I met this question yesterday while solving an I.Q test. It took me around 1 hour to come up with a solution that I'm not certain is the correct one.

Alright, without further ado, here's the puzzle:

enter image description here

What is the number that you should put instead of the dot?



As you can see, every row is divided into 1 big number on the left, followed by four 1-digit numbers on the right.

So my idea is that if you take the sum of the first and second small numbers, then you take the sum of the third and fourth small numbers, and then you take the sum of the digits of both of the sums, then the result will be equal to the sum of the digits of the big number.

For example, take the third row:

The big number: 153

The four 1-digit numbers: 2, 1, 8, 7

The sum of 1st and 2nd = 2 + 1 = 3

The sum of 3rd and 4th = 8 + 7 = 15

The sum of the digits of both of the sums = 3 + 1 + 5 = 9

The sum of the digits of the big number = 1 + 5 + 3 = 9

However, this works for all rows, except the first one at the top. That's the closest I could get to the solution. Should a person find the complete solution, I'll be extremely grateful.

NOTE: it is entirely possible that the person who wrote this puzzle is playing "mind-games" with us, and that he included the first row just to deceive us, thus making the question even harder. But let's not make any assumptions here.

It is also worth noting that the choices that came with the puzzle were as follows:

A) 110

B) 128

C) 164

D) 92

Maybe the idea of the question is to choose the most different choice. the sum of the digits of all of these choices is 11, except for 110:

1 + 2 + 8 = 1 + 6 + 4 = 9 + 2 = 11

1 + 1 + 0 = 2

  • 1
    $\begingroup$ You should provide the source of the puzzle, both to give creator credit and to help the reader make a determination about whether there are "mind-games" at play. If this is a copy of a copy of a copy of a puzzle, it's possible that someone transcribed the first row incorrectly along the way. $\endgroup$ – Blue Jul 24 '19 at 23:59
  • $\begingroup$ well, yes, it is a copy of a copy. I don't know the original writer of this puzzle. The person who gave me this was gathering a lot of I.Q questions from over the internet and putting them together. So I assume the question he found over the internet was also the result of similar gathering. It is possible that someone transcribed the first row incorrectly along the way. $\endgroup$ – Saif Taher Jul 25 '19 at 9:18

Operating under the assumption that we are being deceived by the puzzle maker, i.e. the first row is a ruse, the answer is fairly simple. Take the sum of the last two columns, multiply them by $10$, then add the sum of the first two columns to that number. In row $2$, for example:

$$5+7\rightarrow 120,9+3=12\rightarrow 120+12=132$$

This means, for our currently unsolved row, we'd get $B$, or $128$.

Can you elaborate why the puzzle maker might want to deceive us by giving us a faulty first row?

I'll update this if a more comprehensive solution comes to mind.

  • $\begingroup$ Indeed, I'm not sure about the intentions of the person who wrote this. It's just a possibility that I had in mind. Maybe the first row is not faulty. But if it isn't, I can't really see any solution to this; as that bloody first row always breaks any consistencies that I spot among the others. $\endgroup$ – Saif Taher Jul 24 '19 at 23:53
  • $\begingroup$ Putting that aside, I must say I'm really impressed by how quickly your response was. Well I guess I took that long time because I wanted to find a pattern which the first row didn't violate. Your solution is of course the more rigorous, as it is based on a formula; in contrary to mine, which was merely a vague pattern. $\endgroup$ – Saif Taher Jul 24 '19 at 23:55
  • $\begingroup$ I piggybacked off of your idea to split the numbers by two sets of columns, and then I calculated the columns you noted were less problematic and saw they worked out. Additional confidence given by the fact that I arrived $128$, which was a suggested answer, without looking at the possible answers. $\endgroup$ – scoopfaze Jul 24 '19 at 23:59
  • $\begingroup$ It was suggested by @Blue in a comment above that someone transcribed the first row incorrectly from the original puzzle, which seems very likely, given that if you replace the 9 in the first column by a 2, and reverse the row, your formula will work out the answer correctly $\endgroup$ – Saif Taher Jul 29 '19 at 8:23

If you assume a linear model $y=\sum_{j=1}^4 \beta_j x_j$ with no intercept and ignore the first row, you get a perfect fit $\beta=(1, 1, 10, 10)$, as others have proposed, yielding 128.

If you assume a linear model $y=\beta_0+\sum_{j=1}^4 \beta_j x_j$ with intercept and include the first row, you get a perfect fit $\beta=(744, 56, -91, 252, 171)/26$, yielding $3340/26=128.462$. Close enough?

  • $\begingroup$ In such a question, I'm afraid not. You have to hit the answer clean. $\endgroup$ – Saif Taher Jul 25 '19 at 9:14
  • 2
    $\begingroup$ As clean as the scan. :) $\endgroup$ – RobPratt Jul 25 '19 at 10:19

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