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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 calculation)

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

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.

We had a number of solutions sent in by students at Dulwich School in Shanghai:  Jeffrey and Yuzuki; Jay and Jet; Nathan, Vivian and Joyce; Helen and Hana; Ignatius, Doras, Ella and Maxine; Eddie, Michael, Jakey and Charlie; Daniel, Patrick, Jason, Joey, Max and Paul. Here are some of their ideas:

We first wrote all of the three times tables up to 9 because the letters could not be two digits. Then, I thought that E x 3 must end with 1 and the only 3 times table up 9 that ends with a 1 is 21, and that means E=7 and we the 2 on top of the D, and put 7 into brackets on both Es.
Then we knew that the next number must end with 7, but has to be added to the carried 2, so it ends with 5. The only 3 times table up to nine that ends with 5 is 15, and 15/3=5, so D=5.
Then I repeated the process until we got the answer.

Also:

At first, we thought that A was 3, B was 12, C was 8, D was 5, E was 7, F was 6, G was 2, H was 7, I was 1, and J was 4. But then, we realised that we were wrong because it said that each letter represented only 1 digit. But because we forgot about that rule, we thought that B was 12 but that couldn't be it because 12 had 2 digits.
So we checked with my friend Joyce and she explained how she got the right answer. We realised that we were wrong and we got muddled up with the numbers so now we think A=4, B=2, C=8, D=5, E=7, F=8, G=5, H=7, I=1, and J=4.

Also: 

Process (How we did it):
First, we thought how we could figure out the letters and because there is a 1, we knew that 3 x 7 = 21 and 21 is the only number that ends with 1 in the 3x tables 1-9. After that, we carried the 2 and took it from the 7.  
We then repeated the process, only with different numbers and letters and also for the 2nd problem. 
We also think that there is only one solution for each of the problems.

Thank you for all those interesting ways of working. Well done everyone.