Place letters into the grid such that every row, column, and 2 x 2 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: S O N A T A
R
E
B
S
W
K
R
E
P
Z
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 203
In the illustration we have a sketch of Sir Edwyn de Tudor going to rescue his love, who was held captive by a neighbouring wicked baron.
Sir Edwyn calculated that if he rode at fifteen miles an hour he would arrive at the castle an hour too soon, while if he rode at ten miles an hour he would get there just an hour too late.
Now, it was of the first importance that he should arrive at the exact time appointed, in order that the rescue that he had planned should be a success, and the time of the tryst was five o'clock, when the captive would be taking afternoon tea.
The puzzle is to discover exactly how far Sir Edwyn de Tudor had to ride.
Sir Edwyn De Tudor, Amusements In Mathematics, Henry Ernest Dudeney.
If Sir Edwyn left at noon and rode 15 miles an hour, he would arrive at four o'clock, which is an hour too soon. If he rode 10 miles an hour, he would arrive at six o'clock, which is an hour too late. But if he went at 12 miles an hour, he would reach the castle of the wicked baron exactly at five o'clock, which is the time appointed.
The text above is the answer given in the book, and below is a method of finding the answer.
If we call the distance to the castle, D and use the fact that Time = Distance ÷ Speed, we have:
Travelling at 15 mph:
Time1 = D ÷ 15 (an hour too soon)
Travelling at 10 mph:
Time2 = D ÷ 10 (an hour too late)
The time gap between these two times is 2 hours, therefore
There are two alternate series, starting with the first two numbers, and each formed by doubling the preceding number in its own series and subtracting 2.
