Find the minimum number of instructions Here, I need some help with solving question number 2. The statement of the question is huge and so I refrained from typing it. This is last year's ZIO paper. 
So, I thought that rather than vanishing bacterias and then again increasing it, we could keep $K$ bacterias as it is and decrease the number of instructions $A, B, C$ by 1 each. Then we can try to form the rest of the bacterias $i.e (N-K)$ using the 'modified' instructions. 
I tried but cannot reach $N$ and I also need to develop a generalized algorithm so that all the test cases are solved easily. Thanks in advance. 
This is not an ongoing test.
 A: If $n(A), n(B), n(C)$ are the number of times instructions $A, B, C$ are used respectivly, the answers are
(a) $n(A) = 1, n(B) = 5, n(C) = 2$
(b) $n(A) = 4, n(B) = 2, n(C) = 1$
(c) $n(A) = 7, n(B) = 2, n(C) = 2$
The final answer for each one is the sum of the instructions and the posted solution is in agreement with the above. 
Edit
Sketch of algorithm using (a) as example:
Total amoebas to add is $91$. Amoebas added by each instruction is $A=6, B=11, C=15$. Maximum number of times the largest (i.e. $C$) can be used is $6$. 
Trying $C$ a total of $6$ times gives $6 \times 15 = 90$ amoebas, resulting in remainder $91 - 90 = 1$ amoebas. Can $1$ amoeba be made by a combination of $A$ and $B$? No. 
Then try $C$ a total of $5$ times, giving remainder $16$. Can $16$ be made by a combination of $A$ and $B$? No.
Then try $4$ giving remainder $31$. Can $31$ be made by a combination of $A$ and $B$? No.
Then try $3$ giving remainder $46$. Can $46$ be made by a combination of $A$ and $B$? Yes! $n(A) = 4, n(B)=2, n(C) = 3$. This gives a total of $9$ instructions, but we don't know if this is the minimum number of instructions, so we continue.
Then try $2$ giving remainder $61$. Can $61$ be made by a combination of $A$ and $B$? Yes! $n(A) = 1, n(B)=5, n(C) = 2$. This gives a total of $8$ instructions, but we don't know if this is the minimum number of instructions, so we continue.
And so on. The way to determine if a remainder can be made by a combination of $A$ and $B$, is to use the same procedure as I just outlined but with the largest now being $B$ instead of $C$. 
A: All right, for part a, you start with 23 amoebas and need to finish with 114 for an increase of 91.  A increases the number of amoebas by 6, B by 11, and C by 15.  So you have the equation $6a+11b+15c=91$.  Considering this equation mod 3 yields 
$$2b\equiv1\pmod3,b\equiv2\pmod3$$.  If $b=3k+2$, our equation becomes
$$6a+33k+22+15c=91$$
$$6a+33k+15c=69$$
$$2a+11k+5c=23$$
$k\ge2$ yields no solutions, giving 2 possibilities.  $2a+5c=23$ gives a solution of $a=4,c=3$ for a total of 9 instructions.  If $k=1$, we have $b=5$ and $2a+5c=12$, yielding $a=1,c=2$.  This has $8$ instructions, the best we can do.
Part b might be easier.  Our equation is $6a+14b+15c=67$.  We quickly get $c\equiv1\pmod2$ and $b\equiv2\pmod3$.  Substituting as before,
$$6a+42k_1+28+30k_2+15=67$$
$$6a+42k_1+30k_2=24$$
Our k's are obviously both $0$, yielding $a=4,b=2,c=1$ for a total of $7$ instructions.
This method does not work as well on part c as all 3 coefficients are coprime.  Our equation is $8a+11b+25c=128$.  The best option may be to check all values of $c$, then checking mod 8.  If $c=0$, we have $b\equiv0\pmod8$.  Letting $b$ be as high as possible, we get $b=8,a=5$ for 13 instructions.  If $c=1,8a+11b=103,3b\equiv7\pmod8$.  Multiplying by 3 gives $b\equiv 5\pmod8$.  This gives $(a,b,c)=(6,5,1)$ for 12 instructions.  If $c=2,8a+11b=78,3b\equiv6\pmod8,b\equiv2\pmod8$ for $(7,2,2)$:$11$ instructions.  If $c=3,8a+11b=53,3b\equiv5\pmod8,b\equiv7\pmod8$: no solution.  Finally, if $c=4,8a+11b=28$ which again has no solutions.  The optimal solution is $a=7,b=c=2$.
