SOLUTION: I'm a four-digit number! My 2nd digit is twice greater than my 3rd. The sum of all my digits is thrice greater than my last digit! The product of my 3rd and 4th digits is 12 times
Algebra ->
Customizable Word Problem Solvers
-> Numbers
-> SOLUTION: I'm a four-digit number! My 2nd digit is twice greater than my 3rd. The sum of all my digits is thrice greater than my last digit! The product of my 3rd and 4th digits is 12 times
Log On
Question 479353: I'm a four-digit number! My 2nd digit is twice greater than my 3rd. The sum of all my digits is thrice greater than my last digit! The product of my 3rd and 4th digits is 12 times greater than the ratio of my 2nd to 3rd. What am I? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! 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()
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, check it in the given statements
