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This is similar to the "Four fours" puzzle, but using the digits 2, 0, 1 and 7.

Rules:

  • Use all four digits exactly once
  • Allowed operations: +, -, x, ÷, ! (factorial), exponentiation, square root
  • Parentheses and grouping (e.g. "21") are also allowed
  • Squaring uses the digit 2 so expressions using multiple twos, like $2^2$ or $1^2 + 7^2$, are not allowed
  • Keep the order "2017" in at least 16 expressions (and more if you can!)

Good luck and Happy New Year!

Similar question for 2016

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fitch496
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  • is modulus operator allowed? – Sid Jan 02 '17 at 08:23
  • @Sid No, the modulus operator is not allowed. – fitch496 Jan 02 '17 at 08:39
  • Does the fourth rule also apply to "square rooting", because the square root works more or less the same – Fearsome Statue Jan 02 '17 at 10:48
  • @FearsomeStatue no, it does not apply. You can use the square root any number of times (even though I don't see how that would help you...). It is possible to solve all numbers without using square root at all :-) – fitch496 Jan 02 '17 at 11:05
  • can we round, for example 120/7 -> 17? –  Jan 02 '17 at 23:45
  • @ev3commander No, rounding is not allowed. – fitch496 Jan 03 '17 at 10:55
  • Is grouping after some operations allowed (like "2+0!"=21) and is negating allow (e.g. -7+(2*0-1); the -1 is a subtraction, but the -7 at the start is a negation; sometimes helpful for maintaining 2017, both for the starting digit and if you want to negate the entire expression for the sake of having fun going into the negatives in the expression). – Jakob Pamp Bengtsson Jan 04 '17 at 11:55
  • @JakobPampBengtsson No, grouping after some operations or any "string" operations are not allowed. Negation is allowed. – fitch496 Jan 04 '17 at 12:24

8 Answers8

17

This answer has 29 expressions with the "2017" order. Those NOT in order are denoted by sadness - :(

$1=2*0+1^7$

$2=2^0+1^7$

$3=2+0+1^7$

$4=-2+0-1+7$

$5=-2+(0*1)+7$

$6=(2*0)-1+7$

$7=2^0-1+7$

$8=(2*0)+1+7$

$9=2+(0*1)+7$

$10=2+0+1+7$

$11=2+0!+1+7$

$12=(2+0)*(-1+7)$

$13=(2+0+1)!+7$

$14=(2+0!)!+1+7$

$15=-2+0+17$ (Improved for order by Ivo Beckers)

$16=-((2*0)!)+17$

$17=(2*0)+17$

$18=(2^0)+17$

$19=2+0+17$

$20=2+0!+17$

$21=20+1^7$

$22=-2+ (\sqrt{-(0!-17)})!$ (Improved by Pratheek B!)

$23=(2+0!)!+17$

$24=(2+0!)*(1+7)$

$25=(7-1-0!)^2$ :(

$26=20-1+7$

$27=20+(1*7)$

$28=20+1+7$

$29=27+(1+0!)$ :(

$30=10\sqrt{2+7}$ :(

$31=(2+0!+1)!+7$

$32=2^{-(0!)-1+7}$

FOOLING AROUND (I'm simply curious about how far we can go)

$33=17*2-0!$

$34 = (2+0)*17$ :D

$35=((2+0!)!-1)*7$ (Improved by Christoph!)

$36=(7-1+0)^2$

$37=20+17$ :D

$38=???$

$39=7^2-10$

$40=???$

$41=(2+0!)!*7-1$

Beastly Gerbil
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greenturtle3141
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6

In order solution for 22:

$-2 + \sqrt{-0! + 17}! $

Glorfindel
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Pratheek B
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Okay, I know there's another answer but I'm posting mine before I look at it (honest!). 28/32 in order.

1 = 2^(0*1*7)
2 = 2 + 0*1*7
3 = 2 + 0 + 1^7
4 = -2 + 0 - 1 + 7
5 = -2 + 0*1 + 7
6 = -2 + 0 + 1 + 7
7 = 2*0*1 + 7
8 = (2^0)*1 + 7
9 = 2 + 0*1 + 7
10 = 2 + 0 + 1 + 7
11 = 2 + 0! + 1 + 7
12 = 2*(0 - 1 + 7)
13 = (2 + 0 + 1)! + 7
14 = 2*(0*1 + 7)
15 = -2 + 0 + 17
16 = 2*(0 + 1 + 7)
17 = 2*0 + 17
18 = 2^0 + 17
19 = 2 + 0 + 17
20 = 20 * 1^7
21 = 20 + 1^7
*22 = 21 + 7^0
23 = (2 + 0!)! + 17
24 = (2 + 0!) * (1 + 7)
*25 = (7 - 2)^(0! + 1!)
26 = 20 - 1 + 7
27 = 20 + 1*7
28 = 20 + 1 + 7
*29 = 21 + 0! + 7
*30 = 210 / 7
31 = (2 + 0! + 1)! + 7
32 = 2 * (-0! + 17)

My attempts to press on...

