The units digit of a two-digit number is 1 less than the square of the tens digit. The number reversed is 7 more than twice the original number. Find the original number.
" The units digit of a two-digit number is 1 less than the square of the tens digit "
the only possiblities are
10 , 23 , and 38
A quick check,
83 = 2(38) + 7
the original number is 38
or (the long algebraic way)
Let the unit digit be y
let the x digit be x
then y = x^2 - 1
Original number: 10x + y
reversed number: 10y + x
10y + x = 2(10x + y) + 7
19x -8y + 7 = 0
sub in y = x^2 - 1
8x^2 - 19x + 15 = 0
(x-3)(8x-5) = 0
x = 3 or x = a fraction
so x=3, then y= 9-1 = 8
the original number is 38
To solve this problem, let's use algebraic equations.
Let's assume the tens digit of the two-digit number is "x" and the units digit is "y".
According to the problem, we know that the units digit is 1 less than the square of the tens digit. So, we can write the equation:
y = x^2 - 1 (Equation 1)
We also know that the number reversed is 7 more than twice the original number. The original number can be represented as 10x + y, and its reverse is 10y + x. So, we can write the equation:
10y + x = 2(10x + y) + 7 (Equation 2)
Now, we have a system of equations (Equation 1 and Equation 2) that we can solve simultaneously to find the values of x and y, and hence the original number.
Let's solve this system of equations:
Substitute the value of y from Equation 1 into Equation 2:
10(x^2 - 1) + x = 20x + 2y + 7
Expanding and rearranging the terms:
10x^2 - 10 + x = 20x + 2y + 7
10x^2 - 19x - 2y + 17 = 0
Now, we have a quadratic equation in terms of "x". Let's solve it using factoring or the quadratic formula:
Factorizing the quadratic equation might be challenging, so let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 10, b = -19, and c = 17
x = (-(-19) ± √((-19)^2 - 4 * 10 * 17)) / (2 * 10)
= (19 ± √(361 - 680)) / 20
= (19 ± √(-319)) / 20
The square root of a negative number is not a real number, so we can see that there are no real solutions for "x" in this case. Therefore, there is no two-digit number that satisfies the given conditions.