3
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7 # 4 = 5

1 # 1 = 2

3 # 2 = 5

4 # 3 = ?

Possible answers: 1, 2, 3, 4 or 5

In this puzzle, I tried to take square of each number and subtracting them or adding them, for instance 49 + 16 = 65 (13 * 5), but each time my result or connection of numbers doesn't suit others. I hope you can solve it. Thanks. (source: probably Kuark publications, YOS questions.I have seen this questions on another source, tried to solve myself.)

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  • $\begingroup$ This type of puzzle has an infinite number of solutions. I can always fit a polynomial to agree with both the examples and any of the "possible" answers. $\endgroup$
    – Galen
    Apr 26, 2020 at 17:44
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    $\begingroup$ might be on mod 6 sum $\endgroup$
    – Oray
    Apr 26, 2020 at 17:55
  • $\begingroup$ @Galen, thank you for enlightening me but where can I post these types of questions? I have a lot of them which I spend half an hour on each question and I still can't solve. Which site of Stack Exchange fits to these questions? I really don't know. First, I thought about Mathematics, then I published them here. You mean there are a lot of solutions to them, but these are the questions asked by universities in Turkey. If you have solution, please provide one then, I will gladly upvote them. Thanks again. $\endgroup$
    – garakchy
    Apr 26, 2020 at 19:13
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    $\begingroup$ but these are the questions asked by universities in Turkey. Wow :o I wonder what they're testing; ability to find some pattern maybe. $\endgroup$
    – Galen
    Apr 26, 2020 at 19:16
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    $\begingroup$ Which site of Stack Exchange fits to these questions? To be fair, Puzzling.SE might be the best choice, but as puzzles go, they're problematic in not having a finite set of solutions. $\endgroup$
    – Galen
    Apr 26, 2020 at 19:18

2 Answers 2

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Consider Oray’s answer in the comments:

x # y = z, where z = x + y (mod 6).

Thus:

7 # 4 => 7+4=11, 11 = 5 mod 6
1 # 1 => 1+1=2, 2 = 2 mod 6
3 # 2 => 3+2=5, 5 = 5 mod 6
4 # 3 => 4+3=7, 7 = 1 mod 6.

So the answer is

1.

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  • $\begingroup$ Good point. Your answer is logical but let me wait if another versions of the answer come, too. Thank you. $\endgroup$
    – garakchy
    Apr 26, 2020 at 19:01
  • $\begingroup$ I'm accepting your answer as correct one. Thank you, so much. I have a lot of these type of questions, so if you have solution to my other questions, please provide them. Thanks again. $\endgroup$
    – garakchy
    Apr 26, 2020 at 19:16
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Alternative answer

A # B = Number of consonants in the spelling of A and B

Examples

7 # 4 = Seven # Four = 5
1 # 1 = One # One = 2
3 # 2 = Three # Two = 5

So the answer is

4 # 3 = Four # Three = 5

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1
  • 1
    $\begingroup$ This is a really clever answer as well! Nice work. $\endgroup$
    – El-Guest
    Apr 26, 2020 at 21:07

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