Answer to Question #237451 in Algebra for Mr. Lee

Question

The sum of the squares of the digits in a 3digit number is 98.

The sum of the first digit is increased and the last digit is the

middle digit. If the digits were reversed in order, the number

is increase by 198. What is the unknow number?

Answer 1

x² + y² + z² = 98

x+z=y

zyz+198=zyx

100x + 10y + z + 198 = 100z + 10y + x

99x - 99z + 198 = 0

x - z + 2 = 0

x = z - 2

x + z = y

z - 2 + z = y => 2z - 2 = y

x² + y² + z² = 98

(z - 2)² + (2z - 2)² + z² = 98

z² - 4z + 4 + 4z² - 8z + 4 + z² = 98

6z² - 12z + 8 = 98

3z² - 6z - 45 = 0

3z² - 15z + 9z - 45 = 0

3z(z - 5) + 9(z - 5) = 0

(3z + 9)(z - 5) = 0

z = 5 and -3 but z ≠ -3 digit should be taken in natural number.

z = 5

x = 3

y = 8

now, number is 385