A clock for 2017

Question

Design a clock where each number from 1 to 12 is obtained as an arithmetical operation using each digit of 2017 exactly once: for example, 4 could be made as $2\times 7-10$.

"tick" means checkmark. It took me a while to figure out what it meant because this question is about clocks. — user58432, Apr 14, 2019 at 6:53

2012rcampion · Accepted Answer · 2016-12-13 10:50:15Z

52

With the digits in order:

$$ \begin{align} 1 &= 2 + 0 - 1 ^ 7 \\ 2 &= 2 + 0 \times 1 \times 7 \\ 3 &= 2 + 0 + 1 ^ 7 \\ 4 &= -2 - 0 - 1 + 7 \\ 5 &= 2 \times (0 - 1) + 7 \\ 6 &= 2 \times 0 - 1 + 7\\ 7 &= 2 \times 0 \times 1 + 7 \\ 8 &= 2 \times 0 + 1 + 7 \\ 9 &= 2 + 0 \times 1 + 7 \\ 10 &= 2 + 0 + 1 + 7 \\ 11 &= 2 + 0! + 1 + 7 \\ 12 &= 2 \times (0 - 1 + 7) \\ \end{align} $$

answered Dec 13, 2016 at 10:50

2012rcampion

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4

$\begingroup$ 👏 my applauds... $\endgroup$
– Matas Vaitkevicius
Dec 13, 2016 at 12:52
$\begingroup$ 4 feels like a bit of a cheat to me though :( $\endgroup$
– dcfyj
Dec 16, 2016 at 13:59
1

$\begingroup$ To avoid using the negative on 4 you could do 2+0!+1^7 :) $\endgroup$
– dcfyj
Dec 16, 2016 at 14:04
$\begingroup$ really impressive! $\endgroup$
– mau
Dec 18, 2016 at 16:15

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Marius · Accepted Answer · 2016-12-19 10:10:58Z

I tried to make a digital clock.

$0 = (7 + 1 + 2) \times 0$
$1 = (2 + 7 + 1) ^ 0$
$2 = (7 + 1) \times 0 + 2$
$3 = 7 \times 0 + 2 + 1$
$4 = 2 \times 7 - 10$
$5 = 7 - 2 + 1 \times 0$
$6 = 7 - 1 + 2 \times 0$
$7 = 7 + 1 * 2 \times 0$
$8 = 7 + 1 + 0 \times 2$
$9 = 7 + 2 + 1 \times 0$
$10 = 1 + 2 + 7 + 0$
$11 = 12 - 7^0$
$12 = 12 + 7 \times 0$
$13 = 12 + 7 ^ 0$
$14 = 7 \times 2 + 1 \times 0$
$15 = 7 \times 2 + 1 + 0$
$16 = (7 + 1) \times 2 + 0$
$17 = (7 + 1) \times 2 + 0!$
$18 = (7 + 2) \times (1 + 0!)$
$19 = 10 + 2 + 7$
$20 = 17 + 2 + 0!$
$21 = 7 \times (2 + 1 + 0)$
$22 = 7 \times (2 + 1) + 0!$
$23 = 17 + (2 + 0!)!$ or $(7-2-1)! - 0!$ thanks to stack reader
$24 = 2 \times 7 + 10$

[Edit]
What the hell...lets do it for minutes also (I cheated a bit):

$25 = (7 - 1 - 0!)^2$
$26 = 27 - 1 + 0$
$27 = 27 + 1 \times 0$
$28 = 27 + 1 + 0$
$29 = 27 + 1 + 0!$
$30 = 10 \times \lfloor\frac{7}{2}\rfloor$
$31 = \lceil\log(17!) \times 2\rceil + 0!$ // $\log(17!) = 14.5510$
$32 = (1+0!)^{(7-2)}$
$33 = 17 \times 2 - 0!$
$34 = 17 \times 2 + 0$
$35 = 17 \times 2 + 0!$
$36 = \frac{70}{2} + 1$
$37 = \lfloor\ln {7}^{20}\rfloor - 1$ // $\ln {7}^{20} = (38.9182)$
$38 = \lfloor\ln {7}^{20}\rfloor \times 1$ // $\ln {7}^{20} = (38.9182)$
$39 = \lfloor\ln {7}^{20}\rfloor + 1$ // $\ln {7}^{20} = (38.9182)$
$40 = 10 \times \lceil\frac{7}{2}\rceil$
$41 = \lceil\ln {7}^{21}\rceil + 0 $ // $\ln {7}^{21} = (40.8641)$
$42 = \lfloor\ln {72}^{10}\rfloor$ // $\ln {72}^{10} = (42.76666)$
$43 = \lceil\ln {72}^{10}\rceil$ // $\ln {72}^{10} = (42.76666)$
$44 = \lceil{(\ln 710})^{2}\rceil$ // $({\ln 710})^{2} = (43.1027)$
$45 = \lfloor\log(10!) * 7 - \ln(2)\rfloor $ // $\log(10!) = 6.5597$
$46 = \lceil\log(10!) * 7 - \ln(2)\rceil $ // $\log(10!) = 6.5597$
$47 = 7^2 - 1 - 0!$
$48 = 7^2 - 1 + 0$
$49 = 7^2 + 1 \times 0$
$50 = 7^2 + 1 + 0$
$51 = 7^2 + 1 + 0!$
$52 = \lceil\log(2^{170})\rceil$ // $\log(2^{170}) = (51.1750)$
$53 = \lfloor\ln(17!)\rfloor + 20$ // $\ln(17!) = 33.5050$
$54 = 27 \times (1 + 0!)$
$55 = \lceil\ln(27!)\rceil - 10$ // $\ln(27!) = 64.5575$
$56 = \lfloor\ln(17^{20})\rfloor $ // $\ln(17^{20}) = 56.6642 $
$57 = \lceil\ln(17^{20})\rceil $ // $\ln(17^{20}) = 56.6642 $
$58 = 70 - 12 $
$59 = 7^2 + 10$

