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I have a set of words. If those words follow a certain rule, they become an Equal Word™. Use these examples to find out the rule for an equal word.

$$ % set Title text. (spaces around the text ARE important; do not remove.) % increase Pad value only if your entries are longer than the title bar. % \def\Pad{\P{0.0}} \def\Title{\textbf{ Equal }} % \def\S#1#2{\Space{#1}{20px}{#2px}}\def\P#1{\V{#1em}}\ \def\V#1{\S{#1}{9}} \def\T{\Title\textbf{Words }^™\Pad}\def\NT{\Pad\textbf{Not}\T\ }\displaystyle \smash{\lower{29px}\bbox[yellow]{\phantom{\rlap{rubio.2017.02.04}\S{6px}{0} \begin{array}{cc}\Pad\T&\NT\\\end{array}}}}\atop\def\V#1{\S{#1}{5}} \begin{array}{|c|c|}\hline\Pad\T&\NT\\\hline % \text{ MAN }&\text{ CLEAR }\\ \hline \text{ BALANCED }&\text{ WAVE }\\ \hline \text{ ORDER }&\text{ HIDE }\\ \hline \text{ FINE }&\text{ DJINN }\\ \hline \text{ EQUALS }&\text{ LINEAR }\\ \hline \text{ STAND }&\text{ STAIN }\\ \hline \end{array}$$

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I think the rule here is

The sum of the positions of each letter in the alphabet should be an even number.

For example:
MAN: 13 + 1 + 14 = 28
STAND: 19 + 20 + 1 + 14 + 4 = 58
BALANCED: 2 + 1 + 12 + 1 + 14 + 3 + 5 + 4 = 42

The other words end up to an odd total:
CLEAR: 3 + 12 + 5 + 1 + 18 = 39
WAVE: 23 + 1 + 22 + 5 = 51
STAIN: 19 + 20 + 1 + 9 + 14 = 63

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    $\begingroup$ Hm. If so, HIDE and EQUALS should swap sides. $\endgroup$ – M Oehm Sep 10 '18 at 11:21
  • $\begingroup$ ah yes you're right $\endgroup$ – Pim Schwippert Sep 10 '18 at 11:32
  • $\begingroup$ Even with the typo you got it correct! Well done! $\endgroup$ – Rohit Jose Sep 10 '18 at 15:26

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