Question 577782
In a three digit number , the hundreds digit , the tens digit and the units digit are in descending order and in Arithmetic progression . Each digit of the number is multiplied by the sum of the other 2 digits . The sum of all such results is A. B is the product of the tens digit and the sum of all the digits and A = (4/3)B . Find the number of such numbers ..
<pre> 
Let
h = the hundredths digit
t = the tens digit
u = the units digit
 
Since the hundreds digit , the tens digit and the units digit are in 
descending order and in arithmetic progression, the units digit is
the smallest.  Let d = the common difference in the arithmetic
progression.  So that d will be positive, we take the units digit as
the first term in the arithmetic progression, and the digits are then

u, t = u+d, h = u+2d
<pre> 
Each digit of the number is multiplied by the sum of the other 2 digits. 
</pre> 
h(t+u) = (u+2d)(u+d + u) = (u+2d)(2u+d) = 2uČ+ud+4ud+2dČ = 2uČ+5ud+2dČ 
t(h+u) = (u+d)(u+2d + u) = (u+d)(2u+2d) = 2uČ+2ud+2ud+2dČ = 2uČ+4ud+2dČ
u(h+t) = u(u+2d + u+d) = u(2u+3d) = 2uČ+3ud
</pre> 
The sum of all such results is A. 
<pre>
A = 2uČ+5ud+2dČ + 2uČ+4ud+2dČ + 2uČ+3ud = 6uČ+12ud+4dČ
</pre>
B is the product of the tens digit and the sum of all the digits 
<pre>
B = t(h+t+u) = (u+d)(u+2d + u+d + u) = (u+d)(3u+3d) = 3uČ+3ud+3ud+3dČ = 3uČ+6ud+3dČ 
 
A = {{{4/3}}}B
 
6uČ+12ud+4dČ = {{{4/3}}}(3uČ+6ud+3dČ)
6uČ+12ud+4dČ = 4uČ+8ud+4dČ

Here we have a polynomial equation in
more than one variable in which all terms
are of the same degree.  In such equations
we must be careful. 

2uČ-4ud = 0
2u(u-d) = 0
 u(u-d) = 0
 
u = 0    u-d = 0
           u = d

For the case u = 0, u is independent of d,
and when we substitute u = 0 in

6uČ+12ud+4dČ = 4uČ+8ud+4dČ

we get identity 4dČ = 4dČ.  So we can
proceed to get solutions.
 
t = u+d = 0+d = d 
h = u+2d = 0+2d = 2d

d   h=2d  t=d   u=0    number   
1    2     1     0       210
2    4     2     0       420
3    6     3     0       630
4    8     4     0       840

u = d   

Since u depends on the value of u, we
must find what is allowed in order that

6uČ+12ud+4dČ = 4uČ+8ud+4dČ

be an identity

If u = d

6dČ+12(d)d+4dČ = 4dČ+8(d)d+4dČ
  6dČ+12dČ+4dČ = 4dČ+8dČ+4dČ
          22dČ = 16dČ
           7dČ = 0

Which is true ONLY if d = 0

However if d = 0, and u = d, then 

u and d are both 0, and

t = u+d = 0+0 = 0
h = u+2d = 0+2(0) = 0

But 000 is not acceptable as a three
digit number.

Therefore there are exactly 4 solutions.

Edwin</pre>