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Question 210655: I am a four-digit positive integer such that the sum of my digits is 18 and my digits reversed are exactly four times greater than myself. What number am I?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! a four-digit positive integer such that the sum of my digits is 18 and my digits
reversed are exactly four times greater than myself. What number am I?
:
We know the 1st digit has to be 1 or 2, (multiplying any other number by 4
would give us a 5 digit number)
:
Assuming it's 2, then the last number would be 8 (4*2=8)
:
let the middle digits be x & y:
:
2 + x + y + 8 = 18
x + y = 18 - 10
x + y = 8
:
2000 + 100x + 10y + 8 = the number
:
"digits reversed are exactly four times greater than myself.
4(2000 + 100x + 10y + 8) = 8000 + 100y + 10x + 2
8000 + 400x + 40y + 32 = 8000 + 100y + 10x + 2
400x + 40y + 8032 = 100y + 10x + 8002
400x - 10x + 40y - 100y = 8002 - 8032
390x - 60y = - 30
Simplify, divide by 10
39x - 6y = -3
Multiply the 1st equation by 6 and add
39x - 6y = -3
6x + 6y = 48
-------------- adding eliminates y
45x = 45
x = 1 is the 2nd digit
then, obviously
y = 7 is the 3rd digit
:
2178 is the number
:
4 * 2178 = 8712; sorry about that!!
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