-1
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In the picture below? Once each i guess. I think it's a trick and is not necessarily something that adds up to 9 on the left column. Idk. Haha.

Any hints please?

enter image description here

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3
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[EDIT]
Possible strategy other than brute force.
This is not possible with 100% logical deduction so you will have to make assumptions at one point.
Let's transform this first into an addition because it's easier to explain.
So the equation above translates to

ABC + DEF = 9GH.

pick 2 digits (A & D) that add up to 9.
The position is not important because in addition they are obviously interchangeable.
Let's pick 1 & 8, the obvious case.

Now you have 1BC+8EF=9GH.
Since there is no carry over we can reduce this to BC+EF = GH. and the available digits 2,3,4,5,6,7.
Then try to split them in 2 groups so that 2 digits from the group add up to the third one. This is the simple case, without carry over.
I wasn't able to find such groups.
You can try to find a combination with a carry over.
Then move to the second case. Pick 7 and 2. And then 6 & 3.

After you are done with this, you can consider the case with a carry over from the tenths position. So, as you said, pick 2 numbers that add up to 8. Then move to the equation BC+EF = GH with the remaining digits. [/EDIT]

Here is the brute force approach.
I wrote some PHP code. Ignore the way it looks. It was done fast with stackoverflow help for permutations.

<?php

function pc_permute($items, $perms = array( )) {
    if (empty($items)) {
        $return = array($perms);
    }  else {
        $return = array();
        for ($i = count($items) - 1; $i >= 0; --$i) {
             $newitems = $items;
             $newperms = $perms;
         list($foo) = array_splice($newitems, $i, 1);
             array_unshift($newperms, $foo);
             $return = array_merge($return, pc_permute($newitems, $newperms));
         }
    }
    return $return;
}
$array = array(1, 2, 3, 4, 5, 6, 7, 8);
$permutations = pc_permute($array);
foreach ($permutations as $p)
{
    $one = (int)('9'.$p[0].$p[1]);
    $two = (int)($p[2].$p[3].$p[4]);
    $diff = (int)($p[5].$p[6].$p[7]);
    if ($one - $two == $diff) {
        echo '&nbsp;'.$one.'<br />-'.$two.'<br />----<br />='.$diff.'<br /><br />';
    }
}
?>

This prints out 96 combinations.

 972
-314
----
=658

 972
-614
----
=358

 963
-215
----
=748

 963
-715
----
=248

 954
-216
----
=738

 954
-716
----
=238

 945
-317
----
=628

 945
-617
----
=328

 954
-236
----
=718

 963
-245
----
=718

 945
-327
----
=618

 972
-354
----
=618

 945
-627
----
=318

 972
-654
----
=318

 954
-736
----
=218

 963
-745
----
=218

 981
-324
----
=657

 981
-624
----
=357

 945
-318
----
=627

 981
-354
----
=627

 945
-618
----
=327

 981
-654
----
=327

 945
-328
----
=617

 945
-628
----
=317

 927
-341
----
=586

 927
-541
----
=386

 918
-342
----
=576

 918
-542
----
=376

 981
-235
----
=746

 918
-372
----
=546

 927
-381
----
=546

 918
-572
----
=346

 927
-581
----
=346

 981
-735
----
=246

 954
-218
----
=736

 981
-245
----
=736

 954
-718
----
=236

 981
-745
----
=236

 954
-238
----
=716

 954
-738
----
=216

 918
-243
----
=675

 918
-643
----
=275

 963
-218
----
=745

 981
-236
----
=745

 918
-273
----
=645

 963
-718
----
=245

 981
-736
----
=245

 918
-673
----
=245

 981
-246
----
=735

 981
-746
----
=235

 963
-248
----
=715

 963
-748
----
=215

 936
-152
----
=784

 936
-752
----
=184

 936
-182
----
=754

 972
-318
----
=654

 981
-327
----
=654

 972
-618
----
=354

 981
-627
----
=354

 936
-782
----
=154

 981
-357
----
=624

 981
-657
----
=324

 972
-358
----
=614

 972
-658
----
=314

 945
-162
----
=783

 954
-271
----
=683

 954
-671
----
=283

 945
-762
----
=183

 918
-245
----
=673

 954
-281
----
=673

 918
-645
----
=273

 954
-681
----
=273

 945
-182
----
=763

 945
-782
----
=163

 918
-275
----
=643

 918
-675
----
=243

 936
-154
----
=782

 945
-163
----
=782

 936
-754
----
=182

 945
-763
----
=182

 918
-346
----
=572

 918
-546
----
=372

 945
-183
----
=762

 945
-783
----
=162

 936
-184
----
=752

 936
-784
----
=152

 918
-376
----
=542

 918
-576
----
=342

 954
-273
----
=681

 927
-346
----
=581

 927
-546
----
=381

 954
-673
----
=281

 954
-283
----
=671

 954
-683
----
=271

 927
-386
----
=541

 927
-586
----
=341
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  • $\begingroup$ Any suggestions on how to come up with an answer 1. by hand 2. other than brute force? What do you think of my suggestion? $\endgroup$ – BCLC Apr 11 '17 at 12:37
  • 1
    $\begingroup$ I've edited the answer with suggestions on where to start. $\endgroup$ – Marius Apr 11 '17 at 12:52
0
$\begingroup$

It's

936-152=784

Any alternatives? Lazy to use excel/Google sheets. Not necessarily teaching logic anyhoo. If ever, we don't know if it's some hindsight bias or something idk.

How i got my answer:

Pick some pair that

adds up to 8

and not

9.

Let's say

1 and 7.

(The above pair has the properties described in hidden text prior)

If we're going to

borrow

then box (1,2) must be

smaller than box (2,2). If they respectively are 3 and 5 then box (3,2) is 8; If 2 and 5, then 7

which is not allowed and so on. Luckily the first pair works.

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  • 5
    $\begingroup$ Welcome to the site. Firstly, you're not really supposed to answer your own puzzles, especially five minutes after you post them. Secondly, you should put your answer in a spoiler tag in case anyone else wants to try solving it themselves. $\endgroup$ – F1Krazy Apr 11 '17 at 9:10

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