There are many valid combinations, assuming that the same number (digit) can be used twice or more.

If you start with the second line first you have 16. The first line is a distraction.

KevinJ wrote:

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As far as I am aware, there is only 1 combination.

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Here are 4 possible combinations (the third combination has dissimilar digits):

2 5 5 9 9
3 5 5 8 9
* * * * *
4 4 6 8 8

There are 11 solutions, 3 with dissimilar digits.

KevinJ wrote:

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Did you try to open the spreadsheet at the bottom of my original post?  It would not open for me using any of the numbers below.

My feedback:

Rules

1. The fifth number plus the third number equals fourteen.
2. The fourth number is one more than the second number.
3. The first number is one less than twice the second number.
4. The second number plus the third number equals ten.
5. The sum of all five numbers is 30.

a b c d e
2 5 5 9 9 - Fails on rule 2 and fails on rule 3.
3 5 5 8 9 - Fails on rule 2 and fails on rule 3.
* * * * *
4 4 6 8 8 - Fails on rule 2 and fails on rule 3.

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Doh, I didn't see the "one more" and "one less" conditions. Back to the drawing board ...

Edit:

OK, I got it this time. There is only one solution, as you say.

Let the 5-digit code be abcde.

Analysing each of the 5 statements in turn we get ...

1/ e + c = 14

2/ d = b + 1

3/ a = 2b - 1

4/ b + c = 10

5/ a + b + c + d + e = 30

from 1:

e = 14 - c

from 4:

b = 10 - c

from 2:

d = b + 1 = 11 - c

from 3:

a = 2b - 1 = 2 x (10 - c) - 1 = 19 - 2c

from 5:

a + b + c + d + e = (19 - 2c) + (10 - c) + c + (11 - c) + (14 - c) = 54 - 4c


54 - 4c = 30

The rest is straight substitution.

The first line is a distraction, put the 14 aside. You have 16 to play with.

Two is 1 less than Four, so 4 & 5, to have enough from 16.

One has to be 1 less than twice Two, so 7 & 4.

The three add up to 16.

Two & Three add up to 10, so 4 & 6.

Which leaves 8 to add up to 30.

74658

Having an equation is better, but I did manage to juggle the numbers in my head..... just!

Starting with line two & also thinking of two is 1 less than four was easier the four 1 more than two, for me anyway.

dabbler wrote:

Like most of these algebraic problems presented as puzzles, a lot of confusion depends of wording.

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I'm noticing that I miss entire words when I read and also when I write. This is happening more often. I wonder if it is a sign that something is wrong upstairs. In this particular case I mistook "one more" and "one less" for "more" and "less", which then left me with this:

1/ e + c = 14

2/ d > b

3/ a < 2b

4/ b + c = 10

5/ a + b + c + d + e = 30

That's why I found 11 solutions.