Since it is axiomatic that six odd numbers MUST total an even number, you have to be "constructive" (ie cheat): 3 + 3 + 3 + 1 + 11 = 21, or even more "constructively", let one of the 9s slip so that it becomes a 6 and have: 6 + 5 + 3 + 3 + 3 + 1 = 21. I'm sure there are other equally effective ways of being "constructive"!
I'm not sure if you are limited by purely addition or what the rules are so I have the following answer:Using only 6 numbers, but any function:3*5-5+9+1+1 = 21
3*3*3-(5+(5/5)) 9+3+3+5+1(to the power of 1) ...using 6 numbers logic fullfilled ...here in this answer v use the clause of using all given digits
5+5+5+5+1 That is only 5 numbers being added. There can be no answer for six numbers for the following reasons: The numbers are all odd. Two odd numbers make an even number. Two even numbers make an even number. 21 is odd. Take 6 odd numbers and add them up in pairs. Each pair of odd numbers gives an even number so there are 3 even numbers. Even + Even + Even = Even. So sum of any six odd numbers is even but 21 is not even.
It is not possible to make six odd numbers total an odd number, so this requires a little lateral thinking: 3 + 3 + 3 + 1 + 11, or you could invert one of the nines and make a six... 6 + 5 + 3 + 3 + 3 + 1...
It totally depends on how big each group is. I can't add two numbers if you don't give me any numbers.
3 + 3 + 3 + 1 + 11 = 21
you didnt give me any numbers
Chose any set of 15 numbers that total 990. This group will average 66.
Any pair of consecutive numbers will have an odd total. 10 and 12 are consecutive even numbers that total 22.
To give you whole numbers 7 and 11 sorry if I missed out any
First, the word is remainder, not reminder.In any case, none of the followed numbers would give a remainder of 2 since there are no such numbers!First, the word is remainder, not reminder.In any case, none of the followed numbers would give a remainder of 2 since there are no such numbers!First, the word is remainder, not reminder.In any case, none of the followed numbers would give a remainder of 2 since there are no such numbers!First, the word is remainder, not reminder.In any case, none of the followed numbers would give a remainder of 2 since there are no such numbers!
There are more irrational numbers between any two rational numbers than there are rational numbers in total.