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

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

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#1scriptsizeraise.4ex)} ~{ {large #2} [.5ex] { large #3} } } def box #1#2{ kern.2em raise.7ex{bbox[4pt,border:2pt solid]{kern#1emtinystrut}} llap{sflarge #2kern.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 {2s+6}s + ggg {5s+2}s {=} ,gray {17} } { ggg{1s+1} s + bbb{2s+5} s + ggg{4s+4} s{=} ,gray{17} } $

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

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

         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. – $smallcolor{#3366ff}{textsf{Oleg}}$ $smallcolor{#8888ff}{textsf{Sep 30 '20 at 22:02}}$

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

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

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.

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.

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

.

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

I think it's simple as:

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