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Add the four basic operators × ÷ + −: 10 9 8 7 6 5 4 3 2 1 to get the total 2018. Keep the order; do not combine numbers. Don't use any brackets. Use all four operators at least once.

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  • $\begingroup$ "do not add [...] numbers" Surely you mean "concatenate", not "adding" (using the + symbol) $\endgroup$ – Lolgast Jan 11 '18 at 14:40
  • $\begingroup$ Sorry for my bad English, it must be clear now, right? $\endgroup$ – Jack Jan 11 '18 at 14:45
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    $\begingroup$ Yeah, that's better. $\endgroup$ – Lolgast Jan 11 '18 at 14:46
  • $\begingroup$ I think your intent is clear enough, but "do not combine numbers" prohibits the use of multiplication, addition, etc. @Lolgast's suggestion to use the word concatenate is much better. $\endgroup$ – Lawrence Jan 12 '18 at 9:39
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I wrote a Python script and made sure these are the only solutions:

$10*9*8*7/6/5*4*3+2*1 = 2018$
$10*9*8*7/6/5*4*3+2/1 = 2018$

Here's the Python code I wrote to solve this puzzle:

#!/usr/bin/python3

o=["+","-","*","/"]
op=[0]*9

def con():
    n = ""
    for i in range(10, 0, -1):
        n = n + str(i) + (str(o[op[10-i]]) if i > 1 else "")
    return n

def f(n):
    if n == 9:
        s = con()
        if eval(s) == 2018:
            print(s + " = " + str(eval(s)))
    else:
        for i in range(4):
            op[n] = i
            f(n+1)

f(0)

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  • $\begingroup$ That's good. What's rationale on like these problems? $\endgroup$ – Jack Jan 11 '18 at 14:51
  • $\begingroup$ Nicely done - not sure you need the .0 though $\endgroup$ – CalvT Jan 11 '18 at 14:52
  • $\begingroup$ @CalvT븃 The codes are outputs from the script. $\endgroup$ – iBug Jan 11 '18 at 14:55
  • $\begingroup$ @iBug I doubt hiding the code is necessary - it does not immediately show any solutions, at most the way you could arrive at them, but it would also take a bit of effort understanding it. If people want to do that, they wouldn't really mind a spoiler block either. Also, you should be able to put code in there using regular spoiler blocks and the pre and /pre tags at the start and end. $\endgroup$ – Lolgast Jan 11 '18 at 14:56
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    $\begingroup$ If these answers are the only two that exist, then there are no solutions for 2018 because you haven't used the minus symbol at least once, failing the "Use all four operators at least once" criterion. $\endgroup$ – John Kossa Jan 12 '18 at 23:38
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To answer Bachrach44's comment on iBug's answer, here are all the solutions my script has found so far for the years 2000 onward:

10*9*8*7/6/5*4*3-2-1=2013

10*9*8*7/6/5*4*3-2*1=2014
10*9*8*7/6/5*4*3-2/1=2014

10*9*8*7/6/5*4*3-2+1=2015

10*9*8*7/6/5*4*3+2-1=2017

10*9*8*7/6/5*4*3+2*1=2018
10*9*8*7/6/5*4*3+2/1=2018

10*9*8*7/6/5*4*3+2+1=2019

10-9+8*7*6*5*4/3-2-1=2238

10-9+8*7*6*5*4/3-2*1=2239
10-9+8*7*6*5*4/3-2/1=2239

10-9+8*7*6*5*4/3-2+1=2240

10-9+8*7*6*5*4/3+2-1=2242

10-9+8*7*6*5*4/3+2*1=2243
10-9+8*7*6*5*4/3+2/1=2243

10-9+8*7*6*5*4/3+2+1=2244

10+9+8*7*6*5*4/3-2-1=2256

10+9+8*7*6*5*4/3-2*1=2257
10+9+8*7*6*5*4/3-2/1=2257

10+9+8*7*6*5*4/3-2+1=2258

10+9+8*7*6*5*4/3+2-1=2260

10+9+8*7*6*5*4/3+2*1=2261
10+9+8*7*6*5*4/3+2/1=2261

10+9+8*7*6*5*4/3+2+1=2262

10*9*8*7*6/5/4*3/2-1=2267
10*9/8*7*6/5*4*3*2-1=2267

10*9*8*7*6/5/4*3/2*1=2268
10*9*8*7*6/5/4*3/2/1=2268
10*9/8*7*6/5*4*3*2*1=2268
10*9/8*7*6/5*4*3*2/1=2268