*33 = 2*17 - 0!
34 = (2 + 0)*17
35 = ((2 + 0!)! - 1)*7
36 = 2 * (0! + 17)
37 = 20 + 17
38 = ?
*39 = 7^2 - 10
40 = ?
*41 = ((2 + 0!)! * 7) - 1
42 = (2 + 0!)! *1*7
*43 = ((2 + 0!)! * 7) + 1
44 = ?
45 = ?
46 = ?
*47 = 7^2 - 0! - 1
48 = (2 + 0!)! * (1 + 7)
49 = ((2 + 0!)! + 1) * 7
*50 = 7^2 + 0 + 1
51 = (2 + 0!) * 17

Neil W
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I've been looking for an elegant solution to this problem using octal (base-8) arithmetic. Perhaps someone could help me complete this. There's one I couldn't get the numbers in the right order, and another that I couldn't find any solution. Here's what I've got so far:

enter image description here

Community
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wildBillMunson
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0

Here's my answer, with 29 in order: (I'm working on 25, 29, and 30)

$ 1 = 2 * 0 + 1 ^ 7 $
$ 2 = 2 + 0 * 1 * 7 $
$ 3 = 2 + 0 + 1 ^ 7 $
$ 4 = -2 + 0 - 1 + 7 $
$ 5 = -2 + 0 + 1 * 7 $
$ 6 = -2 + 0 + 1 + 7 $
$ 7 = 2 * 0 * 1 + 7 $
$ 8 = 2 + 0 - 1 + 7 $
$ 9 = 2 + 0 + 1 * 7 $
$ 10 = 2 + 0 + 1 + 7 $
$ 11 = 2 + 0! + 1 + 7 $
$ 12 = 2 * (0 - 1 + 7) $
$ 13 = (2 + 0 + 1)! + 7 $
$ 14 = (2 + 0 * 1) * 7 $
$ 15 = -2 + 0 + 17 $
$ 16 = (2 + 0) * (1 + 7) $
$ 17 = 2 * 0 + 17 $
$ 18 = 2 ^ 0 + 17 $
$ 19 = 2 * 0! + 17 $
$ 20 = 2 + 0! + 17 $
$ 21 = (2 + 0 + 1) * 7 $
$ 22 = -2 + \sqrt{-0! + 17}! $ (Thanks @PratheekB!)
$ 23 = (2 + 0!)! + 17 $
$ 24 = (2 + 0!) * (1 + 7) $
$ 25 = (7 - 2) ^ {1 + 0!} $
$ 26 = 20 - 1 + 7 $
$ 27 = 20 * 1 + 7 $
$ 28 = 20 + 1 + 7 $
$ 29 = 27 + 0! + 1 $
$ 30 = 210 \div 7 $
$ 31 = (2 + 0! + 1)! + 7 $
$ 32 = 2 * (-0! + 17) $

Besides 22, I came up with these by myself.

Oliver Ni
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0

A bit cheeky with the use of a decimal point:

(20+1)/.7 = 30

Or, using the same logic:

210/7 = 30

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MrT
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0

Half answer(all in order)

$1=(2*0*1*7)!$
$2=2+0*1*7$
$3=2+0+1^7$
$4=2+0!+1^7$
$5=(2+0!)!-1^7$
$6=(2+0+1^7)!$
$7=2*0*1+7$
$8=2*0+1+7$
$9=2-0!+1+7$
$10=2+0+1+7$
$11=2+0!+1+7$
$12=(2+0!)!-1+7$
$13=20-1*7$
$14=20+1-7$
$15=-2+0+17$
$16=-2+0!+17$
$17=2*0+17$
$18=2-0!+17$
$19=2+0+17$
$20=2+0!+17$
$21=(2+0+1)*7$
$22=-2+(\sqrt{-0!+17})!$ Thanks @Pratheek B
$23=(2+0!)!+17$
$24=(2+0!+1^7)!$
$**25=(-0!-1+7)^2**$

Matheinstein
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-1

Possible in order solutions for 25 and 29:

2 + (-0! - 1 + 7) = 25
2 + (0! + 1 + 7) = 29

Or is that cheating?

stack reader
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peder541
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