@oleslaw....there was no need for 24, because I got 0, but thanks. :) — Marius, Dec 13, 2016 at 9:18
@Marius Oh, you're right. You can delete it or leave it as you want :P — oleslaw, Dec 13, 2016 at 9:20

2012rcampion · Accepted Answer · 2016-12-13 10:29:49Z

11

$1 = 7 \times 0 + 2 - 1$

$2 = 7 \times 0 + 2 \times 1$

$3 = 20 - 17$

$4 = 7 - 2 - 1 - 0$

$5 = 7 - 2 - 0 \times 1$

$6 = 7 - 1 - 0 \times 2$

$7 = 0 \times 1 \times 2 + 7$

$8 = 0 \times 2 + 1 + 7$

$9 = 0 \times 1 + 2 + 7$

$10 = 0 + 1 + 2 + 7$

$11 = 12 - 7 ^ 0$

$12 = 0 \times 7 + 12$

I'm assuming I'm not allowed to use ^, so give me a few minutes to find an acceptable solution for 11!

edited Dec 13, 2016 at 10:29

2012rcampion

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answered Dec 13, 2016 at 8:23

TheGreatEscaper

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$\begingroup$ Drat! I got ninjaed... $\endgroup$
– Sid
Dec 13, 2016 at 8:25
$\begingroup$ @TheGreatEscaper I believe there is no solution for 11 using only the +, -, * and / operators (checked with brute-force computer program) $\endgroup$
– oleslaw
Dec 13, 2016 at 9:10
$\begingroup$ @oleslaw I'd figured just about that much. Especially when you realise the 0 is basically useless for 11. $\endgroup$
– TheGreatEscaper
Dec 13, 2016 at 9:31

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Bugs · Accepted Answer · 2016-12-13 09:02:25Z

4

$1 = 2*0*7+1$

$2 = 2+0*1*7$

$3 = 2+0*7+1$

$4 = (7+1+0)/2$

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

$6 = -2+0+1+7$

$7 = 2*0*1+7$

$8 = 2*0+1+7$

$9 = 2+0*1+7$

$10 = 2+0+1+7$

$11 = 12-7^0$

$12 = 12-7*0$

edited Dec 13, 2016 at 9:02

Bugs

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answered Dec 13, 2016 at 8:26

Pokemon

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$\begingroup$ Yup, 12-7^0=11 not 12. $\endgroup$
– Pokemon
Dec 13, 2016 at 8:28
$\begingroup$ Yeah, my bad... I saw 12=12-7^0 $\endgroup$
– Sid
Dec 13, 2016 at 8:28

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Sid · Accepted Answer · 2016-12-13 08:24:36Z

3

1=2*7*0+1
2=1*0*7+2
3=7*0+(2+1)
4=2*7-10
5=1*0+(7-2)
6=2*0+(7-1)
7=2*0*1+7
8=2*0+(7+1)
9=1*0+(7+2)
10=0+7+2+1
11=12-7^0
12=0*7+12

answered Dec 13, 2016 at 8:24

Sid

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JMP · Accepted Answer · 2016-12-13 08:44:30Z

3

A couple more for $12$:

$12=20-1-7, 12=(2+0)\times(-1+7)$

answered Dec 13, 2016 at 8:44

JMP

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Andrew Savinykh · Accepted Answer · 2016-12-14 08:22:36Z

Just for entertainment value, if we limit ourselves with just 4 basic operations (+-*/) without even unary minus, and if we agree to use four separate digits 2,0,1,7 without combining them into numbers like 12, we still can get 11 results out of 12!

