Brady is stacking 600 plates in a single stack. Each plate is coloured black, gold or red. Any black plates are always stacked below any gold plates, which are always stacked below any red plates. The total number of black plates is always a multiple of two, the total number of gold plates is always a multiple of three, and the total number of red plates is always a multiple of six. For example, the plates could be stacked with:

- 180 black plates below 300 gold plates below 120 red plates, or
- 450 black plates below 150 red plates, or
- 600 gold plates.

In how many different ways could Brady stack the plates?
(A) 5139
(B) 5142
(C) 5145
(D) 5148
(E) 5151
Solution 1

Let the number of black plates, gold plates, and red plates be $b, g$ and $r$, respectively $(b, g$ and $r$ are whole numbers).

Brady is stacking 600 plates, and so $b+g+r=600$, where $b$ is a multiple of $2, \mathrm{~g}$ is a multiple of 3 , and $\mathrm{r}$ is a multiple of 6 .

Rewrite this equation as $g=600-b-r$ and consider the right side of the equation.

Since 600 is a multiple of 2 , and $b$ is a multiple of 2 , and $r$ is a multiple of 2 (any multiple of 6 is a multiple of 2), then $600-b-r$ is a multiple of 2 (the difference between even numbers is even).

Since the right side of the equation is a multiple of 2 , then the left side, $g$, must also be a multiple of 2 .

We are given that $g$ is a multiple of 3 , and since $g$ is also a multiple of 2 , then $g$ must be an even multiple of 3 or a multiple of 6 .

Similarly, rewriting the equation as $b=600-g-r$ and considering the right side of the equation: 600 is a multiple of 6 , and $g$ is a multiple of 6 , and $r$ is a multiple of 6 , so then $600-g-r$ is a multiple of 6 (the difference between multiples of 6 is a multiple of 6 ).

Since the right side of the equation is a multiple of 6 , then the left side, $b$, must also be a multiple of 6 .

That is, each of $b, g$ and $r$ is a multiple of 6 , and so we let $b=6 B, g=6 G$, and $r=6 R$ where $B, G$ and $R$ are whole numbers.

So then the equation $b+g+r=600$ becomes $6 B+6 G+6 R=600$, which is equivalent to $B+G+R=100$ after dividing by 6 .

Each solution to the equation $B+G+R=100$ corresponds to a way that Brady could stack the 600 plates, and every possible way that Brady could stack the 600 plates corresponds to a solution to the equation $B+G+R=100$.

For example, $B=30, G=50, R=20$ corresponds to $b=6 \times 30=180, g=6 \times 50=300$, $r=6 \times 20=120$, which corresponds to Brady stacking 180 black plates, below 300 gold plates, which are below 120 red plates.

Since $B+G+R=100$, then each of $B, G$ and $R$ has a maximum possible value of 100 .

If $B=100$, then $G=R=0$.

If $B=99, G+R=100-B=1$.

Thus, $G=0$ and $R=1$ or $G=1$ and $R=0$.

That is, once we assign values for $B$ and $G$, then there is no choice for $R$ since it is determined by the equation $R=100-B-G$.

Thus, to determine the number of solutions to the equation $B+G+R=100$, we must determine the number of possible pairs $(B, G)$ which lead to a solution.

For example, above we showed that the three pairs $(100,0),(99,0)$, and $(99,1)$ each correspond to a solution to the equation.

Continuing in this way, we determine all possible pairs $(B, G)$ (which give $R$ ) that satisfy $B+G+R=100$.

| Value of $B$ | Value of $G$ | Number of plates: $b, g, r$ |
| :---: | :---: | :---: |
| 100 | 0 | $600,0,0$ |
| 99 | 0 | $594,0,6$ |
| 99 | 1 | $594,6,0$ |
| 98 | 0 | $588,0,12$ |
| 98 | 1 | $588,6,6$ |
| 98 | 2 | $588,12,0$ |
| 97 | 0 | $582,0,18$ |
| 97 | 1 | $582,6,12$ |
| 97 | 2 | $582,12,6$ |
| 97 | 3 | $582,18,0$ |
| $\vdots$ | $\vdots$ | $\vdots$ |
| $100-n$ | 0 | $6(100-n), 0,6 n$ |
| $100-n$ | 1 | $6(100-n), 6,6(n-1)$ |
| $100-n$ | 2 | $6(100-n), 12,6(n-2)$ |
| $\vdots$ | $\vdots$ | $\vdots$ |
| $100-n$ | $n$ | $6(100-n), 6 n, 0$ |

We see from the table that if the value of $B$ is $100-n$ for some whole number $n \leq 100$, then $G$ can equal any whole number from 0 to $n$ and so there are $n+1$ possible choices for $G$.