10*9*8*7*6/5/4*3/2+1=2269
10*9/8*7*6/5*4*3*2+1=2269

10*9+8*7*6*5*4/3-2-1=2327

10*9+8*7*6*5*4/3-2*1=2328
10*9+8*7*6*5*4/3-2/1=2328

10*9+8*7*6*5*4/3-2+1=2329

10*9+8*7*6*5*4/3+2-1=2331

10*9+8*7*6*5*4/3+2*1=2332
10*9+8*7*6*5*4/3+2/1=2332

10*9+8*7*6*5*4/3+2+1=2333

10*9/8*7*6*5-4-3/2-1=2356

10*9/8*7*6*5-4-3/2*1=2357
10*9/8*7*6*5-4-3/2/1=2357

10*9/8*7*6*5-4-3/2+1=2358

10*9/8*7*6*5-4+3/2-1=2359

10*9/8*7*6*5-4+3/2*1=2360
10*9/8*7*6*5-4+3/2/1=2360

10*9/8*7*6*5-4+3/2+1=2361

10*9/8*7*6*5+4-3/2-1=2364

10*9/8*7*6*5+4-3/2*1=2365
10*9/8*7*6*5+4-3/2/1=2365

10*9/8*7*6*5+4-3/2+1=2366

10*9/8*7*6*5+4+3/2-1=2367

10*9/8*7*6*5+4+3/2*1=2368
10*9/8*7*6*5+4+3/2/1=2368

10*9/8*7*6*5+4+3/2+1=2369

10-9*8+7*6*5*4*3-2-1=2455

10-9*8+7*6*5*4*3-2*1=2456
10-9*8+7*6*5*4*3-2/1=2456

10-9*8+7*6*5*4*3-2+1=2457

10-9*8+7*6*5*4*3+2-1=2459

10-9*8+7*6*5*4*3+2*1=2460
10-9*8+7*6*5*4*3+2/1=2460

10-9*8+7*6*5*4*3+2+1=2461

10-9-8+7*6*5*4*3-2-1=2510

10-9-8+7*6*5*4*3-2*1=2511
10-9-8+7*6*5*4*3-2/1=2511

10-9-8+7*6*5*4*3-2+1=2512

10-9-8+7*6*5*4*3+2-1=2514

10-9-8+7*6*5*4*3+2*1=2515
10-9-8+7*6*5*4*3+2/1=2515

10-9-8+7*6*5*4*3+2+1=2516

10-9+8*7*6*5/4*3*2-1=2520

10-9+8*7*6*5/4*3*2*1=2521
10-9+8*7*6*5/4*3*2/1=2521

10-9+8*7*6*5/4*3*2+1=2522

10-9+8+7*6*5*4*3-2-1=2526

10-9+8+7*6*5*4*3-2*1=2527
10-9+8+7*6*5*4*3-2/1=2527

10+9-8+7*6*5*4*3-2-1=2528
10-9+8+7*6*5*4*3-2+1=2528

10+9-8+7*6*5*4*3-2*1=2529
10+9-8+7*6*5*4*3-2/1=2529
10+9*8*7*6*5/4/3*2-1=2529
10+9*8*7/6*5*4*3/2-1=2529

10+9-8+7*6*5*4*3-2+1=2530
10+9*8*7*6*5/4/3*2*1=2530
10+9*8*7*6*5/4/3*2/1=2530
10+9*8*7/6*5*4*3/2*1=2530
10+9*8*7/6*5*4*3/2/1=2530
10-9+8+7*6*5*4*3+2-1=2530

10+9*8*7*6*5/4/3*2+1=2531
10+9*8*7/6*5*4*3/2+1=2531
10-9+8+7*6*5*4*3+2*1=2531
10-9+8+7*6*5*4*3+2/1=2531

