5
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Assume that the following three equations hold true:

  • $C \circ I \circ D \circ A = B$
  • $O \circ K \circ M \circ G = C$
  • $Y \circ A \circ B \circ C = D$

Find the right hand side of the following equation:

  • $S \circ W \circ A \circ G =~ ?$

Hints:

1.

Convert each letter to its corresponding numerical value.
E.g., $C \circ I \circ D \circ A = B$ is converted to $3 \circ 9 \circ 4 \circ 1 = 2$

2.

Use mathematical operators to make above equations work out.

3.

All the equations follow the same pattern. Just find a pattern, apply it to the last equation and find the result.

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  • 3
    $\begingroup$ There is hardly any overlap between the letters in each equation, which means it is trivial to assign values to each letter and make S + W + A + G equal whatever you want. $\endgroup$ – GentlePurpleRain Nov 22 '15 at 5:48
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    $\begingroup$ @GentlePurpleRain, Yes.... there is a logic to solve this puzzle. This is an interview question. I took almost half an hour to solve it and of course after getting a hint.. $\endgroup$ – H.Modh Nov 22 '15 at 6:25
  • $\begingroup$ Hint: Convert each alphabet to its corresponding numeric value...!! Then solve it using mathematical operators. $\endgroup$ – H.Modh Nov 22 '15 at 7:12
  • $\begingroup$ Would the question be more accurate if listed as $f(S,W,A,G)=?$ or is the usage of three operators significant? $\endgroup$ – Ian MacDonald Nov 23 '15 at 17:34
  • $\begingroup$ Do the letters map to unique numbers (e.g. $B=2 \implies C \ne 2$)? Do all the circles represent the same operation (e.g. all multiplications, or all additions, etc)? $\endgroup$ – Lawrence Nov 24 '15 at 4:10
3
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The answer is

$S \circ W \circ A \circ G = G$

Explanation: After plugging in the values $A=1,~B=2,~C=3,~\ldots,~Z=26$, each of the three example equations $~~~\alpha\circ\beta\circ\gamma\circ\delta=\epsilon~~~$ follows the pattern

$~~~(\alpha+\beta)/\gamma+\delta=\epsilon^2~~~~~$ respectively $~~~~~\epsilon=\sqrt{(\alpha+\beta)/\gamma+\delta}$

Indeed, the three given equations give us the following:

    C o I o D o A = B    yields     3 o  9 o  4 o  1 = 2
    O o K o M o G = C    yields    15 o 11 o 13 o  7 = 3
    Y o A o B o C = D    yields    25 o  1 o  2 o  3 = 4

One easily verifies that

$(3+9)/4+1=2^2~~~$ and $~~~(15+11)/13+7=3^2~~~$ and $~~~(25+1)/2+3=4^2$

This then leads to

$S\circ W\circ A\circ G=G~~~~~$ as $~(19+23)/1+7=7^2$

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  • 7
    $\begingroup$ I don't know if i approve of you only reopening the question after you found the solution and then immediately posting it without giving others the chance. $\endgroup$ – The Dark Truth Nov 23 '15 at 14:00
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    $\begingroup$ There are clearly three operators in the question and you have used four in your solution. This is not valid. $\endgroup$ – Ian MacDonald Nov 23 '15 at 15:24
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    $\begingroup$ @TheDarkTruth: I do not think that your accusation is justified. There were four hours between my vote to reopen this question and the submission of my solution. (You may check this in the review logs.) $\endgroup$ – Gamow Nov 23 '15 at 16:00
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    $\begingroup$ @Gamow: Sorry i only saw your name amongst the 5 reopen votes. I'm not able to find any times as to when the respective vote was given. And since there were only 76 seconds between reopening and your solution post i arrived at an early conclusion. $\endgroup$ – The Dark Truth Nov 23 '15 at 16:05
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    $\begingroup$ @IanMcDonald: The problem statement says nothing about "three operators"; this is just your own private interpretation. The first version of the puzzle had plus-signs instead of the little circles, and explicitly stated that these plus-signs just indicate that the letters have to be combined some way. $\endgroup$ – Gamow Nov 23 '15 at 16:09

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