the digits are reverse of each other.
Let the father's age = 10t+u
Let the son's age = 10u+t
Neither t nor u can be 0, since the first digit of neither age can be 0.
the ages of a father and son add up to 55
(10t+u)+(10u+t) = 55
10t+u+10u+t = 55
11t+11u = 55
Divide through by 11
t+u = 5
the father is older than 11 when the son is born
So the father is more than 11 years older than his son
(I would certainly hope so! There aren't many 11 year-old
fathers.)
10t+u > 10u+t + 11
9t-9u > 11
9(t-u) > 11
t-u > 11/9
t-u > 1&2/9
t-u ≧ 2
From above,
t+u = 5
Solve for t
t = 5-u
Substitute in
t-u ≧ 2
5-u-u ≧ 2
5-2u ≧ 2
-2u ≧ -3
Divide both sides by -2, which reverses the inequality
u ≦ 1.5
u can DISABLED_event_only= 1, since it can't be 0.
t+1 = 5
t = 4
So the father is 10t+u = 10(4)+1 = 40+1 = 41
And the son is 10u+t = 10(1)+4 = 10+4 = 14
the son is currently older than 5.
We knew the son's age was a two-digit number; otherwise
his father's age couldn't be the reverse of the digits
of his age. So we did not need this information.
Edwin