a 3 digit, and 1 digit number have a sum of 147, a different of 133 and a product of 980. find the numbers, and then give their quatient.
Let A, B and C be the three digits of the three-digit number, in order. Let D be the one-digit number.
100A + 10 B + C + D = 147
100A + 10 B + C - D = 133
2D = 14
D = 7 is the one-digit number
7*(100A + 10 B + C) = 980
(100A + 10 B + C)= 140
That tells you that A = 1, B = 4 and C = 0. The quotient of the numbers is 140/7 = 20
To solve this problem, let's first represent the numbers mathematically.
Let's say the 3-digit number is represented as XYZ, where:
X represents the hundreds digit,
Y represents the tens digit, and
Z represents the units digit.
Similarly, let's represent the 1-digit number as A.
According to the given information, we can set up the following equations:
Equation 1: XYZ + A = 147
Equation 2: XYZ - A = 133
Equation 3: XYZ * A = 980
To find the values of X, Y, Z, and A, we can solve these equations simultaneously.
First, let's rearrange Equation 1 and Equation 2 to express Z in terms of X, Y, and A:
Equation 4: Z = 147 - XY - A
Equation 5: Z = 133 + XY - A
Now, we can substitute Equation 4 into Equation 3:
(147 - XY - A) * A = 980
Expanding and rearranging this equation, we get:
A^2 - XYA - 147A + 980 = 0
Since we know A is a single-digit number, we can substitute different values of A (from 1 to 9) into this equation and solve for XY.
After finding the values of XY, X, Y, and A, we can find Z using either Equation 4 or Equation 5.
Finally, we can find the quotient by dividing the 3-digit number by the 1-digit number.
Please note that finding the exact solution requires solving the quadratic equation which may result in complex numbers.