Are you joking? pt.2

Mr. White's age is a number of 2 digits, $x$ and $y$ .

$$ The number we are talking about is $N=xyxyxy$ .

$$ We can write $N=10^4xy+10^2xy+xy=xy(10^4+10^2+1)=10101xy$

$$ $10101=3*7*13*37 \Rightarrow A=(3*7*13*37)xy$

$$ In conclusion his wife is 37 and his 4 son's are respectively 13,7,3 and 1 (don't forget about it) years old $\Rightarrow 1+3+7+13+37=61$

Let Mr White's age be $\overline {ab}$ . Reading the question, we formulate the following equation with "Writing his age 3 times in succession, we get a 6-digit number that is equal to the product of his age, his wife's age and his 4 sons' ages.":

$\overline{ababab} = \overline{ab} \times wabcd$

where w is the age of his wife, and a, b, c, d, are the ages of his sons.

Now, we divide the equation by $\overline{ab}$ to get the following: $10101 = wabcd$

Now, we find the prime factors of 10101, by seeing whether the primes up to about 40 divide 10101. This is why we were told that Mr White was "approximately forty years old", so that we can optimise our search for the prime numbers and assume that he was a typical family man. Thankfully, the largest prime was 37:

$10101 = 3 \times 7 \times 13 \times 37$ . Because these are the prime factors of 10101 (and they are to no power but 1), these must be their ages. Note that we only have three ages here, so one of his sons must be 1.

Thus, the answer is the sum of their ages

$1 + 3 + 7 + 13 + 37 = 61$

Must be nice having most of the ages in your family (bar one) to be prime!

It's pretty easy to see that the product of the ages of the wife and the 4 sons is 10101.

10101 = 101^2 - 100

= 101^2 - 10^2

= (101 -10)(101 +10)

= 91 * 111

= (13 * 7) * (3 * 37) * 1

The conclusion is obvious.

Let Mr Whites age be W, Let his wife's age be L, And his sons ages will be represented by A, B, C, D If we write Mr Whites two digit age three times, we realise we can express the result as 10000W+100W+W or 10101W, so '10101W=WMABCD' Dividing by W, we get 10101=MABCD and 10101=is the product of 1,3,7,13,37(prime factors plus a one to make 5 ages).Now we know the ages of his relations, and the sum is 61.

404040= 2^3 X 3 X 5 X 7 X 13 X 37; product of ages= 40 X S X W. Eliminate 40 from the prime factors of 404040. Remaining factors= 3X7X13X37, SUM of ages will be= 3+7+13+37+1=61

For all those who are not super mathematically inclined - here goes.

First, we want a number ababab such that it has a lot of factors (at least factors that are obvious to us).

So I went for 454545 because it obviously has 5 and 9 as factors.

Next you prime factorise 454545. You get 454545 = 9 5 3 7 13*37.

Hence 45 is the man's age, 37 is his wife's, 13,7,3 and 1 are his 4 son's ages. And 37+13+7+3+1=61