SOLUTION: a 4 digit postitive integer such that the sum of the digits is 18 & my digits reversed are exactly 4 times greater than myself. What number is it?
Algebra.Com
Question 211802: a 4 digit postitive integer such that the sum of the digits is 18 & my digits reversed are exactly 4 times greater than myself. What number is it?
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 the digits is 18 & my digits
reversed are exactly four times greater than myself. What number is it?
:
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 digit of the number would be 8 (4*2=8); the 1st digit of the reversed number
:
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 (x + y = 8) by 6 and add
39x - 6y = -3
6x + 6y = 48
-------------- adding eliminates y
45x = 45
x = 1 is the 2nd digit
then, obviously since x + y = 8:
y = 7 is the 3rd digit
:
2178 is the number
:
4 * 2178 = 8712; the reversed number
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