Hi Math Wizard, could you please help me to solve this problem with your easiest solution:

The sum of the ages of the father and his son is 99. If the age of the son is added to the inverted age of the father the sum is 72. if the inverted age of the son is subtracted from the age of the father, the difference is 22. what are their ages? Ans: 74 & 25

# Age Problem: The sum of the ages of the father and his son is 99

November 10, 2016 - 6:22am

#1
Age Problem: The sum of the ages of the father and his son is 99

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Let the father's age be 10x+y, with x and y being the digits of his age. (Reversing digits you get 10y+x).

Let the son's age be 10a+b, with a an b being the digits of his age. (Reversing digits you get 10b+a).

The clues give us the 3 equations below:

(1) 10a+b + 10x+y=99, or son's age + 10x + y = 99,

(2) son's age + x +10y = 72, and

(3) father's age - reversed son's age = 22, or. -10a-b +10x+y = 22

Subtracting the second equation from the first, we get

9x - 9y = 27. ---> 9(x-y)=27 ---> x-y=3

Subtracting equation (3) from egquation (1), we get

11a+11b=77 ---> 11(a+b)=77 ---> a+b=7

The son's age is not a multiple of 9.

If you divide the son's age by 9,

you get a remainder of 7.

How do I know?

When you divide a number by 9,

the remainder is the sum of the digits,

or the sum of the digits of the sum of the digits,

or ... (keep adding until you get a single digit),

and if you get 9, then the number was divisible by 9.

Since the sum of the ages is a multiple of 9,

when you divide the father's age by 9, the remainder should be 9-7=2.

The sum of the digits of the father's age cannot be 2,

and it cannot be 20 either,

so it must be x+y=11.

Along with x-y=3, that tells you that

x=7 and y=4,

so the father is 74.

We can re-write equation (3) as

74 - reversed son's age = 22, so

74-22 = reversed son's age, or

reversed son's age = 52.

So, son's age = 25.

Thank you so much.. your such a genius..

81 16?

The sum should be 99