maths homework

[fever28]

I know I Should know this but...

5 year olds math home work :-(

you have 5 plates arreanged in a cross, each direction consists of three plates. the middle one being shared.
you also have 15 counters.

the problme, is to put a diffrent amount of counters on each plate so the vertical row adds up to 10, and the horizontal role adds up to 10

although its quite easy to solve, (just look at it and one or two trial and error attempts)

I dont know what the method is for working this out, rathe rthan trial and error.

[SYRAC]

If you have 5 plates and different numbers you must have 1,2,3,4,5. 5 has to go in the centre and then 1,4 go one way and 2,3 the other

[doodler]

As there are 15 counters you put 5 on the middle plate ( 15-10). You now need to find 3 numbers which add to 10.- 5+4+1 and 5+3+2

[fever28]

I know the answer,, im not that dim

I was just wondering if there is a formula.


lets a ssume it wasnt such a simple question how would you work it out

[jh004j7891]

3
3 5 2 = 10
2
=
10

hope it helps ( hope its right)

[fever28]

also same issue but adding up to eight (i know the answer :-) but the "must be 5" in the middle doesnt apply

[jh004j7891]

sorry previous post got misspaced should be 3+5+2 on each line

[kettledrum]

You're quite right it doesn't have to be 5 in the middle, but it does have to be an uneven number - so that when taken away from 15, there is an even number to be divided equally between the horizontal & the vertical lines.

So if 1 counter is in the middle, you have 14 counters to split ... 7 each way..... 2 & 5 and 3 & 4

If 3 counters are in the middle you can divide the remaining 12 into 6 each each.... 1 & 5 and 2 & 4

and if you have 5 in the middle .....

[fever28]

thank you "uneven in the middle" is a bit of logic that works for this and the other questions.

i was just findinhg hard to simplify it for a5 year old, I can just look and see what adds up to 10 in my head quickly but they cant so a nice rule to apply helps a lot cheers.

[kettledrum]

I beg your pardon, Fever28 !

I failed to read your question properly and didn't notice that both the horizontal & vertical rows had to add up to 10. In that case 5 does have to be in the middle.....trial and error is really the only way to tackle this, after sorting out the options by making sure you have an even number of counters to divide between the 2 rows.

Sorry to have misled you but I think you understood and have worked out for youself that it did had to be 5 in the middle in order for each row to add up to 10.

[Muffet]

Try this. You have 15 counters. When placed on the plates, and the rows are added, you need a total of 20, so you've got to find another 5 from somewhere. But the middle plate is included twice, so, if you put the 5 on there, it will be counted twice, and you get your extra 5.

[SYRAC]

Didn't mean to imply you were dim, sorry, but as far a I'm aware there is no formula it is pure logic. 5 plates and 15 counters means 15 has to be made up of 5 different numbers and in this case it has to be 1,2,3,4,5 - there is no other combination, the rest follows logically.