Here is the C# code:

var found = new Tuple<int[],Tuple<Func<Decimal, Decimal, Decimal>,string>[]>[12];
var number = new[] { 2, 0, 1, 7 };
var op = new Tuple<Func<Decimal, Decimal, Decimal>, string>[] 
{
    new Tuple<Func<Decimal, Decimal, Decimal>, string>((x,y) => x + y,"+"), 
    new Tuple<Func<Decimal, Decimal, Decimal>, string>((x,y) => x - y,"-"),
    new Tuple<Func<Decimal, Decimal, Decimal>, string>((x,y) => x * y,"*"), 
    new Tuple<Func<Decimal, Decimal, Decimal>, string>((x,y) => x / y,"/"),
};
foreach (var i in GetPermutations(number, 4))
{
    foreach (var j in GetPermutationsWithRept(op, 3))
    {
        var ii = i.ToArray(); var jj = j.ToArray();
        decimal result = ii[0]; var divideByZero = false;
        for (int k = 0;k < 3; k++)
        {
            if (jj[k] == op[3] && ii[k + 1] == 0)
            {
                divideByZero = true;
                break;
            }
            result = jj[k].Item1(result,ii[k+1]);
        }
        if (divideByZero) continue;
        if (result <= 12 && result >=1 && result == ((decimal)(int)result))
        {
            found[(int)result-1] = new Tuple<int[],Tuple<Func<Decimal, Decimal, Decimal>,string>[]>(ii,jj);
        }
    }
}
PrintResult(found);

And here is the result:

1=(((7*0)-1)+2)
2=(((7/1)*0)+2)
3=(((7-1)-0)/2)
4=(((7-1)-0)-2)
5=(((7/1)-0)-2)
6=(((7+1)-0)-2)
7=(((1*0)/2)+7)
8=(((7-1)-0)+2)
9=(((7/1)-0)+2)
10=(((7+1)-0)+2)
11=Unknown
12=(((7-1)-0)*2)

Implementation of GetPermutations, GetPermutationsWithRept and PrintResult is left as an exercise for the reader.

Anyone would like to write up a code-golf challenge for finding the clock faces ;)?

As a side note the dreadful Tuples look much nicer in C# 7. — Andrew Savinykh, Dec 14, 2016 at 8:25

Bhaskar · Accepted Answer · 2016-12-13 14:05:55Z

0

1 = 1 + 0 * 2 * 7

2 = 2 + 0 * 1 * 7

3 = 1 + 2 + 0 * 7

4 = 7 - (0 + 1 + 2)

5 = 7 - 2 + 0 * 1

6 = 7 - 1 + 0 * 2

7 = 7 + 0 * 1 * 2

8 = 7 + 1 + 0 * 2

9 = 7 + 2 + 0 * 1

10 = 7 + 0 + 1 + 2

11 = 71 % 20

12 = 12 + 0 * 7

, where % is a modulus operator.

answered Dec 13, 2016 at 14:05

Bhaskar

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$\begingroup$ why are some digits bold? $\endgroup$
– Marius
Dec 13, 2016 at 14:13
2

$\begingroup$ @Marius Why are all the other numbers afraid of 7? Because 7 8 9 (eight is a homophone of ate). So 7 must be bold. Or it could be that Bhaskar highlighted the numbers which contribute to the final number, so 0 * 2 * 7 does not. $\endgroup$
– Andrew Morton
Dec 13, 2016 at 19:14
$\begingroup$ @Marius looks like non-zero terms are bold $\endgroup$
– wilson
Dec 14, 2016 at 2:57
1

$\begingroup$ @Marius : As Andrew and wilson say, I highlighted the non-zero terms that contribute to the final results. I thought it would help in reading it quicker. Nvm if it didnt. :) $\endgroup$
– Bhaskar
Dec 14, 2016 at 9:25

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Fᴀʀʜᴀɴ Aɴᴀᴍ · Accepted Answer · 2016-12-15 08:39:35Z

0

$1 = 1 ^ {720}$

$2 = 2^0 + 1^7 $

$3 = 2^1 + 7^0$

$4 = 7 - 2 - 1 - 0$

$5 = 7 - 2^1 + 0$

$6 = (2 + 1)! + (7 × 0)$

$7 = 7 + ((2 + 1) × 0)$

$8 = 2 + 0 - 1 + 7$

$9 = 7 + 2 + (1 × 0)$

$10 = 2 + 0 + 1 + 7$

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

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

answered Dec 15, 2016 at 8:39

Fᴀʀʜᴀɴ Aɴᴀᴍ

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