Once upon a time, there were $3$ brothers living in a small remote village, with no extra-ordinary features. Nonetheless, the old wiseman of the village once told all $3$ brothers that their age combination was so unique that no younger men on Earth could resemble: when the product of any $2$ brothers' ages was divided by the other one's age, the remainder would always be $6$ . Also, the eldest age is less than twice the youngest age.

What would be the sum of all $3$ brothers' ages?

The answer is 55.

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

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