Candy Riddle

Let $x$ be the amount of candies in the boys' bag and $y$ be that in the girls' bag. We can set up the modular equivalences, as followed:

$x \equiv 4 \pmod{7}$ $y \equiv 5 \pmod{8}$ $x+y \equiv 2 \pmod{17}$

Alternatively, we can rewrite $x = 7p + 4$ and $y = 8q +5$ for some non-negative integers $p,q$ . Then there are three scenarios for these quotients $p,q$ .

First of all, if $p=q$ , the modular equivalence will be, as followed:

$x+y \equiv 15p + 4+5 \equiv 2 \pmod{17}$ $7 \equiv 2p \pmod{17}$

Since $2\times 9 \equiv 1 \pmod{17}$ , the modular inverse of $2$ is $9$ per modulo $17$ .

$9\times 2p \equiv p \equiv 9\times 7 \equiv 63 \equiv 12 \pmod{17}$

Because $p$ is non-negative, the minimal value of $p=q$ is $12$ . This results in $x=7\times 12 + 4 =88$ and $y=8\times 12 + 5 =101$ . The total is $88+101 = 189$ .

Now we will examine the other scenarios.

Second, if $p>q$ , then we can set up $p = q + n$ for some positive integer $n$ , and the modular equivalence will be, as followed:

$x+y \equiv 15q + 7n + 9 \equiv 2 \pmod{17}$ $7n+7 \equiv 2q \pmod{17}$ $2q \equiv 7(n+1) \pmod{17}$ $9\times 2q \equiv q \equiv 63(n+1) \equiv -5(n+1) \pmod{17}$ $q+5(n+1) \equiv 0 \pmod{17}$

In order to yield the least possible values of $x,y$ , the resulting number must be $17$ , the least positive multiple. That is $5(n+1) + q = 17$ .

Clearly, the number $5$ acts as a divisor with $n+1$ as the quotient and $q$ as the remainder. Therefore, $n+1 = 3$ ; $n=2$ ; $q=2$ . That will lead to $p = q+n = 4$ . Then $x=7\times 4 + 4 =32$ and $y =8\times 2 + 5 =21$ . The total is $32+21 = 53$ .

Definitely, the first scenario fails to be the least values, and so we move on to the last scenario.

If $q>p$ , then we can set up $q = p + m$ for some positive integer $m$ , and the modular equivalence will be, as followed:

$x+y \equiv 15p + 8m + 9 \equiv 2 \pmod{17}$ $7 \equiv 2p-8m \pmod{17}$ $7 \equiv 2(p-4m) \pmod{17}$ $9\times 7 \equiv 12 \equiv p-4m \pmod{17}$ $4(m+3) - p \equiv 0 \pmod{17}$

Similarly, as previously explained, $4(m+3) - p = 17$ although this time, the remainder is of negative value, so we just need to find the least multiple of $4$ that exceeds $17$ . That is obviously $20$ . Then it is not brutal to calculate that $4\times 5 - 3 = 17$ . Thus, $m+3 = 5$ ; $m=2$ ; and $p = 3$ . That lead to $q = 3+2 =5$ . Then $x=7\times 3 + 4 =25$ and $y =8\times 5 + 5 =45$ . The total is $25+45 = 70$ .

Again, the last scenario also fails to satisfy, making the second scenario the desired solution. Finally, the difference between the candies in both bags equals $32-21 = \boxed{11}$ .