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)   (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 =

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













All digits are between 0 and 9 inclusive



Substitute:

Multiply through by 3

Subtract 28 for all three sides

Divide through by -4 reversing inequalities


The only digit between 1.75 and 0.25 is 1 

Therefore C = 1 and since




A = 8
and
B = C-1 = 1-1 = 0 
and
D =  =  = 4

So "ABCD" = 8014

Edwin
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