Reasoning
If we assign the letters P, C and M to the fruits, and work in pence we have:
(1) M + C = 119
(2) M + P = 145
(3) C + P = 140
Using (2) – (1) we have:
(4) P – C = 26
And (3) + (4) gives:
2P = 166 P = 83
This can then be substituted back into (2) and (3).
??
Puzzle 590
At a recent motorsport gathering, four proud owners were gathered around the silver sports car, each discussing their pride and joy.
Alex was overheard bragging about his brand new Stellar. The person whose surname was Smith went to great lengths to explain to Billie the virtues of his blue Lunar. The person whose surname was Wilshaw, and Dale, were already good friends. The person whose surname was Richards showed everyone a recent photo of his red Orbital. The person whose surname was Stone listened intently while Charlie kept talking about his green Galactic.
Who owns which car?
Reasoning
By (2) the Lunar was Blue, and was owned by Mr Smith.
By (4) the Orbital was Red, and was owned by Mr Richards.
By (5) the Galactic was Green, and was owned by Charlie.
Therefore the Stellar was silver, and by (1) was owned by Alex. Smith Blue Lunar RichardsRed Orbital Charlie Green Galactic Alex Silver Stellar
By (2), Mr Smith wasn't called Billie, so it must have been Billie Richards, and Dale Smith.
By (5) Mr Stone wasn't called Charlie, so it must have been Charlie Wilshaw, and Alex Stone. Dale Smith Blue Lunar Billie RichardsRed Orbital Charlie Wilshaw Green Galactic Alex Stone Silver Stellar
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Puzzle 591
If Malcolm, Laura, John, Alan, Brenda and Kevin like chocolate …
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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