More understandable explanation for Cat-Rabbit-Dog quiz

Question

This is a quiz from Brilliant.org Algebra Fundamentals courses.

Quiz: At Step 40, how many cats will there be in the 7th row?

Answer: 14

Explanation given by Brilliant.org is not clear to me so if someone can give a better explanation it will be helpful.

Explanation by Brilliant.org

In rows where dividing by 3 has a remainder of 1, the pattern is cat-rabbit-dog.

7 when divided by 3 has a remainder of 1, so cats occur at 1, 4, 7, etc., i.e. at points where dividing by 3 has a remainder of 1.

In step 40, the grid is 40 columns wide. The full cat-rabbit-dog pattern happens for $39\div 3=13$ times, but since the pattern at row 7 starts with a cat, column 40 has one extra cat. So there are $13 + 1 = 14$ cats.

Note: A more compact way of saying "dividing $x$ by 3 has a remainder of 1" is $x \mod 3 = 1$.

Answer 1

By reading all the answers I able to understand more and here is my simple explanation.

• Its a sequence.
• 1^st row and 4^th row were same in Step 4, so the pattern is same in every 3^rd row, pattern in 1^st and 4^th row is Cat-Rabbit-Dog-Cat-Rabbit-Dog-...
• In Step n we have n² animals and n animals in each row.
• Question is, in Step 40 how many cats are in row 7^th?
• In Step 40 each row has 40 animals.

Answer 2

In Step 40, there will be $40$ rows each containing $40$ animals. In each row, each group of three consecutive animals consists of one of each type. You can separate the seventh row into the first animal followed by $13$ groups of three. The $13$ groups of three will contain $13$ cats, so the answer will be $14$ if the first animal in the seventh row is another cat, and $13$ otherwise.

Counting down the first column, we see that the first, fourth, seventh, etc. (going up by $3$ each time) are cats. So the answer is $14$, because there is an extra cat at the start of the $7$th row.

What the explanation is trying to add is that you would get the same answer if the row was changed to any other number in the sequence $1,4,7,...$ (but no bigger than $40$). These numbers are the numbers that leave a remainder $1$ when divided by $3$. Other rows - those whose position divides exactly by $3$ or leaves remainder $2$ - will not have a cat at the start, so would have $13$ cats.

Answer 3

Apparently, in step $n$, there is an animal in position $(r,c)$ iff $1\le r,c\le n$, and that animal is a cat iff $r+c\equiv 2\pmod 3$. So for $n=40$ and $r=7$, we want to find the number of $c$ with $1\le c\le 40$ and $7+c\equiv 2\pmod 3$. Those $c$ have the form $c=1+3k$ where $k$ is allowed to range from $0$ (making $c=1$) up to $13$ (making $c=40$). So there are $$14$$ valid choices for $k$.

Answer 4

It strikes me that this is a case where talking about division and remainders clouds the issue. I would explain things as follows:

Every third row is identical, so the seventh row is the same as the fourth is the same as the first.

At the fortieth step, the first row will have $13$ groups of three (cat, rabbit, dog) and one extra cat, for a total of $14$ cats. So the seventh row will also have $14$ cats.