SOLUTION: A four-digit number with all different digits has the sum of its digits is 13. Find the number if the thousands place is double the units place and the tens place is one more

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Question 1037857: A four-digit number with all different digits has the
sum of its digits is 13. Find the number if the thousands
place is double the units place and the tens place is
one more than the hundreds place.

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
A four-digit number with all different digits has the
sum of its digits is 13. Find the number if the thousands
place is double the units place and the tens place is
one more than the hundreds place.

Let the number be "ABCD"

the sum of its digits is 13.
A+B+C+D=13

the thousands place is double the units place
A = 2D or  D = A%2F2

the tens place is one more than the hundreds place.
C = B+1 or B = C-1

Substitute in A+B+C+D=13 and solve for A

A%2BC-1%2BC%2BA%2F2+=+13

A%2B2C%2BA%2F2+=+14

2A%2B4C%2BA+=+28

3A%2B4C+=+28

3A+=+28-4C

A+=+%2828-4C%29%2F3

All digits are between 0 and 9 inclusive

0+%3C=+A+%3C=+9

Substitute:
0+%3C=+%2828-4C%29%2F3+%3C=+9
Multiply through by 3
0+%3C=+28-4C+%3C=+27
Subtract 28 for all three sides
-28+%3C=+-4C+%3C=+-1
Divide through by -4 reversing inequalities
7+%3E=+4C+%3E=+1
1.75+%3E=+C+%3E=+0.25
The only digit between 1.75 and 0.25 is 1 

Therefore C = 1 and since

A+=+%2828-4C%29%2F3
A+=+%2828-4%281%29%29%2F3
A+=+%2824%29%2F3
A = 8
and
B = C-1 = 1-1 = 0 
and
D = A%2F2 = 8%2F2 = 4

So "ABCD" = 8014

Edwin