A number consist of 3 digits.The middle digit is 0 and the sum of the first and last digits is 13.If the digits are reversed the number is increased by 297.Find the original number.MY SOLUTION: let x=the 1st digit & y=last digit x+y=13 i.e eqn1 & pls help
ok. you have
x+y = 13
The value of the number is 100x+y
So, reversed, it's 100y+x. So, your second equation is
100y+x = 100x+y + 297
Now just solve for x and y.
Great it is confusing some how
To solve this problem, we can use a system of equations.
Let's represent the three-digit number as "xyz," where x is the first digit, y is the second digit (which is 0), and z is the third digit.
We are given two conditions:
1. The sum of the first and last digits is 13: x + z = 13 (equation 1).
2. When the digits are reversed, the number is increased by 297: 100z + 10y + x = 100x + 10y + z + 297 (equation 2).
Since y = 0, we can simplify equation 2 to: 100z + x = 100x + z + 297.
Now, let's rearrange equation 1 to express x in terms of z: x = 13 - z.
Substitute this value of x into the simplified equation 2: 100z + (13 - z) = 100(13 - z) + z + 297.
Now, solve for z:
100z + 13 - z = 1300 - 100z + z + 297.
99z + 13 = 1597 - 99z.
99z + 99z = 1597 - 13.
198z = 1584.
z = 1584 / 198.
z = 8.
Now that we have z, we can substitute it back into equation 1 to find x:
x + 8 = 13.
x = 13 - 8.
x = 5.
Therefore, the original number is 508.