The Baker Vs Annoying Customer

For the cake starting prices, suppose that the four integers used in the ratios are $a,a+1,a+2,a+3$ , so we are playing with three ratios, $\tfrac{a+1}{a}$ , $\tfrac{a+2}{a+1}$ , $\tfrac{a+3}{a+2}$ . If $r_{SM}, r_{MB}, r_{SB}$ are the ratios of the costs of small to medium, medium to big and small to big, then $r_{SB} = r_{SM}r_{MB}$ is greater than both $r_{SM}$ and $r_{MB}$ . If the two ratios $r_{SM} = r_{MB}$ then the equation of ratios $r_{SB} = r_{SM}r_{MB}$ must be $\tfrac{(a+3)^2}{(a+2)^2} = \tfrac{a+1}{a}$ , which has no solutions in integers. Thus the two ratios $r_{SM},r_{MB}$ are distinct, and so the equation of ratios $r_{SB} = r_{SM}r_{MB}$ is $\tfrac{a+1}{a} = \tfrac{a+2}{a+1}\tfrac{a+3}{a+2} = \tfrac{a+3}{a+1}$ , and hence $a=1$ is the only positive integer solution. Thus $r_{SM},r_{MB} = \tfrac32, \tfrac43$ in some order, with $r_{SB} = \tfrac21=2$ . There are thus two cases: the small/medium/big prices being $2n,3n,4n$ for some positive integer $n$ , or $3n,4n,6n$ for some positive integer $n$ .

Suppose that prices are of the form $2n,3n,4n$ . Then the opening day prices are $2n-1,3n-1,4n-1$ , and $9n-2$ cannot be achieved, but $9n-1$ can be achieved using these opening day prices. Since $7$ can be achieved using $1,2,3$ , $n \neq 1$ . Since $16$ can be achieved using $3,5,7$ , $n \neq 2$ . Since $25$ can be achieved using $5,8,11$ , $n \neq 3$ . Since $35$ cannot be achieved using $7,11,15$ , $n \neq 4$ . Hence we deduce that $n \ge 5$ . Since $9n-1$ can be achieved using $2n-1,3n-1,4n-1$ , we can find nonnegative integers $a,b,c$ such that $a(2n-1) + b(3n-1) + c(4n-1) \; = \; 9n-1$ , and hence $(2a + 3b + 4c - 9)n \; = \; a+b+c-1$ Thus $a+b+c-1 \ge 5(2a+3b+4c-9)$ , so we have $2a + 3b + 4c \ge 10 \hspace{2cm} 9a + 14b + 19c \le 44$ Thus $9(10 - 3b - 4c) \le 18a \le 2(44 - 14b - 19c)$ , and so $b + 2c \le -2$ . Thus there are no solutions to these inequalities in nonnegative integers. This case is impossible.

Thus the prices must be of the form $3n,4n,6n$ , making the opening day prices $3n-1,4n-1,6n-1$ , and $13n-2$ cannot be achieved, but $13n-1$ can be achieved using the opening day prices. Since $11$ can be achieved using $2,3,5$ , $n \neq 1$ . Since $24$ can be achieved using $5,7,11$ , $n \neq 2$ . Suppose now that $n \ge 4$ . Since $13n-1$ can be achieved, we can find nonnegative integers $a,b,c$ such that $a(3n-1) + b(4n-1) + c(6n-1) = 13n-1$ , and hence $(3a + 4b + 6c - 13)n \; = \; a + b + c - 1$ Thus we deduce that $a+b+c-1 \ge 4(3a + 4b + 6c - 13)$ , and hence $3a + 4b + 6c \ge 14 \hspace{2cm} 11a + 15b + 23c \le 51$ Thus $11(14 - 4b - 6c) \le 33a \le 3(51 - 15b - 23c)$ , and so $b + 3c \le -1$ . Thus there are no nonnegative integer solutions to these inequalities. Thus the only option left is that $n=3$ .

The original prices are $9,12,18$ , and the opening day prices are $8,11,17$ . The Frobenius number of $8,11,17$ is $37 = 8 + 11 + 17 + 1$ , so the answer is $\boxed{36}$ .

