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.
<pre><b>
Let the number be "ABCD"
</pre>
the sum of its digits is 13.
<pre>
A+B+C+D=13
</pre>
the thousands place is double the units place 
<pre>
A = 2D or  D = {{{A/2}}}
</pre>
the tens place is one more than the hundreds place.
<pre>
C = B+1 or B = C-1

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

{{{A+C-1+C+A/2 = 13}}}

{{{A+2C+A/2 = 14}}}

{{{2A+4C+A = 28}}}

{{{3A+4C = 28}}}

{{{3A = 28-4C}}}

{{{A = (28-4C)/3}}}

All digits are between 0 and 9 inclusive

{{{0 <= A <= 9}}}

Substitute:
{{{0 <= (28-4C)/3 <= 9}}}
Multiply through by 3
{{{0 <= 28-4C <= 27}}}
Subtract 28 for all three sides
{{{-28 <= -4C <= -1}}}
Divide through by -4 reversing inequalities
{{{7 >= 4C >= 1}}}
{{{1.75 >= C >= 0.25}}}
The only digit between 1.75 and 0.25 is 1 

Therefore C = 1 and since

{{{A = (28-4C)/3}}}
{{{A = (28-4(1))/3}}}
{{{A = (24)/3}}}
A = 8
and
B = C-1 = 1-1 = 0 
and
D = {{{A/2}}} = {{{8/2}}} = 4

So "ABCD" = 8014

Edwin</pre>