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?