Question

fill in the boxes
\(\huge \tt \begin{align} \color{black}{ \fbox{-}+ \fbox{-}+ \fbox{-}+ \fbox{-}+ \fbox{-}= \fbox{30}}\end{align}\)

u can use
\(\large\tt \begin{align} \color{black}{\{1,3,5,7,9,11,13,15\}}\end{align}\)
u can repeat the numbers

Answer 1

cool problem and latex xD

Answer 2

lol

Answer 3

test it with the largest number first

Answer 4

interesting

Answer 5

(15-9)+..that kind of thing?

Answer 6

or actually can we use factorials?

Answer 7

I can see it with factorials easy haha

Answer 8

no only one or two digit number

Answer 9

mhmmmmm

Answer 10

yes but how ?

Answer 11

i give u a rule

Answer 12

I don't understand I don't think adding 5 odd numbers will give even

Answer 13

i tried lol

Answer 14

u have 5 places
your answer is30
so in each of them should be 6 ok?

Answer 15

Impossible given the numbers, unless there's a trick to this

Answer 16

for example u can choose 9&3 instead of 6&6
then u can find it

Answer 17

there has to be some extra rules

Answer 18

if there no solution, prove it @iambatman

Answer 19

it has no answer
be sure

Answer 20

because u have 5 places with unit numbers

Answer 21

and u have an even number in your answer

Answer 22

I hope rotating the numbers is not allowed :P
\[\huge \tt \begin{align} \color{black}{ \fbox{6}+ \fbox{6}+ \fbox{6}+ \fbox{6}+ \fbox{6}= \fbox{30}}\end{align}\]

Answer 23

cool

Answer 24

Actually no what I said was right lol

Answer 25

There is no answer

Answer 26

exactly iambatman

Answer 27

more mathematically \[\large \tt \begin{align} \color{black}{ \fbox{odd}+ \fbox{odd}+ \fbox{odd}+ \fbox{odd}+ \fbox{odd} \stackrel{?}{=} \fbox{30}}\end{align}\]

\[\large \tt \begin{align} \color{black}{ ( \fbox{odd}+ \fbox{odd})+ (\fbox{odd}+ \fbox{odd})+ \fbox{odd} \stackrel{?}{=} \fbox{30}}\end{align}\]

\[\large \tt \begin{align} \color{black}{ ( even )+ (even )+ \fbox{odd} \stackrel{?}{=} \fbox{30}}\end{align}\]

\[\large \tt \begin{align} \color{black}{ (even )+ \fbox{odd} \stackrel{?}{=} \fbox{30}}\end{align}\]

Answer 28

Nice haha.

Answer 29

You could quickly just find the average and realize right away without having to do all this stuff haha xD

Answer 30

also this way
\(\Large \tt \begin{align} \color{black}{5(2n-1) (mod 2) \\
=10n-5) (mod 2) \equiv 1}\end{align}\)
its odd
where \(\large n\in \mathbb{z} \)
so no solution

Answer 31

nice :) but there is a small catch in your argument you're assuming all numbers are same

Answer 32

Checkmate!

Answer 33

every odd positive integer can be represented as \((2n-1)\) ?

Answer 34

Yes

Answer 35

so i think it includes all odd numbers

Answer 36

5(2n-1) does not represent sum of "5 arbitrary odd numbers"

it represents adding "one arbitrary" odd number 5 times

Answer 37

ok this should be it
\(\large\tt \begin{align} \color{black}{\left( (2n_1-1) (mod~ 2)+(2n_2-1) (mod ~2)\\+(2n_3-1) (mod~ 2)+(2n_4-1) (mod ~2) \\+(2n_5-1) (mod ~2) \\
\right)(mod~ 2) \equiv 1}\end{align}\)
\(\{n_1,n_2,n_3,n_4,n_4,n_5\}\subset \mathbb{Z}\)