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This problem looked more difficult than I think it really was, although it did require a great deal of logical thinking. Abdullah from Higher Bebington Junior School emailed us with his solution. He says:

For each problem I first looked to find a number that would make the units column accurate, then I substituted the number for the answer in the tens column and then continued the process until the calculation was complete.

This is a very good way of going about solving the problem, well done Abdullah. Here are Abdullah's final answers:

1 4 2 8 5 7

x 3

4 2 8 5 7 1

and

2 8 5 7 1 4

x 3

8 5 7 1 4 2

Joshua from Tattingstone School tells us how he approached the problem in more detail:

I wrote out single digit multiples of three up to 9 because each letter was one digit. I noticed that the numbers 1 to 9 only appeared once in the units column of the answers. I looked at the question and realised that 3 x e had to be 21 because it was the only answer ending in 1. This meant that e had to be 7. I carried the 2 and took it from 7 (the other e) and got 5. So d x 3 had to end in 5 which meant d had to be 5 because 5 x 3 = 15. I then repeated the process for the rest of questions 1 and 2.

This is very clearly explained - thank you Joshua.

Goh Wei Hao from Corporation Primary School in Singapore thinks that these are the only solutions. He says:

There is no multiple of three that has 1 in the ones place except for 7. (In the first sum)

There is no multiple of three that has 2 in the ones place except for 4. (In the second sum)

This is certainly true (and Joshua mentions this too), although we would need to take the explanation a bit further to justify that there aren't any other solutions.

Well done also to the following who sent correct solutions:

Ang Boon, Ang Li, Bay Jia, Goh Bee, Khoo Kian, Tan Li, Tan Yang and Teoh Wei all from Corporation Primary School in Singapore

Harry from St Mary's Stansted

Katrina and Helen from British School Manila

Tutku and Adil from Istanbul