That is, when $B=100$, there is 1 choice for $G$, when $B=99$, there are 2 choices for $G$, when
$B=98$, there are 3 choices for $G$, and so on.

Each additional decrease of 1 in $B$ gives 1 additional choice for $G$ until we arrive at $B=0$ $(n=100)$, which gives $n+1=100+1=101$ possible choices for $G(0,1,2,3, \ldots, 100)$.

Therefore, the total number of solutions to $B+G+R=100$ is given by the sum $1+2+3+\cdots+99+100+101$.

Using the fact that the sum of the first $m$ positive integers $1+2+3+\cdots+m$ is equal to $\frac{m(m+1)}{2}$, we get $1+2+3+\cdots+99+100+101=\frac{101(102)}{2}=5151$.

Since each of these solutions corresponds to a way that Brady could stack the plates, there are 5151 ways that Brady could stack the plates under the given conditions.

## Solution 2

In a given way of stacking the plates, let $b$ be the number of groups of 2 black plates, $g$ be the number of groups of 3 gold plates, and $r$ be the number of groups of 6 red plates.

Then there are $2 b$ black plates, $3 g$ gold plates, and $6 r$ red plates.

Since the total number of plates in a stack is 600 , then $2 b+3 g+6 r=600$.

We note that the numbers of black, gold and red plates completely determines the stack (we cannot rearrange the plates in any way), and so the number of ways of stacking the plates is the same as the number of ways of solving the equation $2 b+3 g+6 r=600$ where $b, g, r$ are integers that are greater than or equal to 0 .

Since $r$ is at least 0 and $6 r$ is at most 600 , then the possible values for $r$ are $0,1,2,3, \ldots, 98,99,100$.

When $r=0$, we obtain $2 b+3 g=600$.

Since $g$ is at least 0 and $3 g$ is at most 600 , then $g$ is at most 200 .

Since $2 b$ and 600 are even, then $3 g$ is even, so $g$ is even.

Therefore, the possible values for $g$ are $0,2,4, \ldots, 196,198,200$.

Since $200=100 \times 2$, then there are 101 possible values for $g$.

When $g=0$, we get $2 b=600$ and so $b=300$.

When $g=2$, we get $2 b=600-3 \times 2=594$ and so $b=297$.

Each time we increase $g$ by 2 , the number of gold plates increases by 6 , so the number of black plates must decrease by 6 , and so $b$ decreases by 3 .

Thus, as we continue to increase $g$ by $2 \mathrm{~s}$ from 2 to 200, the values of $b$ will decrease by $3 \mathrm{~s}$ from 297 to 0 .

In other words, every even value for $g$ does give an integer value for $b$.

Therefore, when $r=0$, there are 101 solutions to the equation.

When $r=1$, we obtain $2 b+3 g=600-6 \times 1=594$.

Again, $g$ is at least 0 , is even, and is at most $594 \div 3=198$.

Therefore, the possible values of $g$ are $0,2,4, \ldots, 194,196,198$.

Again, each value of $g$ gives a corresponding integer value of $b$.

Therefore, when $r=1$, there are 100 solutions to the equation.

Consider the case of an unknown value of $r$, which gives $2 b+3 g=600-6 r$.

Again, $g$ is at least 0 and is even.

Also, the maximum possible value of $g$ is $\frac{600-6 r}{3}=200-2 r$.

This means that there are $(100-r)+1=101-r$ possible values for $g$. (Can you see why?) Again, each value of $g$ gives a corresponding integer value of $b$.

Therefore, for a general $r$ between 0 and 100, inclusive, there are $101-r$ solutions to the equation.

We make a table to summarize the possibilities:

| $r$ | $g$ | $b$ | \# of solutions |
| :---: | :---: | :---: | :---: |
| 0 | $0,2,4, \ldots, 196,198,200$ | $300,297,294, \ldots, 6,3,0$ | 101 |
| 1 | $0,2,4, \ldots, 194,196,198$ | $297,294,291 \ldots, 6,3,0$ | 100 |
| 2 | $0,2,4, \ldots, 192,194,196$ | $294,291,288 \ldots, 6,3,0$ | 99 |
| $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ |
| 98 | $0,2,4$ | $6,3,0$ | 3 |
| 99 | 0,2 | 3,0 | 2 |
| 100 | 0 | 0 | 1 |

Therefore, the total number of ways of stacking the plates is

$$
101+100+99+\cdots+3+2+1
$$

We note that the integers from 1 to 100 can be grouped into 50 pairs each of which has a sum of $101(1+100,2+99,3+98, \ldots, 50+51)$.

Therefore, the number of ways that Brady could stack the plates is $101+50 \times 101=5151$.

ANswer: (E)