Alex is a keen dog admirer and over the years has had a number of dogs.
Alex has had an Alsatian, a Dalmatian, a Poodle, and a Great Dane, but not necessarily in that order.
Alex had Jamie first.The Dalmatian was an adored pet before the Great Dane.Sammy, the Alsatian, was the second dog Alex loved.Whitney was owned before the Poodle.Jimmy was not a Great Dane.
Can you match the dogs to their names and find the order in which Alex had them?
Answer # Name Breed
1 Jamie Dalmatian
2 Sammy Alsatian
3 Whitney Great Dane
4 Jimmy Poodle
Reasoning
(Clue 1) Jamie was first, and (Clue 3) Sammy the Alsatian was second: # Name Breed
1 Jamie
2 Sammy Alsatian
3
4
(Clue 4) Whitney was owned before the poodle, leaving Jimmy last: # Name Breed
1 Jamie
2 Sammy Alsatian
3 Whitney
4 Jimmy Poodle
(Clue 2) the Dalmatian was before the Great Dane: # Name Breed
1 Jamie Dalmatian
2 Sammy Alsatian
3 Whitney Great Dane
4 Jimmy Poodle
??
Puzzle 6
Can you place a different 4-letter word into each of the brackets to create two longer words. drift[----]wind home[----]bench foot[----]mother court[----]stick moth[----]roomquarter[----]ground space[----]yard wild[----]span your[----]less back[----]fighter
In the following example, the word 'book' creates the words 'cookbook' and 'bookcase'.
cook[book]case
Hint
The answer is 4 digits long, so what must G equal?
Answer
423 + 675 = 1098.
Reasoning
Remembering that:
even + even = even
odd + odd = even
even + odd = odd
To discuss individual letters, it's easiest to represent the sum as:
A B C
D E F +
————————
G H I J
A + D has to be over 9, which means that G = 1.
B + E = I, is even + odd = odd, which means that we can't have a carry from C + F (otherwise it would have been even + odd + 1, which is even).
The 1 has already gone, so the smallest possible value for either C or F is 3, which means that the other can't be 7 or 9 (otherwise we'd have a carry).
Therefore, C and F are 3 and 5, but we don't know which is which. But we do now know that J = 8.
A + D = H, is even + even = even, which means that we can't have a carry from B + E. Therefore, E can't be 9, as this would force a carry. So E = 7.
I is the only remaining odd number, so I = 9.
Which means that B = 2.
Neither A nor D can be 0 (otherwise we would have two of the same digit). So, H = 0.
Therefore, A and D are 4 and 6 (but we don't yet know which is which).
Since the top row's digits have to add to 9, A can't be 6, so A = 4, making C = 3.
