Beef, chicken or duck? So many choices

Since we can determine the exact total cost by decomposing (b pounds beef, c pounds chicken, d pounds duck) into a linear combination of (1 pound beef, 2 pound chicken, 0 pound duck) and (1 pound beef, 0 pound chicken, 2 pound duck),

$(b, c, d) = \lambda_1 (1, 2, 0) + \lambda_2 (1, 0, 2)$

$b = \lambda_1 + \lambda_2$

$c = 2 \lambda_1$

$d = 2 \lambda_2$

Note: $\lambda_1, \lambda_2$ do not need to be integers.

Thus, $b = \frac {c+d} {2}$

Since b is an integer, c and d must be of the same parity. Also, note that b is completely determined by c and d.

There are 11 even integers and 10 odd integers from 0 to 20, inclusive. Thus, the number of such possible triples is $11 \times 11 + 10 \times 10 = 221$ .

Since $2$ pounds of chicken and $2$ pounds of beef cost the same, $c$ and $d$ are symmetrical. For all $0 \leq b \leq 10$ , $c$ and $d$ can take the values $0 \leq c,d \leq 2b$ each exactly once which is $2b+1$ values. Sum up the cases for $0 \leq b \leq 10$ we have $\sum^{10}_{b=0}2b+1=121$ . For all $11 \leq b \leq 20$ , $c$ and $d$ can take the values $2b-20 \leq c,d \leq 20$ each exactly once which is $21-2(b-1)$ values. Sum up the cases for $11 \leq b \leq 20$ we have $\sum^{20}_{b=11}21-2(b-1)=100$ . Add these two cases and we have the total $221$ .

Let 1 pound beef,chicken,duck cost x,y,z cents respectively Here x+2y=800 given equation 1 Also x+2z=800 given equatiin 2 Equation 2-1 results as chicken 1 pound cost=duck 1 pound cost Or y=z Now for any (b,c,d) Total cost is bx+cy+dz=cost Also using y=z we get bx+(c+d)y=cost As we know x+2y=800 So c+d=2b Case 1 for b belongs to 0 to 10 So for b=k, k<=10 so c+d=2k number of solution =2k+1, valid from k=0 to k=10 applying sumation we get number of solution is 121 Case 2 For b between 11 to 20 C+d=2k, now as c and d are independently less than 20 so Replacing c by p+2k-20 And d by q+2k-20 So p+q=40-2k number of solution is 41-2k or 40-(2k-1) Applying sumation over k from 11 to 20 or sum of AP WITH FIRST TERM 19 AND LAST TERM 1 WITH 10 TERMS We get sum =100 Adding case 1 and case 2 we get 121+100=221

Let’s say that 1 pound of beef = B cents

1 pound chicken = C cents

1 pound duck = D cents

==> B + 2C = 648 we can see that RHS(648) is a multiple of 2. and 2C is also a multiple of 2. So, B must be a multiple of 2. So, let B = 2K

==> 2K + 2C = 648 ===> K+C = 324 ===> C = 324-K ==>2k + 2D = 926 ====> K+D = 463 ===> D = 463 – K

==> b pounds of beef, c pounds of chicken and d pounds of duck will cost: b B + c C + d D = b (2K) + c C + d D = b (2K) + c (324 – K) + d (463 – K) = K (2b – c – d) + 324c + 463d

but we need to know the exact value of this. For that to be true, the above equation should be independent of the variable K: =>2b – c -d = 0 => 2b = c + d

for b=0 –> c+d=0 ==> 1 solution (c = 0) for b=1 –> c+d=2 ==> 3 solutions (c=0 or 1 or 2) for b=2 –> c+d=4 ==> 5 solutions … and so on, till b=10 –> c+d = 20 ==> 21 solutions

but for b=11 -> c+d = 22 ==> 19 solutions (c=2 or 3 or …. or 19 or 20) b=12 –> c+d = 24 ===> 17 solutions (c=4 or 5 or .. or 19 or 20) …and so on till b=20 –> c+d = 40 ==> 1 solution

Thus, total solutions = (1+3+5 + …. + 19 + 21) + (19 + 17 + … + 1) = 121 + 100 = 221

Let x, y and z be the prices of beef, chicken and duck in cents respectively. So $x + 2y = 648$ and $x + 2z = 926$ . It is obvious that this equation cannot be solved (there are 2 statements and 3 variables). We have $y = 324 - \frac{x}{2}$ and $z = 463 - \frac{x}{2}$ .

