A number problem!

Inspired by Lillypink I thought i'd add a number problem before i go to bed.

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?

Good Luck. Solution later when i wake up.

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

Wow.... Wasn't expecting someone to solve this so quickly and thoroughly too...

I better come up with more....

Quote:

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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

---

o_O

7 minutes and solves, dang

Quote:

---

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

---

Wanna be my tutor? Pays 10K an hour;)

Quote:

---

Originally Posted by Gluttony:464191

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

---

I remember doing this problem in 8th grade for extra credit lol.

Quote:

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Originally Posted by Fortress

o_O

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Omg. I hate math.