Carol E. answered • 08/14/13

Skilled Editor, Math and Literacy Skills Coach, Writer, Test Prep

Let first number of {} = 26 and add 13 to every number after 26 stopping at 52, and eliminate any number not ending in an even integer and ending in 0 to get the sum set. Stop at 998 to get 52 as the sum of each 3-number sets. 52/13= 4. There is just 26,39, and 52 as possibilities. 998+998=52 while 999+999=54 and not divisible by 13. And 54 is the highest number one can whereby add adding single digits in a series of three.

Carol E.

I am wondering too since I can't figure why 54 divisible by 17 isn't a possible winninumber red. there seems to be missing criteria. If 13 is necessary, then wheArdis the statement that makes us eliminate 54 as a possible sum to use.

08/14/13

Gabor R.

huh?

08/15/13

Miss. U.

Im not sure I understand what you're trying to say. :/

08/15/13

Miss. U.

Thank you for your answer. Unfortunately, I have still have not gotten what I needed. I am still not aware of how.. if you add all the lucky numbers... the sum is divisible by 13. Do you think you can help me prove that with a little more clarification?

Thanks! :)

08/14/13