If 1 + 9 + 8 = 1,

What is 2 + 8 + 9?

Looking forward to your answers ;)

  • 1
    $\begingroup$ Welcome to puzzling SE. This far from being a good puzzle. You are asking people in here to try to guess a pattern you have in mind offering a single example. You should either show more examples following the same pattern or add more clues about the pattern you have in mind. Also, not sure if the riddle tag is the best one on this case. $\endgroup$
    – Marius
    Commented Oct 24, 2016 at 13:24
  • $\begingroup$ Oh I'm sorry. Wha tag would you use in this case? $\endgroup$ Commented Oct 24, 2016 at 13:25
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    $\begingroup$ Ok guys, stop removing his reputation points like jerks. He just joined the community 5 minutes ago... give him a chance and perhaps a few polite advises. And to speak in a language the point greedy folks might understand, he won't be able to upvote anyone if he doesn't get at least a few reputation points... $\endgroup$ Commented Oct 24, 2016 at 13:38
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    $\begingroup$ @TristandeJager Your puzzle was nice and interesting, only important thing for next time would be to add at least 1 more example that fit the logic so we can extrapolate similarities and eliminate irrelevant solutions. $\endgroup$ Commented Oct 24, 2016 at 13:44
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    $\begingroup$ @stackreader Yes I understand, I won't let you down next time ;) $\endgroup$ Commented Oct 24, 2016 at 13:45

2 Answers 2


It could be


Since 1 + 9 + 8 = 1

One + Nine + Eight initial letters spell ONE

2 + 8 + 9 gives Two + Eight + Nine spells TEN

  • $\begingroup$ You got it right Tom! Sorry for the ambiguous puzzle all. $\endgroup$ Commented Oct 24, 2016 at 13:36
  • $\begingroup$ Welcome to Puzzling! Thanks for the quick response and am sure you can make another puzzle - tough with a unique answer! $\endgroup$
    – Tom
    Commented Oct 24, 2016 at 13:42

There can be many answers to this in the current situation of the question -

Here goes one -

Answer is

2+9+8 = 2


one nine eight = 1 (Number of t's in the digits)
two eight nine = 2 (Number of t's in the digits)

Again, its too broad and ambiguous.

  • $\begingroup$ This is an interesting approach. Not the answer I had in mind, but you are not wrong! My fault in this case because of the broad question. $\endgroup$ Commented Oct 24, 2016 at 13:46
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    $\begingroup$ No problems. All the best for the next one! Keep puzzling $\endgroup$
    – Techidiot
    Commented Oct 24, 2016 at 13:47

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