Look at this: 4 + 4 + 4. That is three lots of four.

Is there a shorter way to write three lots of four, without adding it all out one bit at a time?

Two pans, and each pan holds one way of writing the same amount. When both pans hold the same value, the beam hangs level.

Now the same idea again, with a different number of groups. Watch whether the beam still hangs level.

Today we lay counters into equal rows and write the multiplication that matches. First we make 3 rows of 4. Then 2 rows of 6. Then 5 rows of 2. Last of all, 3 rows of 3. Each time, count the groups, count how many are in each, and say both the repeated addition and the multiplication.

Today we balance the beam by finding the missing number. Work out how many equal groups are hiding in each puzzle.

First, two where the group size or number of groups is missing:

• 5 + 5 = 2 × ?
• 2 + 2 + 2 = 3 × ?

Then two where the missing part is inside the addition:

• 3 × 4 = 4 + 4 + ?
• 3 × 4 = 6 + ?  (two steps) — first work out what 3 × 4 makes, then find what you add to 6 to reach it.