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 ..
Options available are
a)4 b)3 c)2 d) More than 4
Found 2 solutions by Edwin McCravy, richard1234:
Answer by Edwin McCravy(20056)   (Show Source):
You can put this solution on YOUR website! 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 ..
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
Each digit of the number is multiplied by the sum of the other 2 digits.
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
The sum of all such results is A.
A = 2uČ+5ud+2dČ + 2uČ+4ud+2dČ + 2uČ+3ud = 6uČ+12ud+4dČ
B is the product of the tens digit and the sum of all the digits
B = t(h+t+u) = (u+d)(u+2d + u+d + u) = (u+d)(3u+3d) = 3uČ+3ud+3ud+3dČ = 3uČ+6ud+3dČ
A =  B
6uČ+12ud+4dČ = (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
Answer by richard1234(7193)  (Show Source):
You can put this solution on YOUR website! Suppose the digits are a, a-d, a-2d. Then,
 + (a-d)(2a-2d) + (a-2d)(2a-d))
Additionally, (3a-3d) = 3a^2 - 6ad + 3d^2) . The problem tells us that A = (4/3)B so
 \Rightarrow 2a^2 - 4ad = 0) (upon working out all the algebra)
This factors nicely to  = 0)
Either a = 0 or a-2d = 0 (a = 2d). a cannot equal 0 because we have a three digit number (either the digits are negative or the number is 000) so a = 2d. If this is the case, then the only possible numbers are
a=2, d=1 --> 210
a=4, d=2 --> 420
a=6, d=3 --> 630
a=8, d=4 --> 840
4 possible numbers, answer is A.
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