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This question already has an answer here:

I have a sequence of numbers - 1,3,5,7,9,11,13,15 I want to get the sum 30 by choosing 3 numbers from this sequence of numbers . The numbers can be repeated. Thank you

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marked as duplicate by Will, Marius, manshu, Deusovi, Community Jul 14 '16 at 8:31

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my choice is

11+13+3! = 11+13+6 =30

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As all these numbers are odd, the sum of three of them is odd as well. Is this a lateral-thinking puzzle? For example,

do some of the commas represent decimal points? 1.3+5.7=7, or with some similar trick we may achieve the sum of two being odd.

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  • $\begingroup$ ya it is kind of lateral thinking! $\endgroup$ – gStephin Jul 14 '16 at 7:24
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There are many possible tweaks for this well known question

One of them Could be

11 + 3! + 13 = 30

or

()+15+15

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As adding three odd numbers would never give an even number...there must be a trick...

hence this is my best guess.

3!+9+15=30

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Also possible to do it these ways:

3*7 + 9

or

3*5 + 15

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