SOLUTION: 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

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) = 2u�+ud+4ud+2d� = 2u�+5ud+2d�
t(h+u) = (u+d)(u+2d + u) = (u+d)(2u+2d) = 2u�+2ud+2ud+2d� = 2u�+4ud+2d�
u(h+t) = u(u+2d + u+d) = u(2u+3d) = 2u�+3ud

The sum of all such results is A.

A = 2u�+5ud+2d� + 2u�+4ud+2d� + 2u�+3ud = 6u�+12ud+4d�

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) = 3u�+3ud+3ud+3d� = 3u�+6ud+3d� A = B 6u�+12ud+4d� = (3u�+6ud+3d�) 6u�+12ud+4d� = 4u�+8ud+4d� 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. 2u�-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 6u�+12ud+4d� = 4u�+8ud+4d� we get identity 4d� = 4d�. 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 6u�+12ud+4d� = 4u�+8ud+4d� be an identity If u = d 6d�+12(d)d+4d� = 4d�+8(d)d+4d� 6d�+12d�+4d� = 4d�+8d�+4d� 22d� = 16d� 7d� = 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,

Additionally, . The problem tells us that A = (4/3)B so

(upon working out all the algebra)

This factors nicely to

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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