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Natasha and Izzy from Finney Primary School say:

Our solution was that we gave each domino a number going left to right (and top to bottom) . The first domino 1, second 2 and so on. The first solution was the odd numbers (in order 1, 2 so on) were:

odd: 1,4,5,6. even: 2,3,7,8,9.

So, in other words the dominoes which have an odd number of spots are:

And the rest have an even number of spots. Natasha and Izzy say that none of the dominoes was left over. Rachel from Brynteg School explains that:

All whole numbers are even or odd. The numbers on the dominoes are even or odd so there are no numbers left.

Jack and Peter answered the second part of the problem. They wrote:

For the pairs equalling five you have the 0:3 + 0:2, 0:1 +
1:3, 1:1 + 1:2. The ones left over are, 2:2, 2:3 and 3:3.

Other ways include, 2:2 + 0:1, 1:1 + 0:3, 0:2 + 1:2 then all
the others are left over.

There are always some dominoes left over but there are two
that are always left over: 3:3 and 3:2 because they both equal five
or over.

Well done, Jack and Peter. There are other ways to make pairs which total 5 and Thomas from Reading explains why:

You could pair the dominoes up to total a number of five spots
in three ways at any one time by pairing a three with a two or a
one with a four. There are two dominoes which add up to two and
they could each go with either one of the dominoes which add up to
three. There are also two dominoes with a total of four spots and
either one of those could go with the domino with only one spot on
it.

So here are all the possible ways of
grouping the dominoes to make 5:

2,2 and 0,1 0,3 and 0,2 1,2 and 1,1 or

3,1 and 0,1 0,3 and 0,2 1,2 and 1,1 or

2,2 and 0,1 0,3 and 1,1 1,2 and 0,2 or

3,1 and 0,1 0,3 and 1,1 1,2 and 0,2