Originally Posted by
Gluttony
I remember seeing this in a math forum:
I'm a four-digit number!
a b c d
My 2nd digit is twice greater than my 3rd.
b = 2c
c = .5b
The sum of all my digits is thrice greater than my last digit!
a + b + c + d = 3d
The product of my 3rd and 4th digits is 12 times greater than the ratio of my 2nd to 3rd.
c*d = 12(b/c)
multiply both sides by c
c^2d = 12b
replace c with .5b
(.5b)^2d = 12b
.25^b^2d = 12b
multiply both sides by 4
b^2d = 48b
Divide both sides by b
bd = 48
Let's try: b=6, d=8, only pair of single digit factors
then
c = .5(6)
c = 3
a + b + c + d = 3d
a + 6 + 3 + 8 = 3(8)
a + 17 = 24
a = 24 - 17
a = 7
The number: 7638
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