Reasoning
Since no digit is duplicated, neither number can end in 1, otherwise, the last digit of the answer would already have been used.
Neither number can end in 5 because the answer would then end in 5 or 0.
So the first number can only be 62, 63, 64, 67, or 68.
We can now look at what the second number can end with, and we find that …
if the first number was 62, the second number can only end in 4 or 7.
Why …
not 1 as previously explained
not 2 because we've already used that in the 62
not 3 because 62 x *3 would end in 6, which is already in the 62
not 5 as previously explained
not 6 because we've already used that in the 62
not 8 because 62 x *8 would end in 6, which is already in the 62
We can repeat this for the other possible first numbers and find that …
if the first number was 62, the second number can only end in 4 or 7.
if the first number was 63, the second number can only end in 4, 7, or 8.
if the first number was 64, the second number can only end in 2, 3, 7, or 8.
if the first number was 67, the second number can only end in 2, 3, or 4.
if the first number was 68, the second number can only end in 3 or 4.
Let's check these in turn …
If the first number was 62, the only possible values are:
62 x 14 = 868
62 x 34 = 2108
62 x 54 = 3348
62 x 74 = 4588
or
62 x 17 = 1054
62 x 37 = 2294
62 x 57 = 3534
62 x 87 = 5394
Only 4 calculations are required for each option, as we didn't need to check 62 x 84 as we already know that the answer would end in 8, or 62 x 47 as we already know that the answer would end in 4.
All of these answers fail as they don't contain all of the digits.
Similarly, we can look at 63, then 64, etc. For each of the numbers, we have to check the possible endings, but in each case, only 4 calculations are required (as seen in the examples above).
We (quickly?) find 64 x 58 = 3712.
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Puzzle 2
Can you find every occurrence of the word JULY that appears in this grid (horizontally, vertically, or diagonally)?
The hint will reveal the number of times it occurs, but where are they?
Note: this puzzle is not interactive, and the letters cannot be selected.
Hint
How many statements would be true if the culprit was Alex?
Answer
Emery committed the crime.
Reasoning
If we assume that Alex was the culprit, we can see that there would be 6 true statements, but we know there were exactly 5 true statements, so it can't be Alex.
We can repeat this will all five suspects and quickly see that only Emery being the culprit would result in exactly 5 statements being true.
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Puzzle 4
Every year, just before my birthday, I open my money box to see how much money I have saved.
Imagine my surprise when I found exactly £49.58 made up of equal numbers of 2p, 5p, 10p and 20p.
