W.G. re the ages of his sons:
“Each of their ages is one more than thrice the sum
of its digits.”
Assuming it is true, what are the ages?
Source: Charles W. Trigg, San Diego, Ca.
Let us suppose, for the sake of argument, that the age of at least one of the sons consist of 3 digits.
Since the maxim digit in each of unit, tens and hundreds place is 9, it follows that his maximum age must be 3(9+9+9)+1 = 82 which is <100. This is a contradiction.
Let us suppose that at least one of the sons has a single digit age, which is A(say).
Then, by the given conditions, we must have:
A=3A+1
=> A= -1/2, which is a contradiction.
Therefore, each of the ages of the sons must have two digits.
Let the common form of their ages be denoted by 10x+y
Then, by the given conditions:
10x+y =3(x+y)+1
=> 7x=2y+1
Now RHS <= 2*9+1 =19, so that x<= 19/7
or, x=1,2
x=1 gives 2y+1=7, so that: y=3
x= 2 gives 2y+1=14=> y=13/2, which is a contradiction.
So, 10x+y=13, and we can easily check that 13=3(1+3)+1
Then, APPARENTLY:,
EITHER : WG has precisely one son aged 13
OR: WG has more than one son aged 13.
However, we observe the fact that
(i) WG was asked about the ages of his "sons"
(ii) WG had replied, "Each of their ages ------"
This leads us to infer that:
o WG has more than one son and each of their ages is 13.
Edited on May 26, 2022, 3:23 am
Puzzle Solution
