1+7)⌉ x (3!+4!)
91 : 1 + ⌈√(7)⌉ x (3!+4!)
92 : 1 x ⌈√(7!)⌉ − 3 + 4!
93 : 1 + ⌈√(7!)⌉ − 3 + 4!
94 :
95 :
96 : (1 + 7) x 3 x 4
97 : -1 + ⌈√(7!)⌉ + 3 + 4!
98 : 17 + (3 ^ 4 )
99 : 1 + ⌈√(7!)⌉ + 3 + 4!
100 : 1 ∗ ⌊√(7!)⌋ + 3! + 4!

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25
  • $\begingroup$ (floor(1 - sqrt(sqrt(7)))) + 3 + 4! = 27 $\endgroup$
    – one23456789
    Commented Oct 16, 2022 at 0:48
  • $\begingroup$ ((floor(1 * sqrt(sqrt(7)))) + 3) * 4! = 96 $\endgroup$
    – one23456789
    Commented Oct 16, 2022 at 0:52
  • $\begingroup$ I hope the 2 comments up of this one helped. $\endgroup$
    – one23456789
    Commented Oct 16, 2022 at 0:55
  • 1
    $\begingroup$ Thanks! @Evargalo $\endgroup$
    – Stevo
    Commented Oct 18, 2022 at 6:17
  • 1
    $\begingroup$ @lukas.j I would believe that if that was the case, the OP would have already told me in the comments instead of helping and liking this answer... But then hi! Welcome to PSE! $\endgroup$
    – Stevo
    Commented Oct 18, 2022 at 9:55
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Complete list.

Note that these solutions do not use the unary minus as it is not listed in the OP's list of allowed operators:

1: 1 - 7 + 3 + 4
2: 1 × 7 - (3 + √4)
3: 1 + 7 - (3 + √4)
4: 1^7^3 × 4
5: 1^7^3 + 4
6: 1 - 7 + 3 × 4
7: 1 + 7 + 3 - 4
8: 1 × 7 - 3 + 4
9: 1 + 7 - 3 + 4
10: 1 + 7 + 3! - 4
11: 1^7 + 3! + 4
12: 1^7 × 3 × 4
13: 1 + 7 + 3 + √4
14: 1 × 7 + 3 + 4
15: 1 + 7 + 3 + 4
16: (1^7 + 3) × 4
17: 1 × 7 × 3 - 4
18: 1 + 7 × 3 - 4
19: 1 × 7 + 3 × 4
20: 1 + 7 + 3 × 4
21: 1 - 7 + 3 + 4!
22: (1 + 7 + 3) × √4
23: 1 × 7 × 3 + √4
24: 1 - 7 + 3! + 4!
25: 1 × 7 × 3 + 4
26: 1 + 7 × 3 + 4
27: 1^7 × 3 + 4!
28: (1 + 7) × 3 + 4
29: 1 + 7 - 3 + 4!
30: 1^7 × 3! + 4!
31: 1 × 7 + 3! × 4
32: 1 + 7 + 3! × 4
33: 17 × ⌈ √3 ⌉ - ⌊ √√4 ⌋
34: 1 × 7 + 3 + 4!
35: 1 + 7 + 3 + 4!
36: 1 + 7! / 3! / 4!
37: 1 × 7 + 3! + 4!
38: 1 × 7 × 3! - 4
39: 1 + 7 × 3! + 4
40: 1 × (7 + 3) × 4
41: 1 + (7 + 3) × 4
42: 1 × 7 × 3 × √4
43: 1 + 7 × 3 × √4
44: (1 + 7 + 3) × 4
45: 1 × 7 × 3 + 4!
46: 1 + 7 × 3 + 4!
47: 1 + 7 × 3! + 4
48: (1 + 7) × 3 × √4
49: 1 × 7 × (3 + 4)
50: 1 + 7 × (3 + 4)
51: 17 + 34
52: 1 × (7 + 3!) × 4
53: 1 + (7 + 3!) × 4
54: ⌈ (1 + 7 + 3) × √4! ⌉
55: 17 × 3 + 4
56: (1 + 7) × (3 + 4)
57: 1 + 7 × (3! + √4)
58: 17 × ⌈ √3 ⌉ + 4!
59: 1 × 7 + ⌊ 3!^√√4! ⌋
60: (1 + 7 - 3)! / √4
61: 1 + ⌈ 7^√3 ⌉ + ⌊ √√√√4!! ⌋
62: 1 + ⌈ 7^√3 ⌉ + ⌈ √√√√4!! ⌉
63: 1 × 7 × 3^√4
64: 1 + 7 × 3^√4
65: 17 × ⌈ √3 ⌉ + ⌈ √√√√4!! ⌉
66: 1 - 7 + 3 × 4!
67: 1 + 7 × 3! + 4!
68: 17 × ⌈ √3 ⌉ × √4
69: 1 + ⌈ 73 - √4! ⌉
70: 1 × 7 × (3! + 4)
71: 1 + 7 × (3! + 4)
72: (1 + 7) × 3! + 4!
73: 1^7 + 3 × 4!
74: ⌈ 173 / √√√√√√4!! ⌉
75: 1 - 7 + 3^4
76: 1 + 73 + √4
77: 1 × 73 + 4
78: 1 + 73 + 4
79: 1 × 7 + 3 × 4!
80: 1 + 7 + 3 × 4!
81: 1^7 × 3^4
82: 1^7 + 3^4
83: 1 × ⌈ √7! + 3 × 4 ⌉
84: 1 × 7 × 3 × 4
85: 1 + 7 × 3 × 4
86: ⌊ 173 / √4 ⌋
87: ⌈ 173 / √4 ⌉
88: 1 × 7 + 3^4
89: 1 + 7 + 3^4
90: 1 + ⌈ 73^√√√√√4 ⌉
91: 1 × ⌈ 73^√√√√√√4! ⌉
92: 1 + ⌈ 73^√√√√√√4! ⌉
93: ⌊ 17 × √(3! + 4!) ⌋
94: ⌈ 17 × √(3! + 4!) ⌉
95: 1 × ⌈ √7! + 3! × 4 ⌉
96: (1 + 7) × 3 × 4
97: 1 + (7 - 3)! × 4
98: 1 + 73 + 4!
99: 1 + ⌈ √7! ⌉ + 3 + 4!
100: 1 × (7 + 3)^√4

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1
  • $\begingroup$ I like that you try to avoid unary-. 83 = sqrt(1+7!) +3*4; 95 = sqrt(1+7!) +3!*4 (no rounding needed) $\endgroup$
    – Retudin
    Commented Oct 19, 2022 at 9:00

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