# Logical puzzle from a math homework for Russian fifth-year school students

The daughter of a relative of mine, a fifth-year school student in Russia, got the following math homework from her teacher: Here's my translation:

No. 26. Students were solving a task in which they had to figure out the missing numbers:

(DIAGRAM)

They came up with different answers:

(THREE DIAGRAMS)

Find the rules used by the students to fill out the squares, and come up with a fourth solution.

Explaining the first two solutions is easy. In the first solution, the student assumed that in each horizontal line, the sum of the first two numbers equals the third one. In the second solution, the student assumed that the difference between the numbers in each vertical row is the same.

The girl couldn't explain the third diagram and asked her parents, who then asked me. Unable to help them, I'm posting this question.

My question: Is there a logical explanation of the third solution?

Looking at the third solution, I noticed that the numbers in each horizontal line add up to the same number, 80, but how could one get 80 from the original diagram in the first place? Also, the digits in each horizontal line add up to the same number, 17, but, again, how could one get 17 in the first place? It seems I stumbled upon false leads.

Being a professional physicist, I feel ashamed to tell the girl's parents that I can't solve the problem. I find it likely that the teacher made a mistake in the third diagram, but I'm afraid to bet on that possibility, so I hope you can suggest a valid logical explanation of the third diagram and thereby help me save my face.

• With your observation, there are multiple solutions, right? You can keep changing 2 and 25 by adding 1 to each. So 3,26 , 4,27 , 5,28 etc should work same way! So one can guess any such combination. – DrD Sep 29 '20 at 20:32
• A very simple solution is rot 13 (Guvegrra va gur gbc ebj naq gjragl gjb va gur obggbz!) – DrD Sep 29 '20 at 20:34
• @DrD That's precisely the issue. There's no logic in preferring 2, 25 to 3, 26. That's why I discard the explanation that the numbers in each horizontal line add up to the same number. That is, the knowledge that the numbers in each horizontal line add up to the same number is not enough to fill out the squares. – Sandra Sep 29 '20 at 20:39
• The first two solutions do not require an arbitrary number like 80. – Sandra Sep 29 '20 at 20:54
• I would argue that this is not in fact a "logical" puzzle at all, so much as a creative one. – Tanaya Sep 30 '20 at 20:01

While it feels unlikely that the following is the intended explanation it kind of works. Perhaps enough for saving face?

One more observation one could make is that the products $$2 \times 26 \times 52$$ and $$11 \times 25 \times 44$$ are both perfect squares. Together with the constraint that the sums be the same this would be the smallest such solution.

Another very simple one which only suffers from slightly unconventional symmetry would be

That not only the total digit sums are the same but, in fact, they come in pairs: $$44 \sim 26$$,$$52 \sim 25$$ and $$11 \sim 2$$. Again, this would have to be paired with another constraint to make it unique.

Btw., I like this assignment because it makes students aware of the general idiocy of pretending there is a best (let alone unique) solution to this kind of problem.

• Welcome to Puzzling Stack Exchange! Click here puzzling.stackexchange.com/tour to get started! – user71418 Oct 2 '20 at 2:09
• I am upvoting to this comment to let you out of the unlucky 666 – user69881 Oct 3 '20 at 9:29
• Also because its really good answer – user69881 Oct 3 '20 at 9:29
• @TroyD Thank heavens... – Albert.Lang Oct 3 '20 at 10:41

I realize this may not be a particularly satisfying answer, but I think you're overthinking this. Remember this is a young child's homework problem. While it may not have been expressed particularly clearly, the goal is plainly not to find a single definitive answer, but to fill out the boxes using an identifiable consistent pattern.

As you surmised, Child 1 made the rows into addition problems. Child 2 made each column have a shared difference. Child 3 made the sum of the top row equal the sum of bottom row (with an arbitrary sum!). This is intended to show the test-taker what kind of answer they are seeking --not a logically unique answer, but a defensible one.

Given the numbers chosen, I would surmise that they wanted to give the test-taker an easy possibility for another possible pattern --to make each row a multiplication problem. (2 x 26 = 52; 11 x 4 = 44). You could also give them all a common product with 208 and 88.

• You mean each column as a multiplication – Dr Xorile Oct 2 '20 at 18:24
• @DrXorile - I definitely did not. How would that even work? Do you mean give them all a common product, by filling in 208 and 88? I guess that works too – Chris Sunami supports Monica Oct 2 '20 at 19:30

I very much like both of your explanations for #3, OP Sandra, especially the inventive second one with digit sums, as they fit well with the basic arithmetic natures of #1 and #2.

