Question

ok, I have a trick problem..help....
Is it possible to place 44 coins into 10 different containers in such a way that no two containers have the same number of coins?

Answer 1

maybe, how big are the containers?

Answer 2

it doesn't say

Answer 3

1+2+3+4+5+6+7+8+9+10 = 45 coins in order for each of them to be different

Answer 4

but; if there are no coins in the first one .. hmmm

Answer 5

its 55 aint it

Answer 6

44

Answer 7

55-11-= 44 so it might be possible

Answer 8

take one from each container to get 44 with the first one empty

Answer 9

ugh, i almost got it right

Answer 10

0
1
2
3
4
5
6
7
8
9
----
45 coins so hmmm we are one extra

Answer 11

im going with no

Answer 12

thank you

Answer 13

yeah, i cant see it happening; but i could be wrong .... just saying

Answer 14

I've figured it the way you did but somehow I think it could be done
we have these every month and every single problem has had a solution soo...

Answer 15

idk

Answer 16

{44} {} {} {} {} {} {} {} {} {}

{42} {2} {} {} {} {} {} {} {} {}

{39} {2} {3} {} {} {} {} {} {} {}

{35} {2} {3} {4} {} {} {} {} {} {}

{30} {2} {3} {4} {5} {} {} {} {} {}

{24} {2} {3} {4} {5} {6} {} {} {} {}

{17} {2} {3} {4} {5} {6} {7} {} {} {}

{9} {2} {3} {4} {5} {6} {7} {8} {} {}

{8} {2} {3} {4} {5} {6} {7} {8} {1} {}

just cant see it