5
$\begingroup$

Can anyone solve this puzzle? I have tried many times but I could not find any symmetry or sequence.

puzzle

\begin{array} {|c|c|c|} \hline 3&25&5\\ \hline 16&9&6\\ \hline 125&36&18\\ \hline 81&49&??\\ \hline \end{array}

Options:
A: 14
B: 16
C: 19
D: 21

$\endgroup$

closed as off-topic by Rubio Feb 28 '18 at 1:49

  • This question does not appear to be about creation and solving of puzzles, within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 4
    $\begingroup$ Where is the puzzle? $\endgroup$ – Quintec Feb 27 '18 at 15:12
  • 3
    $\begingroup$ @shyam acharya where is the puzzle from? $\endgroup$ – puzzledPig Feb 27 '18 at 15:23
  • 12
    $\begingroup$ Is the puzzle deliberately skewed? Or should it just be a standard matrix? $\endgroup$ – stuart stevenson Feb 27 '18 at 15:58
  • 1
    $\begingroup$ I'm putting this question on hold until proper attribution of its original source is provided. This looks like you're asking us to solve a puzzle you found elsewhere. For content that you did not create yourself, please provide attribution - at minimum you need to let us know where this came from, and any additional context you can provide is usually a big help to solvers. Posts which use someone else's content without disclosing where it came from are generally deleted. $\endgroup$ – Rubio Feb 28 '18 at 1:49
  • $\begingroup$ @puzzledPig This is IQ question which have been asked in the bank entry exam last year. $\endgroup$ – shyam acharya Feb 28 '18 at 2:47
16
$\begingroup$

I think it's

14

Why?

$3^1$ | $5^2$ | $1 \cdot 5 = 5$
$4^2$ | $3^2$ | $2 \cdot 3 = 6$
$5^3$ | $6^2$ | $3 \cdot 6 = 18$
$9^2$ | $7^2$ | $2 \cdot 7 = 14$
(I'm taking the exponent of the first column and multiply it with the base of the second column)

Of course,

the last row could be $3^4$ | $7^2$ | $4 \cdot 7 = 28$ but that wasn't one of the possible options ...

$\endgroup$
  • 2
    $\begingroup$ Your solution is also slightly troubling in explaining why $16 = 4^2$ rather than $2^4$ or $16^1$... $\endgroup$ – Chris Feb 27 '18 at 17:26
  • $\begingroup$ True. I tried to go for the solution with the least complicated rules, Occam's Razor style. $\endgroup$ – Glorfindel Feb 27 '18 at 19:41
  • $\begingroup$ Opposing the believers of the half-divine $21$, we can arrive at 14 by taking $\min_\text{columns}\left(\text{#prime factors}\right) \cdot \sqrt{\text{second row}}$. $\endgroup$ – Carl Löndahl Feb 27 '18 at 20:05
  • $\begingroup$ @glorfindel: the trouble is you can't actually state a rule that explains your logic. Your way of writing that first value is key to your logic but you have no rule to explain why you write it the way you do - thus to my mind your logic is incomplete. Oceans razor is all very good but you said yourself the only reason it works is because one of the two values you got wasn't an option. If it wasn't multiple choice how would you have decided on the correct answer? $\endgroup$ – Chris Feb 27 '18 at 22:52
4
$\begingroup$

I think it is

21

Because

It is almost the same reasoning as Glorfindel but a little bit different. We look at the first line, it has $3^1|5^2|5$ and we know that $5=1\cdot 5$ where $1=\min(3,1)$. The second line has $2^4|3^2|6$ and $6=2\cdot 3$ where $2=\min(4,2)$. Now third line has $5^3|6^2|18$ and $18=3\cdot 6$ where $3=\min(5,3)$. Now for the last line $3^4|7^2|?$. so $?=7\cdot \min(3,4) = 7\cdot 3 = 21$

An important comment

It is also true that we are taking the maximum between the number and it's power in the second column. (e.g the first line $3^1|5^2| 5 = \min(1,3)\cdot \max(5,2)$) - Credit to Chris.

$\endgroup$
  • $\begingroup$ To add to the symmetry the second column uses the max which makes this answer slightly nicer. Also you have typoed the second value in the first row when quoting it. $\endgroup$ – Chris Feb 27 '18 at 17:25
3
$\begingroup$

I think it's

$21$

First observe

the prime factorization of each number.

Row 1

$3$ | $5 \cdot 5$ | $5$

Row 2

$2 \cdot 2 \cdot 2 \cdot 2$ | $3 \cdot 3$ | $2 \cdot 3$

Row 3

$5 \cdot 5 \cdot 5$ | $2 \cdot 3 \cdot 2 \cdot 3$ | $2 \cdot 3 \cdot 3$

Row 4

$3 \cdot 3 \cdot 3 \cdot 3$ | $7 \cdot 7$ | ?

Here's the pattern I noticed

The count of factors in the third column is the min of the count of factors of the first $2$ columns. The factors in the third are factors that appear in the first $2$ columns. So that means the final number will have $2$ factors and consist of $3$'s and $7$'s. So: $9$, $21$ or $49$. From the options given, it must be $21$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.