HELP ASAP Maths homework Y3 I can't work it out let alone dd

[lexcat]

Am I missing something as dd gave up and I don't seem to be able to do it.
There are four calculations in this diagram, three across and on down.

Use the numbers 1 to 9. Put a different number in each box. Each calculaton must be correct.

[ ]-[ ]=[ ]
.........X
[ ]&divid;[ ]=[ ]

[ ]+[ ]=[ ]

Gave her the 8 and she was away.
Thanks for help.

If she needs to explain how to do it - you start with the division - it can only be 4/2 or 6/2 or 6/3 or8/2 because all other possibilities either have repeated digits or answers greater than 9.
Then look at the vertical sum - the possibilities are 2x3,3x2,2x4 or 4x2
When you start trying them the only ones that you can put in the vertical sum are 3x2=6 (with 8/4=2 as the division) or 4x2=8, with 6/3=2 as the division. All the others lead to repeated digits.
If you try the first of these it looks ok until you try to fit in the other numbers - the bottom row would have to be 5+1= 6 but you can't make 3 from 9 & 3 without using the 3 twice.
So it has to be 4x2=8 for the vertical & the three horizontal ones are 9-5=4, 6/3=2 and 7+1=8

just like sudoku really. Or there's that sudoku with addition as well isn't there.

sittin here convince marthasmama got the answer wrong and came up with same answer
have working here though and can see why it is a year 3 homework. Its about prime numbers...

start with cxf=I
Remove no from list that are prime and only consider no which need two different no multipled
leaves 8=24 and 6=23

then consider D/E=F
remove prime and no which when d/e generates another e value
leaves 8/2=4, 8/4=2, 6/3=2, 6/2=3

Logic says F is 2 as common in possiblities for both sums.

roughly know where 4,3,8 and 6 go

leaves 1,5,7 and 9
c is either 3 or 4 can only get 4
thus a=9 b=5
c=4, I=8 d=6 e=3
g and h are interchangable and either g=1 h=7 or g=7 h=1

ie: A - B = C X D / E = F

G + H = I

Becomes

9 - 5 = 4 x 6 / 3 = 2

1 = 7 = 8

You don't need to understand prime numbers to solve it.

If you've got 123456789
and need to find three different numbers out of that list that multiply/divide together, the smallest number cannot be 1, because that would imply the other two numbers are the same.

Then just work through the possibilities
2x3 = 6 - works
2x4 = 8 - works
2x5 = 10 - too big - stop with 2 times table

and also
3x4 = 12 - too big

2x3=6 and 2x4=8 are therefore definitely the only possibilities.

Both sums share the number 2, so that must go in the intersection of the multiplication and division sum. You then have two spots which you KNOW are for 8 and 6 and two spots for 3 and 4, so you plug them in, try to see if it works.

point and if i had read the posts above in my overtired state i would have realised that the answer had alread been explained