Choco Balls Riddle

Since $85$ is not a multiple of $7$ or $9$ , this amount can not be put in terms of one type of box alone. Neither can it be put in terms of $14$ and $21$ because it would need to be a multiple of $7$ . Thus, the small box must be involved no matter what.

Then suppose $85 \equiv 7n \pmod{9}$ .

Then $85 \equiv 4 \equiv 7\cdot 7 \equiv 49 \pmod{9}$ .

Thus, $n = 7$ , and there is only one way to put $49$ Choco balls in terms of the bigger boxes:

$49 = (4\cdot 7) + (3\cdot 7) = 2\cdot 14 +21$ .

Hence, we need to buy $2$ middle boxes and $1$ large box, which leaves us $4$ small boxes, for $85-49 = 36 = 4\cdot 9$ .

Finally, the number of boxes = $4+2+1 = \boxed{7}$ .

The remainder after dividing 85 by 7 is 1.

$85 \mod 7 = 1$

The remainder after dividing 9 by 7 is 2.

$9 \mod 7 =2$

(The other boxed do not contribute, since 14 and 21 are divisible by 7)

It takes multiplying by 4 to get the right remainder:

$9\times4 \mod 7=2\times4 \mod 7=8 \mod 7 =1$

After taking off 4 boxed with 9 chocolates each, we are left with 49 to distribute between boxes size $2\times7$ and $3\times7$ .

$49/7=7$

7 distributed in chunks of 2 or 3 can be written as $7=2+2+3$ , two boxes of 14 each, one with 21.

Let , number of square boxes, hexagonal boxes and triangular boxes are $x,y,z$ respectively.

Then $9x+14y+21z=85 \\ 7(2y+3z)=85-9x$

It implies $7 | 85-9x \implies 7 | 1-9x \implies 7| 1-2x$

So, appropriate solution of $x=4$ .There are many many other solutions but this will cross the value $85$

$7(2y+3z)=85-36=49 \\2y+3z=7$

Here the only solution is $y=2; z=1$

So, total boxes required to buy $4+2+1=7$