$$\begingroup \def \ans #1#2#3{ ~~~\raise1.3ex{\sf#1\scriptsize\raise.4ex)} ~{ {\large #2} \\[.5ex] { \large #3} } } \def \box #1#2{ \kern.2em \raise.7ex{\bbox[4pt,border:2pt solid]{\kern#1em\tiny\strut}} \llap{\sf\large #2\kern.9em} \kern-.2em } \def \gray #1{ \color{gray}{#1} } \def \ggg #1{ \box {1.9}{\gray{#1} } } \def \bbb #1{ \box {1.9}{ {#1} } } \def \gg #1{ \box 1{\gray{#1} } } \def \bb #1{ \box 1{ {#1} } } \def \g #1{ \box 1{\gray{#1}\kern.3em} } \def \b #1{ \box 1{ {#1}\kern.3em} } \def \s #1{ \gray{\raise.3ex{\normalsize \!\: {#1} \!\: }} } \ans{3}{ \g { 2}\s + \gg {26}\s + \gg {52}\s {=} \,\gray {80} } { \gg{11} \s + \bb{25} \s + \gg{44} \s{=} \,\gray{80} } \kern2em \ans{3}{ \ggg {2\,~~}\s + \ggg {2\s+6}\s + \ggg {5\s+2}\s {=} \,\gray {17} } { \ggg{1\s+1} \s + \bbb{2\s+5} \s + \ggg{4\s+4} \s{=} \,\gray{17} }$$

That an actual student might guess an arbitrary sum for either of these does seem plausible.

Then again,   $$\raise-1.4ex{\ans{3\!\:\raise-.06ex'}{\g { 2}\s\times \gg{26}\s {=}\gg {52}} { \gg{11} \s\times \b{ 4} \s{=} \gg{44}} } \endgroup$$

differs from #3 by just a single number that might after all be an erratum.

Indeed, the first example is on addition, the second is on subtraction, so it’s totally natural to have the third one on multiplication. – $$\small\color{#3366ff}{\textsf{Oleg}}$$ $$\small\color{#8888ff}{\textsf{Sep 30 '20 at 22:02}}$$

• I'm also 100% sure that that was just a typo, and they wanted to put there 4 rather than 25. Indeed, the first example is on addition, the second is on subtraction, so it's totally natural to have the third one on multiplication. – Oleg Sep 30 '20 at 22:02
• Your comment adds good reasoning, @Oleg , and i hope you don't mind its being copied into the answer. Please feel free to edit (or undo). – humn Oct 1 '20 at 2:42
• Rather than an erratum in printing, it could have been on the part of the student. I could see a young student with attention problems putting 2 in the top row thinking it is multiplication but not completing the bottom row before going on to other questions. When they come back to this one the top row is filled in, so they just make the sum of the bottom row the same. OP says "There's no logic in preferring 2, 25 to 3, 26" but a first thought of multiplication is a reason for preferring 2. – Dragonel Oct 1 '20 at 15:35
• @Dragonel I don't think there were any actual students... This is a problem for fifth graders, so it's just a backstory to make the problem more interesting. Same as you would write "Ed has \$1.50, how many apples can he buy if each one costs 7 cents" etc. – Ruslan Oct 1 '20 at 15:46
• My programmer brain saw the puzzle and instantly filled in 13 and 22. 13x2=26 and 26x2=52. 11x2=22 and 22x2=44. – Mooing Duck Oct 1 '20 at 18:02

Alternate 4th solution:

13    26    52
11    22    44

13=13*1    26=13*2    52=13*2*2
11=11*1    22=11*2    44=11*2*2


and the formula is

x*2^i
where i=0,1,2,3,4...
and x is prime


.

I think it's simple as:

2 + 26 + 52 = 80
11 + 25 + 44 = 80


On the bottom sets of six numbers each, the answers to 1) and 3) have been given. For 2), on the top row we have 19 + 26 + 52 = 97 and the bottom row 11 + 18 + 44 = 73. By subtraction, 97 - 73 = 24. We put 24 in the bottom row where the number is missing. By subtracting 19 - 18 we get 1. We put 1 in the top row where the number is missing. By addition we have 1 + 26 + 52 = 79. Also by addition of the bottom row we have 11 + 24 + 44 = 79

• This seems much less satisfying than the existing answer to 2: subtracting the bottom row from the top row yields 8, 8, and 8. – Misha Lavrov Sep 30 '20 at 16:10

as the system is very open to any kind of solution I think you can find a lot of fourth solution.

my rules for the third diagram: insert another row between:

 2  26  52
+9  -1  -8
11  25  44


so the sum of each row is constant but assigned differently by the delta in the inserted row.

in the same way I could complete the original diagram to the same sum in all column:

85 26 52
11 70 44


(each column gives 96)

or you want a system similar to the first solution:

78 =  26 + 52
11 = -33 + 44

• Why would you create another row out of thin air? That's not in line with the previous solutions. – bobble Oct 1 '20 at 19:18
• to explain the rule – Bernd Wilke πφ Oct 1 '20 at 19:48
• You could justify literally any rule by adding a row of your choosing in. The point is to make a rule that explains the given grid. – bobble Oct 1 '20 at 19:51