DAY 03
Comment
The pattern with 100 dots is a 10 by 10 square.
Answer
45.
Readers' methods
Kelly N, Blythe NE24:
The blue dots make a triangle with 9 dots along the base and height.
Two triangles like this can be made into a 9 dot by 10 dot rectangle,
which is 90 dots, so one triangle contains 45 dots.
Andrew F, Stoke Newington N16:
The square numbers contain triangular numbers.
In a 2*2 there is 1 blue dot, in a 3*3 there are three blue dots;
1+2.
By the 10*10 square the pattern is 1+2+3+4+5+6+7+8+9.
This equals 45 blue dots.
J Lord, Cheshunt EN7:
I looked at the diagonal from the top left of the square to the
bottom left and saw there were equal numbers of blue and white
dots on either side of the line of diagonals.
Then I worked out the pattern for a 10 by 10 square:
9+8+7+6+5+4+3+2+1=45.
So there would be 45 dots on either side of the diagonal.
Tom H, Sevenoaks TN13:
This is because when there is a square of one hundred dots in
total, the sides wil be ten dots. When looking at the Examples,
on the left hand side column, there is always one less than the
total number of dots that are blue. This progresses down to no
blue dots on the right hand side column.
Therefore, calling the number of dots of one of the columns n,
then the number of blue dots is always
(n-1) + (n-2) + (n-3)... and so on, until (n-n) which equals zero.
So, for the ten by 10 square, there are
(10-1) + (10-2) + (10-3).....+ (10-10).
This can be written more easily as 9+8+7+6+5+4+3+2+1+0.
This gives us the result of 45.
Bernd F, Aulendorf, Germany:
In the sample patterns I recognized the blue triangles forming
the triangle number pattern.
For a n x n square the blue triangle is (n-1) dots wide and (n-1)
dots high.
The triangle number for (n-1) is 1+2+3...+(n-1).
In this case you get 1+2+3 ..+9=45
Solution:
There are 45 blue dots in the 100x100 dots pattern.
Pupils from Holgate Comprehensive School, Hucknall, Nottinghamshire:
100 ÷ 2 = 50; 50 - 5 = 45
n(n-1)/2
I drew a diagram and counted the squares.
nsquared/2 - n/2, where n is the side length.