More fun in 2016, Part 22

Let the number of lilies, roses and daisies be $x,y,z$ . According to the given information we get $2$ equations: $x+y+z=2016......(1)\\ 3x+2y+\frac{z}{2}=2016.....(2)$ Substituting value of $z$ from $(1)$ in $(2)$ and rearranging for $x$ , we get: $x=\dfrac{2016-3y}{5}$ And since x,y,z are positive integers( Obviously!!) and we have to maximize x we can think of making the numerator a multiple of $5$ as large as possible for us.. That occurs for $y=2$ and its value is $402$ . Substituting $x=402$ and $y=2$ in $(1)$ we get $z=\boxed{1612}$ .... which are required no. of Daisies....
Ya... I know it's kinda Trial and Error... But it's not that too bad... :)

This question is similar to this question , but the numbers becames ridiculously larger. (mutatis mutandis)

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For consistency, we convert all the measurement to units of kopeks. So we have the following two points

1) Vladimir wants to spend a total of 201600 kopeks on exactly 2016 flowers.

2) 1 lily costs 300 kopeks, 1 rose costs 200 kopeks, and 1 daisy costs 50 kopeks.

Let $L,R$ and $D$ denote the number of lillies, roses and daisies Vladimir bought respectively. Then from the first point, we have

$L +D + R = 2016 \qquad (1)$

And from the second point, knowing that Vladimir spent a total of 2016 kopeks, we have $300L + 200R + 50D = 201600$ , or equivalently (divide both sides by 50):

$6L +D + 4R = 4032 \qquad (2)$

So we have two diophantine equations. And because we know that Nadezhda loves lilies, that implies that Vladimir bought a non-zero amount of Lilies. In other words, $L > 0$ . And of course, $D,R\geq0$ .

Taking the difference of the second equation and the first equation, we have

$5L + 3R = 2016 \qquad (3)$

What's left to check is to find the maximum possible value of $L$ . We could just partition the number into the sum of two numbers: one is a multiple of 5 and the other is a multiple of 3, with the numerical value of the latter number minimized. A quick check shows that it's just $2016 = 2010 + 6$ . Thus $L = 2010/5 = 402, R = 6/3 = 2$ . Thus from $(1)$ , we have $D = 2016 - 402 - 2 = \boxed{1612}$ .

Here is a systematic solution, copied over from here

We are given the equations

$L+R+D=2016$

$3L+2R+D/2=2016$

The reduced row-echelon form (rref) of this system is

$L-1.5D=-2016$

$R+2.5D=4032$

meaning that $L=-2016+1.5D$ and $R=4032-2.5D$

Now we want $L,R,D$ to be integers, which means that $D$ is an even integer.

Also, $L$ and $R$ must be non-negative, which means that $-2016+1.5D\geq 0$ and $4032-2.5D\geq 0$ , or, $1612.8\geq D\geq 1344$

Since $L$ is an increasing function of $D$ , we want to maximize $D$ . Comrade Vladimir will buy $\boxed{1612}$ daisies, the largest even number $\leq 1612.8$ .