I will only use ratios r in the range 0<r<1

Suppose that the original prices were $p_1, p_2, p_3$ with $p_1<p_2<p_3$ , then the three ratios are $\{\frac{p_1}{p_2}, \frac{p_2}{p_3}, \frac{p_1}{p_3}\}$

Let the 4 consecutive numbers be $x, x+1, x+2$ and $x+3$ , then the possible fractions of 2 consecutive numbers are $\{\frac {x}{x+1}, \frac{x+1}{x+2},\frac{x+2}{x+3}\}$

First, we must construct a mapping between these sets.

Picking the smallest fraction from each set we must have $\frac{p_1}{p_3} = \frac{x}{x+1}$ .

The product of the other two fractions is $\frac{p_1}{p_2} \frac{p_2}{p_3} =\frac{ p_1}{p_3}$ so that we must have $\frac{x+2}{x+3} \frac{x+1}{x+2} = \frac{x}{x+1}$ which simplifies to

$(x+1)^2 = x(x+3)$ $\Leftrightarrow x^2+2x+1 = x^2+3x$ $\Leftrightarrow x=1$ so that the fractions must be $\frac12, \frac23, \frac34$

Now there are two possibilities:

Case A

$\frac{p_1}{p_2}= \frac{2}{3}$ and $\frac{p_2}{p_3}= \frac{3}{4}$ so that there is a positive integer k such that $p_1=2k,  p_2=3k , p_3=4k$

Discount prices are $p_1-1=2k-1, p_2-1=3k-1 , p_3-1=4k-1$

The amount on the counter is $9k-2$

Case B

$\frac{p_1}{p_2}= \frac{3}{4}$ and $\frac{p_2}{p_3}= \frac{2}{3}$ so that there is a positive integer k so that $p_1=3k, p_2=4k, p_3=6k$

Discount prices are $p_1-1=3k-1, p_2-1=4k-1 , p_3-1=6k-1$

The amount on the counter is $13k-2$

Conclusion In case B, with $k=3$ , the discount prices are $8,11,17$ . Construction of a sum of 37 is not possible, but I could construct 38 through 45 and hence all higher numbers by adding multiples of 8.

So the answer is $8+11+17=\boxed{36}$

Lets label the cake prizes (Big, Medium, small) = $ $A, B, C$ , with $A \geq B \geq C$ .

Clearly, if there are any equalities, the ratio would be $1:1$ and invalid.

$\frac{A}{C} = \frac{c_1+1}{c_1} = 1 + \frac{1}{c_1}, \frac{B}{C} = \frac{c_2+1}{c_2} = 1 + \frac{1}{c_2},$ so $c_2 > c_1$ .

For there to be only 4 consecutive integers between the 3 ratios, 1) $c_2 = c_1 + 1$ OR 2) $c_2 = c_1 + 2$ .

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1) $C = k(k+1) n, B = k(k+2) n, A = (k+1)^2 n$

$\frac{A}{C} = \frac{k+1}{k}$

$\frac{B}{C} = \frac{k+2}{k+1}$

$\frac{A}{B} = \frac{(k+1)^2}{k(k+2)} = \frac{k^2 + 2k + 1)}{k^2 + 2k}$

These are 4 consecutive numbers if $k^2 + 2k = k(k+2) = k - 1$ OR $k + 2$ .

$k(2k+1) \geq (k+2)$ with equality iff $k = 1$ .

So $(A,B,C) = (4,3,2) n$ .

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2) i) Consider the case when $(c_2, c_1)$ = 2:

$C = 2k(k+1) n, B = k(2k+3) n, A = (k+1)(2k+1) n$

$\frac{A}{C} = \frac{2k+1}{2k}$

$\frac{B}{C} = \frac{2k+3}{2k+2}$

$\frac{A}{B} = \frac{(k+1)(2k+1)}{k(2k+3)} = \frac{2k^2 + 3k + 1)}{2k^2 + 3k}$

These are 4 consecutive numbers if $2k^2 + 3k = k(2k+3) = 2k$ OR $2k + 1$ OR $2k + 2$ .

But $k(2k+3) \geq (2k+3)$ . So no solutions.

ii) $(c_2, c_1)$ = 1: (With $2 \nmid k$ )

$C = k(k+2) n, B = k(k+3) n, A = (k+1)(k+2) n$

$\frac{A}{C} = \frac{k+1}{k}$

$\frac{B}{C} = \frac{k+3}{k+2}$

$\frac{A}{B} = \frac{(k+1)(k+2)}{k(k+3)} = \frac{k^2 + 3k + 2)}{k^2 + 3k} = \frac{k \frac{k+3}{2} + 1}{k \frac{k+3}{2}}$ .

