3 Brothers Riddle

Let $a<b<c$ be the ages of all brothers respectively. We can be sure that there would be no cases of twins, where some ages equal, because it will result in a remainder of $0$ for some cases, which is not under such constraint. Also, given the remainders of $6$ for all modulos $a,b,c$ , we can conclude that $6<a<b<c$ as well.

Furthermore, from the wiseman's wording, we can set up modular equivalences, as followed:

$ab \equiv 6 \pmod{c}$ $bc \equiv 6 \pmod{a}$ $ca \equiv 6 \pmod{b}$

Then suppose that $b-a =x$ and $c-b = y$ for some positive integers $x,y$ . That will make \(c-a = x+y). We can rewrite the above equivalences into these:

\[-(x+y)(-y) \equiv y(x+y) \equiv 6 \pmod{c}\] $x(x+y) \equiv 6 \pmod{a}$ $y(-x) \equiv -xy \equiv 6 \pmod{b}$

Rearranging the terms so that there remainders are zeros:

$y(x+y)-6 \equiv 0 \pmod{c}$ $x(x+y)-6 \equiv 0 \pmod{a}$ $xy+6 \equiv 0 \pmod{b}$

Given $x,y$ be positive integers, for the upper equivalences, $y(x+y) - 6 \geq 0$ and $x(x+y) - 6 \geq 0$ because $x,y \geq 1$ , making $y(x+y), x(x+y) \geq 2$ . Then $y(x+y)-6, x(x+y)-6 \geq -4$ , leaving only remainders of $-1, -2, -3, -4$ , none of which can be divisible by modulo $a,b,c > 6$ .

Now let us transform those equivalences into ordinary Diophantine equations, for some non-negative quotients $q$ :

$y(x+y)-6 = cq_1$ $x(x+y)-6 = aq_2$ $xy+6 = bq_3$

From the wiseman's clue, since no younger men could possess these constraints, we are attempting to find the lowest possible quotients. Yet, one issue remains: whether $q$ is allowed to be zero for the upper two equations.

Suppose $x(x+y) - 6 = 0$ ; $x(x+y) = 6 = 2\times 3 = 1\times 6$ . Given $x+y > x$ , there are two possible scenarios: $x=2; y=1$ or $x=1; y=5$ . For the former one, it will result in $y(x+y) -6 = -3 <cq_1$ , which is contradicted. For the latter scenario, it will result in $y(x+y)-6 = 24 = cq_1$ , but $xy+6 = 11= bq_3$ . Thus, $b$ must be $11$ , leading to $c = 11+5 =16$ , but $24$ is not a multiple of $16$ , thus contradicted as well.

Similarly, if $y(x+y) = 6 = 2\times 3 = 1\times 6$ , for $y=2; x=1$ , $x(x+y) -6 = -3 <cq_1$ and for $y=1; x=5$ , it will result in $b=11$ and $a=11-5 = 6$ . However, $a>6$ , so it is contradicted as well.

Hence, all quotients must be positive, and the lowest possible values occur when all quotients equal $1$ .

Let us explore whether these Diophantine equations will hold true:

$y(x+y)-6 = c = a+x+y$ _1 $x(x+y)-6 = a$ _2 $xy+6 = b = a+x$ _3

By subtracting equations (1)-(2) and (1)-(3), we will obtain:

$(x+y)(y-x) = x+y$ $y^2 -12 = y$

Hence, it would be obvious that:

$y-x =1$ $y^2 -y -12 = (y-4)(y+3) = 0$

Given $x,y$ as positive integers, we can conclude that $y=4$ and $x=3$ . Therefore, $a = 3(3+4) -6 =15$ . Then $b=15+3 = 18$ , and $c=18+4=22$ . Since all work with this set-up, there is no need for investigate other higher quotients.

Checking the answers, we can see that:

$15\times 18 = 22\times 12 + 6$ $18\times 22 = 15\times 26 + 6$ $22\times 15 = 18\times 18 + 6$

The wiseman's claim was true.

Finally, $a+b+c = 15+18+22 = \boxed{55}$ .