SOLUTION: I am a four-digit number. The number of my thousands is greater than my tens by four. My tens are two times the number of my ones. Hundreds have I one. I can be named by two di

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: I am a four-digit number. The number of my thousands is greater than my tens by four. My tens are two times the number of my ones. Hundreds have I one. I can be named by two di      Log On


   



Question 200248: I am a four-digit number.
The number of my thousands is greater than my tens by four.
My tens are two times the number of my ones.
Hundreds have I one.
I can be named by two different numbers.

Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
Let the number be THDO. Where T is the thousands digit, H the hundreds, etc
Given
T+=+D+%2B+4 T<= 9
D+=+2%2AO
H+=+1
Since T is <= 9, then D must be <=5
D is 2*O, so D must be even.
What are the even numbers less than 5? 0, 2, 4
Do D can be 0, 2 or 4
D = 2 * 0.
So O can be 0, 1 or 2
Possible numbers are
THDO
4100
6121
8142