I will start with listing the members of each set
Alison (Multiples of 5): {5,10,15,20,25,...}
Becky (Triangular numbers): {1,3,6,10,15,21,...}
Sam (Even but not multiples of 4): {2,6,10,14,18,22,...}
Matt (Multiples of 3 but not multiples of 9): {3,6,12,15,21,24,...}
Some of the the 2 digit numbers that belong in two of the sets are:
21 belongs in the sets of both Matt and Becky. 10 also belongs in the sets of Alison and Becky.
Some of the the 2 digit numbers that belong in three of the sets are:
15 belongs in the sets of Alison, Becky and Matt.
To find the smallest numbers that belongs in all 4 sets I first found the list of the first 100 triangular numbers. I also realised it has to end with 0 because it has to be both divisible by 5 and even. Therefore from the list of triangular numbers I took the numbers that end with 0 and check whether they are multiples of 4 and divisible by 3 and 9. For example, I had to exclude 120 as it is divisible by 4. I also excluded 190 as it is not divisible by 3. The number that met all the requirements was 210:
1. It is divisible by 5 and even but not a multiple of 4.
2. It is triangular (as it was on my list of triangular numbers).
3. It is divisible by 3 but not by 9.
List the numbers that satisfy both Alison's and Sam's statements:
{10,30,50,70,90,110,130,150,...}
One way to describe it would be: start at 10 and go up by 20.
Another would be to list odd numbers and put a '0' to at the end of each. For example:
1,3,5,7,9,... but we put a '0' at the end: 10,30,50,70,90,110,...
List the numbers that satisfy both Alison's and Matt's statements:
{15, 30, 60, 75, 105, 120, 150, 165}
We could describe it like this: start with 15 and add 15 skipping multiple of 9.
Another way to describe it: start with 15 and alternate the addition of 15 and 30.