# Problem of the Day

Marissa is responsible for watering 30 trees that have been planted in a circular pattern. Each day, she selects the first tree to water at random and then proceeds around the circle counter-clockwise, watering every tree until all 30 trees have been watered. Which of the following is a possible value for ?

A. 3
B. 7
C. 9
D. 27
E. 28

B. 7

See the Solution

### Solution

[latexpage]

This solution hinges on the concept of least common multiples(LCMs). If you are unfamiliar with what LCMs are or how to find them, you should watch our foundation video, “Least Common Multiples and Greatest Common Factors” first.

The issue here is that if Marissa is not careful with her choice of \$n\$, she will end up watering the same trees over and over again. If \$n=30\$, for example, she will just water that first tree repeatedly. To see this clearly, think about a case with a more manageable number of trees. Suppose she only has 4 trees to water. If she waters one, the \$4^{th}\$ one after that would be the same tree, as would the \$8^{th}\$, \$12^{th}\$, and so on. The \$n^{th}\$ tree after the first one will always be that same tree as long as \$n\$ is a multiple of 4.

If we apply this lesson to the situation with 30 trees, Â it is clear that she will need to water all 30 trees before she hits a multiple of 30. Let’s see what would happen if we let \$n=9\$ like in choice ‘C’. She would water the first tree and then tree number 9, 18, 27, 36, 45, 54, 63, 72, 81, 90. However, 90 is a multiple of 30, so the \$90^{th}\$ tree is just the first tree that she watered. Once she repeats that first tree, every \$9^{th}\$ tree will also be a repeat. To see how this relates to LCMs, let’s rewrite the above list as explicit multiples of 9. I’ll include the first tree as 9(0) and I will omit the last tree, 9(10), because it is a repeat:

9(0), 9(1), 9(2), 9(3), 9(4), 9(5), 9(6), 9(7), 9(8), 9(9)

That’s ten unique trees that she waters. To water all 30, though, she would need the number in the parentheses to get up to 29 before she starts repeating trees. If \$n=9\$ worked, then the first tree she repeats would be 9(30). However, that is notÂ the first repeat because the LCM of 9 and 30 is 9(10)=90. From this line of reasoning, we can deduce that for any value of \$n\$ that allows her to water all 30 trees, it must be true that the LCM of \$n\$ and \$30\$ is \$n(30)\$. If you check all of the answer choices, you will see that this is only true of \$n=7\$. This is essentially because 7 and 30 have no common factors other than 1. Â In general, the LCM of \$a\$ and \$b\$ is \$ab\$ if and only if the greatest common factor of \$a\$ and \$b\$ is 1.