Algebra II Review Word Problem?
The sum of the digits of a three digit number is 9. When the order of the digits is reversed, the newly formed number is 396 greater than the original number. If the leftmost digit is one-third of the middle digit, what is the number?
If you can please explain the work for me thanks in advance.
Comments
There are only two possibilities 135 and 261 that meet condition 1 and condition 3. 135 also meets condition 2 so its 135.
If the first digit of the original number is a, the second b and the third c, then the original number is
100a + 10b + c
The reversed number is 100c + 10b + a
100a + 10b + c + 396 = 100c + 10b + a
99a -99c + 396 =0
99(c-a) = 396
c-a = 396/99 = 4, => c = a+4
We also know that a + b + c = 9 , and the leftmost digit of the number (a) is 1/3 of the middle number (b), 3a = b
a + b + c = 9
a + 3a + a+4 = 9
5a = 5
a = 1
c = 5
b = 3
The original number was 135, the middle number is 3
The number is 135.
I started by looking at the "leftmost digit is one-third of the middle digit.." So, the number can be 135 and 261, since 1 is one-third of 3, and 2 is one-third of 6.
135 reversed is 531, and 531-135=396.
261 reversed is 162, and 162 is obviously not 396 more than 261, so 135 is the answer.
The number is 135.
_________
Let the number be abc.
We have:
a + b + c = 9
a = b/3
(100c + 10b + a) - (100a + 10b + c) = 396
From the second equation:
a = b/3
b = 3a
From the third equation:
(100c + 10b + a) - (100a + 10b + c) = 396
100(c - a) + (a - c) = 396
100(c - a) - (c - a) = 396
99(c - a) = 396
c - a = 396/99 = 4
c = a + 4
Plug these values back into the first equation.
a + b + c = 9
a + 3a + (a + 4) = 9
5a + 4 = 9
5a = 5
a = 1
Now solve for b and c.
b = 3a = 3*1 = 3
c = a + 4 = 1 + 4 = 5
The number is 135.
if the left digit is x,middle digit=3x ,last digit is y
the sum of all digits x+3x+y=9
4x +y=9..................(1)
the original number =100x+30x+y=130x+y
the reversed number=100y+30x+x=31x+100y
hence 31x +100y=130x+y +396
i.e. 99x-99y=-396
x-y=-4.......................................(2)
solving equations
x=1 y=5
the number=130*1+5=135
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A+B+C=9.
100A+10B+1C
100C+10B+1A-396=100A+10B+1C.
99C+0B-99A-396=0.
99C=99A+396.
divide by 99.
C=A+4.
A*3=B.
substitute
(A)+(3A)+(A+4)=9.
5A+4=9.
5A=5.
A=1.
1*3=B.
3=B.
C=1+4.
C=5.
135.
1+3+5=9. checked.