James has a set of digits from 1 to 9 with no repeats.   He chooses five of them and makes a five-digit number that satisfies all of the following conditions:

(1) The fourth digit is double the sum of the third and fifth digits.

(2) The sum of the first and fifth digits equals the sum of the third and
      fourth digits.

(3) The second digit is the fourth digit minus the third digit.

(4) The sum of the four unused digits is an even number.

(5) The first digit is the ten-thousands digit and the fifth digit is the
      units or ones digit.

What is the number that James formed?

Solution to the Problem:

James' number is 67183.

Let a = 1st digit, b = 2nd digit, c = 3rd digit, d = 4th digit, and e = 5th digit.
Then statements 1, 2, and 3 become:
(1) d = 2 (c + e) or d = 2c + 2e.
(2) a + e = c + d
(3) b = d - c
From statement (1), the fourth digit must be 6 or 8 since
6 = 2 (1 + 2) or 8 = 2 (1 + 3)
Now use statement (3) to get b.   So the four possibilities are:
    a         b         c         d         e    
            5         1         6         2    
            4         2         6         1    
            5         3         8         1    
            7         1         8         3    


Now use statement (2) to get a.
a + e = c + d; therefore, a = c + d - e.   So the four possibilities are:
    a         b         c         d         e    
    5         5         1         6         2    
    7         4         2         6         1    
    10         5         3         8         1    
    6         7         1         8         3    

We can eliminate case #1 because it produces two digits that are the same.
We can eliminate case #3 because a = 10 and that is not a single digit.
So, the only possibilities left are 74261 and 67183.
The number 74261 leaves the digits 3, 5, 8, and 9.  But their sum is not even (sum = 25).
The number 67183 leaves the digits 2, 4, 5, and 9.   Their sum IS even (sum = 20).

Correctly solved by:

1. James Alarie Flint, Michigan
2. John Funk Ventura, California
3. Trey Briggs Mountain View High School,
Mountain View, Wyoming
4. Adam Niemi Mountain View High School,
Mountain View, Wyoming
5. Mike Christy Manhasset, New York
6. Jacob Harmon Mountain View High School,
Mountain View, Wyoming
7. Austin Hale Mountain View High School,
Mountain View, Wyoming
8. Levi Anderson Mountain View High School,
Mountain View, Wyoming
9. David and Judy Dixon Bennettsville, South Carolina