Note that the only way we can determine the exact cost of b pounds of beef, c pounds of chicken and d pounds of duck is that the x variables in the expressions cancel out. So $b - \frac{c}{2} - \frac{d}{2} = 0$ .

Hence $c + d = 2b$ . Since any even value of c + d produces a valid solution for b, we only have to calculate the number of ordered pairs (c, d) such that c + d is even. Either both c and d are odd or both c and d are even. Therefore there are $10 \cdot 10 + 11 \cdot 11 = 221$ such ordered triplets (b, c, d).

p b=2n, p c=324-n, p c=463-n are the prices. The total price is bp b + cp c + dp d = 2bn + 324c - cn + 463 - dn in which n cancels out iff c+d = 2b Then, just do casework on b: 2b + 1 solutions for each value of b from 0 to 10 since c and d are bounded below by 0, and 21-2b solutions for each value of b from 11 to 20 since c and d are bounded above by 20. This can be easily computed in your head knowing the trick that the sum of the first n odd numbers is n2, so in this case we just have 102 + 112 quick computation.

The space of solutions is generated by the additions of the vectors m (1, 2, 0) and n (1, 0, 2) where m and n are integers between 0 and 20. Note that these vectors are independent, since m (1, 2, 0) + n (1, 0, 2)=(0, 0, 0) iff m=n=0. That means that for a pair (m, n) the addition m (1, 2, 0) + n (1, 0, 2) gives a unique vector (b, c, d). So, first we have to find out how many pairs of integers m, n with 0<=m, n<=20 there are, such as 0<= m + n <=20. Secondly we must find out how many choices for 0<=2m<=20 and 0<=2n<=20 we can get. In the first case there are 11 * 11 =121 pairs and in the second one 10 * 10 =100 choices. Adding them we get 221.

$x+2y= 648, x+2z=926, bx+cy+dz$ . If $b=1$ we can get 3 sets $(1,1,1) (1,0,2),(1,2,0)$ . We get $(1,1,1)$ by adding the 2 given equations. For $b=2$ , we get 5 values. For $b=3$ , we get 7 values and it increases in the same way up to $b=10$ where we get 21 values. For $b=11$ , we get 19 values and it starts to decrease and again comes to 3. Since $3+5+7+...+21+19+17..+3=219$ and there are two more values $(0,0,0),(20,20,20)$ so the answer is $221$ .

[Edits for clarity - Calvin]

Let the price of 1 pound of beef be $B$ , the price of 1 pound of chicken be $C$ , and the price of 1 pound of duck be $D$ . We are given the value of $B+2C$ , and the value of $B+2D$ . Note that we cannot determine the exact solution $(B, C, D)$ . For example, $(B, C, D) = (0, 324, 463)$ and $(2, 323, 462)$ are both possible solutions to the simultaneous equations.

We claim that one can determine the exact cost of $b$ pounds of beef, $c$ pounds of chicken and $d$ pounds of duck if, and only if, $2b=c+d$ . If: Since $2b=c+d$ , the cost of $bB+cC+dD$ is $\frac {c}{2} (B+2C) + \frac {d}{2} (B+2D)$ , which we can determine. Only if: Set $a= 2b - c - d$ . Since we can determine the cost $bB+cC+dD$ , we can determine the cost of $bB+cC+dD-\frac {c}{2}(B+2C)-\frac {d}{2}(B+2D) = (b- \frac {c}{2} - \frac {d}{2})B = \frac {a}{2}B$ . Since we already established that we cannot determine the cost of 1 pound of beef, we must have $a=0$ , hence $2b=c+d$ .

Now, we calculate the number of integer solutions to $2b=c+d$ subject to $0 \leq b, c, d \leq 20$ . For $0 \leq b \leq 10$ , we can set $c \$ to be any integer value from $0$ to $2b$ and $d = 2b-c$ . This would satisfy $0 \leq c, d \leq 20$ , so there are $2b-0+1=2b+1$ solutions for every $0 \leq b \leq 10$ . For $11 \leq b \leq 20$ , we can set $c$ to be any integer value from $2b-20$ to 20 and $d= 2b - c$ . This would satisfy $0 \leq c, d \leq 20$ , so there are $20 - (2b-20) + 1 = 41-2b$ solutions for every $11 \leq b \leq 20$ . Thus, the total number of solutions is

$1 + 3 + 5 + \ldots + 19 + 21 + 19 + 37 + \ldots + 3 + 1 = 11^2 + 10^2 = 221$