Which numbers? (2)
I am thinking of three sets of numbers less than 101. Can you find all the numbers in each set from these clues?
Problem
This problem is similar to Which Numbers? (1), but slightly more tricky. You may like to try that one first.
I am thinking of three sets of numbers less than $101$. They are the blue set, the red set and the black set.Can you find all the numbers in each set from these clues?
These numbers are some of the blue set: $26, 39, 65, 91$, but there are others too.
These numbers are some of the red set: $12, 18, 30, 42, 66, 78, 84$, but there are others too.
These numbers are some of the black set: $14, 17, 33, 38, 51, 57, 74, 79, 94, 99$, but there are others too.
There are sixteen numbers altogether in the red set, seven numbers in the blue set and fifty numbers in the black set.
These numbers are some that are in just one of the sets: $6,10, 15, 24, 33, 48, 56, 65, 75, 93$, but there are others too.
These numbers are some that are in two of the sets: $12, 13, 36, 54, 72, 96$, but there are others too.
The only number in all three of the sets is $78$.
These numbers are some that are not in any of the sets: $5, 8, 22, 27, 44, 49, 63, 68, 82, 86, 100$, but there are others too.
There are twelve numbers that are in two of the sets.
There are $41$ numbers that are not in any of the sets.
You can download a sheet of all this information that can be cut up into cards.
Can you find the rest of the numbers in the three sets?
Can you give a name to the sets you have found?
Getting Started
You could start by putting the information you do know onto a hundred square.
You might record them on a hundred square like this:

You can find a special hundred square here.
Student Solutions
In a similar way to Which Numbers? (1) the solutions we had tended to identify correctly two of the sets but struggled with the third.
Joshua of Crookhill Primary School said:
For the red group it is all the multiples of $6$.
For the blue group it is $+ 13$ every time.
and for the black we have no idea what so ever!
Sophie and Jo of Huish Primary continued:
The blue set's give away numbers are $26, 39, 65$ and $91$. We first looked at the end digits and saw they were going up by $3$ each time. We then knew it was a multiple of somthing with a $3$ on the end. We then knew they were going up by $10$ each time. We added the $10$ and the $3$ together to get $13$. So the blue set is going up by $13$ each time: $\{13,26,39,52,65,78,91\}$. There are $7$ numbers in the blue which is the same as on the sheet.
The red set's give away numbers are $12, 18, 30, 42, 66, 78, 84$. We knew they were even, so it would be in either the $2$s, $4$s, $6$s or $8$s. We narrowed it down to the $6$s and the $2$s. The $2$s has $50$ numbers less than $101$, so we knew it was the $6$s. There were $16$ numbers in the red set like it said on the sheet: $\{6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96\}$.
The black set's give away numbers are $14, 17, 33, 38, 51, 57, 74, 79, 94, 99$. We thought a long time about what it could be. As we looked closer we realised that the $10$s digit was always odd. We also realised that there are $50$ numbers with an odd $10$s digit before $101$. So the black set is all the $10$s, $30$s, $50$s, $70$s and $90$s.
Do you agree with Sophie and Jo?
Teachers' Resources
Why do this problem?
This problem requires learners to see the connections between numbers in a set and so find the rest of the set. They will need to make and test hypotheses, and justify their reasoning.
Possible approach

Key questions
Possible extension
Learners could do the much harder Ben's Game or make up their own clues for sets of numbers for others to try.
Possible support