Euler was the first to prove that this puzzle had no solution.
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Puzzle 18
Place letters into the grid such that every row, column, and 2x2 block has letters (in any order) that form a common word. Each letter is only used once, and no letter is repeated in the rows / cols / blocks.
Letters allowed: W A R M T H
V
O
L
S
I
B
O
A
E
L
Note: this puzzle is not interactive, and the squares cannot be clicked.
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.
This makes the sum 423 + 675 = 1098.
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Puzzle 20
Using all of the letters A to Z, each once only, complete these common words. There are currently 2 different answers, can you find them both? -e-elb-ds-i-lswa--fo--er-ya-ur--u-dy-uo-e-e-a-eed--sc-r-yl-p--lls
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