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We had lots of correct solutions to this
problem and the majority identified the Green and Blue sets as
multiples of $5$ and $9$ respectively. A few question marks arose
when looking at the Red set.
Max of St Mary's Catholic High School
said:
$\eqalign{
Red &= 10,19,28,37,46,55,64,73,82\cr
Green &=
5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100\cr
Blue &= 9,18,27,36,45,54,63,72,81,90,99}$
So the red set is $9$ times table but adding $1$ on every
time.
The green set is the $5$ times table. There are $20$ numbers
in this set less than $101$.
The blue set is the $9$ times table. There are $11$ numbers in this
set less than $101$.
Max correctly identifes the Green and Blue
sets. He also noticed that there was a common difference of $9$ in
the red set. However we would need to justify why $91$ and $100$
are not in this set.
Mr Harrison's maths group of Northdown
Primary school had a different suggestion for the Red
set:
We got a hundred square grid and crossed out all the numbers
they gave us at the beginning. We tried to find all the patterns,
for example the red set were in a diagonal line.
We looked for gaps in the patterns. We found:
Red set: (digits sum $=10$)
$\{19,28,37,46,55,64,73,82,91\}$
Green set: (Multiples of $5$)
$\{5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100\}$
Blue set: (Multiples of $9$)
$\{9,18,27,36,45,54,63,72,81,90,99\}$
The red set has $9$ numbers and there are
no other possible numbers less than $101$ which satisfy the
condition Mr Harrison's maths group imposed.
There was a big divide between those of you
who agreed with Max and those who agreed with Mr Harrison's maths
group.
Megan of Crookhill Primary agreed with Mr
Harrison's maths group:
First I plotted the hints on a grid and looked at how many there
were in each colour group. Then I looked at the pattern on the
board to gain more information. To find out the Green set we
knew the numbers that they gave us were already in the $5$ times
table. On the grid there were two lines of $10$ numbers which made
$20$ in that group so we saw that it was Green set.
Then we solved the Red set. The first thing we spotted was the
pattern on the page so then we looked really carefully at the
numbers then we solved it. The digits all add up to ten; that was
the hardest to spot I think The easist to spot was the Blue set. I
enjoy doing my $9$ times table that's probably why so I didn't need
to work anything out.
Well done Megan, and I enjoy my $9$ times
table as well.