# Sum other numbers

Begin with a flagrantly erroneous summation and a woefully vacant substitution table.


234
+   5                   Digit        2    3    4    5    6    7    8
-------           Substitute digit    _    _    _    _    _    _    _
5678



How can the substitution table be filled out to correct this summation?

This is almost too easy if you just follow these guidelines.

• Assign 7 unique substitute digits from 0 through 9 for digits 2 through 8 in the table (one digit per digit)

• Replace digits in the summation by their substitutes in the table (no other kinds of edits, as the summation and table should be taken at face value)

• All numbers and digits are decimal (no notation tricks are involved)

• No leading zeros in the total or either summand

• The summation has a unique solution

Added:   Regular pretty much forces the resultant summation. allows the guidelines to attain it.

• Should all numbers be positive integers? – Techidiot Nov 1 '16 at 8:40
• All positive integers, @Techidiot, only digits are being substituted (ohhh, guess you're used to someone sounding impolite when addressing you by name) – humn Nov 1 '16 at 8:41
• Haha :D No problems ("Apparently, this user prefers to keep an air of mystery about them.") :p BTW, this looks like more of a lateral thinking than arithmetic :) – Techidiot Nov 1 '16 at 8:47
• One final question, by no leading zeros I assume you mean, no leading zeros in the number being added or resulted right? – Techidiot Nov 1 '16 at 9:15
• If Ryan27 has the right answer, then wow, that's clever. +1. – Deusovi Nov 1 '16 at 17:13

Making an assumption:-

That if a substitute digit is itself in the lookup table, it will be replaced again.

 Digit               2    3    4    5    6    7    8
Substitute digit    3    4    9    1    7    8    0 

The Summation becomes:

999 + 1 = 1000 because:
2->3->4->9,
3->4->9,
4->9,
5->1,
5->1,
6->7->8->0,
7->8->0,
8->0

Process:

First, as mentioned in the question, there is one possible summation. It must be 999 + 1 = 1000 as a 3 digit number plus a 1 digit number must equal a 4 digit number, and the first digit of the 4 digit number has to be the same as the 1 digit number.

Then, knowing that 6,7,8 must equal 0 we can first assign any one of those digits the substitute digit of zero, lets choose 8.

Since 0 is now used (and the question states the substitute digits must be unique) in order for 6 or 7 to equal 0, the only substitute digit we can assign is 8 (since 8 = 0).

This same logic is then applied to 2,3,4 since they all need to equal 9

• +1 for this. I think this should be the answer. Recursive replacement. Let me try to edit. – Techidiot Nov 1 '16 at 12:09
• Added an explanation @humn. Thanks for the fun puzzle. – Ryan27 Nov 1 '16 at 19:17
• Thank you for making the puzzle complete – humn Nov 1 '16 at 19:19

 My first guess is as we are adding a 1-digit number to a 3-digit number,
and the result is a 4 digit number, the first digit of the result must me a 1.
So 5 -> 1
There is only one 3-digit number that becomes a 4-digit number when you add 1 to it.
The number 999. So 234 stands for 999.
So 2 -> 9
And 3 -> 9
And 4 -> 9
And since the result is 1000, 5678 stands for 1000.
So 6 -> 0
And 7 -> 0
And 8 -> 0
So in the end, the table is:
2    3    4    5    6    7    8
9    9    9    1    0    0    0
Which gives the summation:
999
+   1
-----
1000


• You are not using unique digits in the subs table. Check other two answers -> With unique numbers and without them. There is nothing new here I guess. – Techidiot Nov 1 '16 at 15:09
• +1 from me. This is the exact logic that I used when working through the puzzle. Just 1 lateral-thinking step from here and you arrive at the same answer I did. – Ryan27 Nov 1 '16 at 15:13

Only thing I could think of (lateral)(Before the edits to original question) ->

• Well, if I didn't make a mistake, there are a few more combinations with this. It at least found yours (1,8,4,6,2,5,0), so I hope it's correct. If it is, there is no way this is the intended solution. – user14478 Nov 1 '16 at 9:38
• ^vote with a note: This is what makes the lateral-thinking tag so much fun. You could almost argue that   (space) is indeed in the substitution table. – humn Nov 1 '16 at 9:38
• @humn Will take some practice to figure out those kind of things to argue ;) – Techidiot Nov 1 '16 at 9:40
• @LukasRotter - Yes, there are many. – Techidiot Nov 1 '16 at 9:41
• Then maybe @humn should edit the question clarifying that there is only one unique solution for this? Otherwise this might be too broad IMO. – user14478 Nov 1 '16 at 9:42

Well...

 999
+  1
----
1000


No one can say that I substituted more than one digit for each of the initial digits. Therefore, each of the initial digits has a unique substitute.

• That way of thinking allows many possible answers though. – stack reader Nov 1 '16 at 11:29
• @stackreader Does it? No leading zeros and unique substitution for 5. – Verence Nov 1 '16 at 11:29
• Sorry then, I'm afraid I don't understand your explanation. Maybe it's just me. – stack reader Nov 1 '16 at 11:33
• ^vote with a note: This solution highlights 3 "unique" meanings of "unique": i) single (not what is meant in the puzzle statement); ii) different from others (as is meant in the puzzle statement); iii) light-heartedly ironic (as is meant by "3 "unique" meanings" in this statement). By the way, the ^vote is also for a part of the puzzle solved correctly here. – humn Nov 1 '16 at 11:36