## April 2017 Logic & Maths Problems – Solutions

1. 7

2. 48 (the differences start at 2 and keep doubling)

3. (a) 11 (b) 111 (c) 1111

4. 3

5. 8

6. 29 days

7. 1987

8. 1432

9. 7. yellow, yellow, yellow, light-blue
In the second row there was one colour change from the first row; red was taken out and replaced with yellow… and this guess received one more keypeg. This means yellow is in and red is out. And if yellow is in, obviously it must be in column 3.
There was one colour change from the second and fourth rows… dark-blue was taken out and replaced with brown. Since the number of keypegs stayed the same, this means they are either both in or both out. So if they are both in then green and light-blue must both be out, according to the information received with the fourth guess. (Since it must be yellow and brown that are drawing the two keypegs.) But how can green and light-blue both be out if two black keypegs were received in the third guess? They can’t so this means that dark-blue and brown must both be out, not in.
So, red is out, dark-blue is out, brown is out
Yellow is in.
One of green or light-blue is in but not both.
A double of green is out (Row 3)
So the four colours must be either:
yellow, light-blue, light-blue, light-blue
yellow, yellow, yellow, green
yellow, yellow, light-blue, light-blue
yellow, yellow, yellow, light-blue
A triple with light-blue just doesn’t work.
If green is in it can’t be in columns 3 or 4. A simple check indicates no solution fits with it in either columns 1 or 2 either so this colour combination is not correct.
If light-blue is in twice one must be in column 4 and the other must be in column 2 since light-blue can’t be in column 1 (row 1) or column 3. But yellow, light-blue, yellow, light-blue would not receive a black keypeg in row two, which it did.
The only combination left happens to work. yellow, yellow, yellow, light-blue fits each guess.

10. a. 8 b. 24 c. 24 d. 8