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Puzzle 33
Can you find a five-digit number that has no zeros, and no repeated digits, where:
The first digit is a prime number. The second digit is the fifth digit minus the first digit. The third digit is twice the first digit. The fourth digit is the third digit plus three. The fifth digit is the difference between the first digit and the fourth digit.
Don't forget that 1 isn't prime, the prime numbers start with 2, 3, 5, …
Hint
Start with the possible answers where the first digit is a prime number, and then look at the third digit.
Answer
23,475.
Reasoning
By (1), the first digit is prime:
2----
3----
5----
7----
By (3), the third digit is twice the first digit, so we can eliminate 5---- and 7----:
2-4--
3-6--
By (4), the fourth digit is the third digit plus three:
2-47-
3-69-
By (5) the fifth digit is the difference between the first digit and the fourth digit:
2-475
3-696
We know from the introduction that no digit is repeated, so we can eliminate 3-696. And, by (2) the second digit is the fifth digit minus the first digit:
23475
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).
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