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Question 253088: find a three-digit positive integer such that the sum of all digits is 14, twice the hundreds digit plus the tens digit equals the ones digit, and, if the digits are reversed, the new number plus the original number equals 1090?
Found 2 solutions by richwmiller, ankor@dixie-net.com:
Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website! a+b+c=14
2a+b=c
100a+10b+c+100c+10b+a=1090
248 is the original number
a=2
b=4
c=8
check
248+842=1090
2+4+8=14
2*2+4=8
4+4=8
ok
Apparently, there is another set of solutions which I missed.
941 and 149 do add up to 1090
and 1+4+9=14
2a+b=8
But 2*1+4=6 and not 8
So I didn't miss a solution after all.
Answer by ankor@dixie-net.com(22740)  (Show Source):
You can put this solution on YOUR website! find a three-digit positive integer such that the sum of all digits is 14, twice the hundreds digit plus the tens digit equals the ones digit, and, if the digits are reversed, the new number plus the original number equals 1090?
:
Let x = the 100's digit
Let y = the 10's digit
Let z = units
:
The original 3 digit number = 100x + 10y + z
:
Write an equation for each statement:
:
"find a three-digit positive integer such that the sum of all digits is 14,"
x + y + z = 14
:
" twice the hundreds digit plus the tens digit equals the ones digit,"
2x + y = z
2x + y - z = 0
:
Use these two equation to eliminate y
2x+ y - z = 0
x + y + z = 14
-----------------Subtraction eliminates y
x - 2z = -14
which is
x = 2z - 14
:
At this point we can see the no. of positive integers that satisfy this equation
are only two
z = 9, x = 4, then y = 1
z = 8, x = 2; then y = 4
:
check the 2nd set of solutions (248) in the statement:
"if the digits are reversed, the new number plus the original number equals 1090?"
842 + 248 = 1090
:
The other set of solutions will not add up to 1090
:
the 3 digit number = 248
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