First of all, need to thank Ankita and Mohit again for the puzzle…wonderful one. Got lots and lots of answers, and as it turns out, there are quite a few answers. Proper treatment of the answer requires a video but unfortunately I do not have my iPad with me this weekend, and hence the answer is only in text.

Before we get to the answers, some special mentions:

– Alok Kuchlous for writing a program to provide me all the answers where 3 digits + 3 digits = 4 digits, and the requirements of the puzzle are met

– Anirudh Baddepudi for being the only one who gave an answer for 4 digits + 2 digits = 4 digits

– Anku Jain for being the first one to send an answer

– Mittal family (Alok, Parul, Smiti and Muskaan) for making this puzzle a family bonding event

– Sanjeev+Armaan Dugar, Ankita and Adhish are the other notable mentions with correct answers.

The question required a fair bit of trial and error, but still one could reason the following – Either x and y both need to be 3 digits each and z being 4 digits, or x & y can be 4 and 2 digits each (with z again as 4 digits) in a special scenario [explained below in Anirudh’s answer].

**4 digits + 2 digits = 4 digits (reproducing Anirudh’s answer):**

**1978+65=2043**

I found this puzzle quite intriguing and very challenging. I first realised that there has to be one number which is in the thousands in this, and automatically assumed that the thousands digit will be 1. This gave the the following digits left:

0, 2, 3, 4, 5, 6, 7, 8, 9. Then, I tried to play around with having two 3 digit numbers adding up to the four digit number, but it was quite frustrating realising that this doesn’t work. Hence, I tried something simpler, with having a four digit number as ‘x’ and a two digit number as ‘y’ in the equation. Therefore, 9 must be in the hundreds place in ‘x.’

Then, I decided to try make the problem simpler, trying it with having a sum of 2 ‘two’ digit numbers, such that its sum when added to 1900 with give me three different digits. For this, the sum of these two ‘2’ digit numbers must be more than a hundred. So looking at various sums, and rules with which the ending digits will be different (and with intuition that the two numbers will be consecutive or have a relationship like that as these questions usually will have an answer of that kind), I came up with this:

78+65=143. Plugging this in to the original form, I got:

**1978+65=2043.**

**3 digits + 3 digits = 4 digits (reproducing Alok Mittal’s answer)**

Here is some logic:

– The digit in the thousands place in z needs to be “1” as addition of two 3 digit numbers will be less than 2000 and since z is a four digit number it has to be at least 1000

– if zero is in one of the numbers being added, it can’t be at unit position (else the other unit number will repeat) or at hundreds position (else even with a carry, the result will be same as other hundreds place number). So it has to be in tens position. In such a case clearly the units sum has to produce a carry. Quickly one can run through cases here and there doesn’t seem to be a solution of this nature

– Zero is in the result number. The corresponding position numbers being added then have to produce a carry, and then its a hit and trail on those

Here is a short program that Alok Kuchlous wrote (which I cannot understand), and also all the answers:

————————– puzzle.rb ————————————–

numbers = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

numbers.permutation.each do |p|

x = p[0..2].join(”).to_i

y = p[3..5].join(”).to_i

z = p[6..9].join(”).to_i

puts “#{x.to_s} + #{y.to_s} = #{z.to_s}” if x < y && x + y == z

end

——————————————————————————–

> ruby puzzle.rb | grep 0

Solutions:

246 + 789 = 1035

249 + 786 = 1035

264 + 789 = 1053

269 + 784 = 1053

284 + 769 = 1053

286 + 749 = 1035

289 + 746 = 1035

289 + 764 = 1053

324 + 765 = 1089

325 + 764 = 1089

342 + 756 = 1098

346 + 752 = 1098

347 + 859 = 1206

349 + 857 = 1206

352 + 746 = 1098

356 + 742 = 1098

357 + 849 = 1206

359 + 847 = 1206

364 + 725 = 1089

365 + 724 = 1089

423 + 675 = 1098

425 + 673 = 1098

426 + 879 = 1305

429 + 876 = 1305

432 + 657 = 1089

437 + 589 = 1026

437 + 652 = 1089

439 + 587 = 1026

452 + 637 = 1089

457 + 632 = 1089

473 + 589 = 1062

473 + 625 = 1098

475 + 623 = 1098

476 + 829 = 1305

479 + 583 = 1062

479 + 826 = 1305

483 + 579 = 1062

487 + 539 = 1026

489 + 537 = 1026

489 + 573 = 1062

624 + 879 = 1503

629 + 874 = 1503

674 + 829 = 1503

679 + 824 = 1503

743 + 859 = 1602

749 + 853 = 1602

753 + 849 = 1602

759 + 843 = 1602

Hope you enjoyed the puzzle!