10+9-8+7*6*5*4*3+2-1=2532
10-9+8+7*6*5*4*3+2+1=2532

10+9-8+7*6*5*4*3+2*1=2533
10+9-8+7*6*5*4*3+2/1=2533

10+9-8+7*6*5*4*3+2+1=2534

10+9+8*7*6*5/4*3*2-1=2538

10+9+8*7*6*5/4*3*2*1=2539
10+9+8*7*6*5/4*3*2/1=2539

10+9+8*7*6*5/4*3*2+1=2540

10+9+8+7*6*5*4*3-2-1=2544

10+9+8+7*6*5*4*3-2*1=2545
10+9+8+7*6*5*4*3-2/1=2545

10+9+8+7*6*5*4*3-2+1=2546

10+9+8+7*6*5*4*3+2-1=2548

10+9+8+7*6*5*4*3+2*1=2549
10+9+8+7*6*5*4*3+2/1=2549

10+9+8+7*6*5*4*3+2+1=2550

10+9*8+7*6*5*4*3-2-1=2599
10*9-8+7*6*5*4*3-2-1=2599

10+9*8+7*6*5*4*3-2*1=2600
10+9*8+7*6*5*4*3-2/1=2600
10*9-8+7*6*5*4*3-2*1=2600
10*9-8+7*6*5*4*3-2/1=2600

10+9*8+7*6*5*4*3-2+1=2601
10*9-8+7*6*5*4*3-2+1=2601

10+9*8+7*6*5*4*3+2-1=2603
10*9-8+7*6*5*4*3+2-1=2603

10+9*8+7*6*5*4*3+2*1=2604
10+9*8+7*6*5*4*3+2/1=2604
10*9-8+7*6*5*4*3+2*1=2604
10*9-8+7*6*5*4*3+2/1=2604

10+9*8+7*6*5*4*3+2+1=2605
10*9-8+7*6*5*4*3+2+1=2605

10*9+8*7*6*5/4*3*2-1=2609

10*9+8*7*6*5/4*3*2*1=2610
10*9+8*7*6*5/4*3*2/1=2610

10*9+8*7*6*5/4*3*2+1=2611

10*9+8+7*6*5*4*3-2-1=2615

10*9+8+7*6*5*4*3-2*1=2616
10*9+8+7*6*5*4*3-2/1=2616

10*9+8+7*6*5*4*3-2+1=2617

10*9+8+7*6*5*4*3+2-1=2619

10*9+8+7*6*5*4*3+2*1=2620
10*9+8+7*6*5*4*3+2/1=2620

10*9+8+7*6*5*4*3+2+1=2621

Note that they always come in groups of 3 with the middle one having twice as many possibilities, which makes sense: You'll always end in either -1, *1, /1 or +1, which result in x-1, x, x and x+1, respectively. These are (should be at least) all the solutions up to the year 2710, I guess that'll do for a while. Based on these results, we could get a question like this next year, then not get any for over 2 centuries, and then we should prepare for a truckload of these questions.

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1
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The solution is

$10*9*8*7/6/5*4*3+2*1$

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  • $\begingroup$ The solution, or A solution? Did you prove it's the only one? (just curious) $\endgroup$ – Florian Bourse Jan 11 '18 at 14:46
  • $\begingroup$ Is there a solution with all four operators using at least once? (I mean using "-") $\endgroup$ – Jack Jan 11 '18 at 14:46
  • $\begingroup$ Since we're not allowed to use brackets, i don't think there is a solution using all four operators. $\endgroup$ – Alix Eisenhardt Jan 11 '18 at 14:51
  • $\begingroup$ @Jack See my answer, it's likely not. $\endgroup$ – iBug Jan 11 '18 at 14:51
-1
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For 2019 it's:

10∗9∗8∗7/6/5∗4∗3+2+1

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  • 1
    $\begingroup$ Welcome to Puzzling! This was not what was asked, but maybe you want to post a new question with this as a solution? $\endgroup$ – Glorfindel Jan 6 at 17:58

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