The sum of the digits of a three-digit number is 11. If the digits are reversed, the new number is 46 more than five times the old number. If the hundreds digit plus twice the tens digit is equal to the units digit, then what is the number?

Answer:731 Explanation for problem:"The sum of the digits of a three-digit number is 11."x + y + z = 11The three digit number = 100x + 10y + zThe reversed number = 100z + 10y + x" If the digits are reversed, the new number is 46 more than five times the old number."100z + 10y + x = 5(100x + 10y + z) + 46100z + 10y + x = 500x + 50y + 5z + 46combine on the right0 = 500x - x + 50y - 10y + 5z - 100z + 46499x + 40y - 95z = -46"the hundreds digit plus twice the tens digit is equal to the units digit,"x + 2y = zx + 2y - z = 0 Three equations, 3 unknownsx + y + z = 11x +2y - z = 0-----------------Addition eliminates z2x + 3y = 11From the 2nd equation statement, we know that the 1st original digit has to be 12(1) + 3y = 113y = 11 - 23y = 9y = 3 is the 2nd digitthen1 + 3 + z = 11z = 11 - 4z = 7137 is the original numberCheck solution in the 2nd statementIf the digits are reversed, the new number is 46 more than five times the old number."731 = 5(137) + 46731 = 685 + 46