Word Problem with not enough information?

Question

I am trying to solve what is supposed to be a simple word problem in a riddle book, however I cannot seem to come up with enough info to set up a system of equations. Am I going about this the wrong way?

All it says before this is "There were two Sers, Ser Manfryd and Ser Raynard, the two families had more in common than they thought."

I set up the equations: with vars as:

• d = dad

  m = mom

  s = son

  (s-1) = daughter

I got

d + m + s + (s-1) = 100

(d^2)= (m^2) + (s^2) + (s-1)^2

After that I have no clue where to go. Any hints?

Answer 1

Suggestions:

From your first equation, $d+m=101-2s$. From your second equation, $d^2-m^2=2s^2-2s+1$.

Therefore $2s^2-2s+1=d^2-m^2=(d+m)(d-m)=(101-2s)(d-m)$.

So $d-m=\frac{2s^2-2s+1}{101-2s}$.

Try values of $s$ unitl you get an integer for $d-m$ (remembering that $s$ is the son's age should help figuring out reasonable values to try).

After a while you get (integer) values for $s$ and $d-m$, from which you should be able to find all the ages.

The other family can be dealt with similarly.

Answer 2

My method is fairly similar to paw88789's, but doesn't require guessing.

Let $z= 2s-1$, so that $d+m = 100-z$, and

$$ d-m = \frac{z^2+1}{2(100-z)} $$

Then we need $z^2+1 \equiv 0 \pmod{100-z}$. But $z^2+1 = 10001-(100-z)(100+z)$, so we need $10001\equiv 0 \pmod{100-z}$. We find the prime factorization $10001 = 73 \cdot 137$, so our only choices for $z$ are $27$ and $99$, but $99$ does not give answers consistent with the framing of the problem, so $z = 27$, $s=14$, $d+m = 73$, and $d-m = 5$.

Mod $4$ considerations in the second family tell us that all of the ages are even, so we can divide all the ages by two to get an almost identical setup - we end up having to find the prime factorization $2501=41 \cdot 61$ instead.

Note on computation: Finding primes $p$ that divide $10001$ is equivalent to finding primes $p$ such that $10^4 \equiv -1 \pmod{p}$, that is, primes such that $10$ has order $8 \pmod{p}$, so (by Fermat's Little Theorem) we need $p \equiv 1 \pmod{8}$. So the only primes we have to check less than $100$ are $17,41,73,89,97$.