These are 4 consecutive numbers if $k \frac{k+3}{2} = k$ OR $k + 1$ OR $k + 2$ .

$k \frac{k+3}{2} > k$ .

$k \frac{k+3}{2} = k + 1 \implies k(k+3) = 2(k+1) \implies k^2 + k - 2 = 0 \implies k = 1$ .

$k \frac{k+3}{2} = k + 2 \implies k(k+3) = 2(k+2) \implies k^2 + k - 4 = 0$ , which has no integral solutions.

So $(A,B,C) = (6,4,3) n$ .

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Possibilities are $(A,B,C) = (4,3,2) n$ OR $(A,B,C) = (6,4,3) n$ .

Let $A' = A - 1$ etc, $N = A' + B' + C'$ .

If the difference between 2 sizes is 1, $N + 1$ is reachable.

For $N + 2$ to be reachable:

a) $xB' + yC' = N + 2$ $x,y \geq 0$ .

b) $A' + 2B' = N + 2$

c) $A' + yC' = N + 2$ , $y > 1$

d) $2A' = N + 2$

e) $2A' + B' = N + 2$

f) $2A' + C' = N + 2$

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b) $B' = C' + 2$

1) $n=2$ : $(A',B',C') = (7,5,3)$

but $3 + 3 + 5 + 5 = N + 1$ .

2) $n=2$ : $(A',B',C') = (11,7,5)$

but $5 + 5 + 7 + 7 = N + 1$ .

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c) $(y-1)C' = B' + 2$

$C' < B'$ ,

1) $2C' - B' = n - 1$

$n = 3$ : $(A',B',C') = (11,8,5)$

but $5 \times 5 = N + 1$ .

2) $2C' - B' = 2n - 1$ X

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d) $A' = B' + C' + 2$

$A' - B' - C' = 1 - n$ X

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e) $A' = C' + 2 \implies A' = B' + 1$ X

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f) $A' = B' + 2$

1) $n=2$ X

2) $n=1$ X

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finally a) $xB' + yC' = A' + 2$ where $x,y \geq -1$ .

i) $x = -1$ : $yC' = A' + B' + 2$ :

1) $(2n-1)y = 7n$

$y = 4$ : $n = 4$ : $(A',B',C') = (15,11,7)$

34 isn't reachable but neither is $38$ . X and any larger $y$ and $n$ will be smaller, and these have already been tested.

2) $(3n-1)y = 10n$

$y = 4$ : $n = 2$ X Any larger $y$ and $n$ will be smaller.

ii) $x = 0$ : $yC' = A' + 2$ :

1) $(2n-1)y = 4n + 1$

$y = 3$ : $n = 2$ X

2) $(3n-1)y = 6n + 1$

$y = 3$ : $n < 2$ X

iii) $x = 1$ : $yC' = A' - B' + 2$ :

1) $(2n-1)y = n + 2$

$y = 1$ : $n = 3$ X

2) $(3n-1)y = 2n + 2$

$y = 1$ : $n = 3$ : $(A',B',C') = (17,11,8)$

$36 = 17 + 11 + 8$

$37 X$

$38 = 2 \times 11 + 2 \times 8$

$39 = 17 + 2 \times 11$

$40 = 5 \times 8$

$41 = 17 + 3 \times 8$

$42 = 2 \times 17 + 8$

$43 = 11 + 4 \times 8$

$44 = 36 + 8$

$45 = 2 \times 17 + 11$

subsequent combinations can be made by adding $8$ to existing ones.

iv) $x \leq 2$ : $yC' \leq A' - 2B' + 2 < 0$ for 1) and 2), so $y = -1$ :

$xB' = A' + C' + 2$

1) $(3n - 1)x = 6n$ :

$x = 3$ : $n = 1$ X

2) $(4n - 1)x = 9n$ :

$x = 3$ : $n = 1$ X.

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This is all the possibilities, hence the only solution is $(A',B',C') = (17,11,8)$

Originally this is $18, 12, 9$ .