@PacManBoi

The answer is J since it is in the order of months.

Could you post how you did it? Considering you said "easier way," you should already have a solution. How did you do it?

As a hint for a quicker way to do it:
I would first use the sum of factors formula:

Let

$$n=p_1 ^{a_1}p_2^{a^2}p_3^{a_3} \ldots p_k^{a_k} $$

(right now latex isn't displaying properly; sometimes it will have a line of math in plain text below the math in latex. Hope you still understand what I'm trying to show you)

The sum of the factors of

$$n $$

is

$$(p_1^0+p_1^1+p_1^2+ \ldots + p_1^{a_1})(p_2^0+p_2^1+p_2^2+ \ldots + p^{a_2})\ldots (p_k^0+p_k^1+p_k^2 + \ldots + p_k^{a_k}) $$

I would then prime factorize each of the answer choices and then apply the formula.

Hope this helps! 😊

blue: 25%
white: 25%
red:25%
yellow:25%

@JoyfulSapling That's actually really cool

Explanation:
Of the 6 cookies Pat chooses, there are three types: chocolate chip, oatmeal, and peanut butter cookies. We also note that there can be 0 of a type. We can use the stars and bars formula to divide 6 objects into three categories (n is the number of objects, which is cookies in this case, and k is the number of categories):

n+k-1 choose k-1
= 6+3-1 choose 3-1
= 8 choose 2
= 28

That is why the answer is 28.

Please find below the proof for the stars and bars formula.
Let's use the cookies problem from above to help us with the proof. Right now, n=6 (6 cookies) and k=3 (3 types of cookies).
Putting 6 cookies into 3 categories looks like this:
costume1.png
There are 8 objects total in that picture (6 cookies + 2 bars).
Let's say we remove all objects and hold them. There are 8 spots to put our 8 objects. We can choose 2 spots of these 8 spots to put our bars, and the cookies go into the remaining 6 spots. Therefore, there are 8 choose 2 = 28 different combinations of 6 cookies.

So, in the formula n+k-1 choose k-1, n+k-1 represents the total number of spots to put the objects and the bars, and k-1 represents the number of bars needed to split the objects into k categories (for example, in our problem, 2 bars were needed to split the 6 cookies into 3 categories).

Yes very fun and totally not pain and suffering. 🙂

@smilingtadpole-1 In this problem, we do not care whether 11 has one or two 1s. What matters is the fact that 11 contains a 1. When you solved this problem, you assumed that because 11 had two ones, it should be counted twice, which is incorrect. I hope this helps!(also charge your device lol)

Thanks for the explanation. I guess I didn’t think that the card had numbers on both sides. But it makes sense now.

-1x-2+3-4+(5x6-7^2)- -18=?

@versatilemole i could give as many solutions as i can to this problem, and then people can add on solutions and count how many ways we can make 100.

Not really 🙂 Notice how Elmo has three blocks. It wouldn't be $$\binom{9}{3}$$ because you use a separate block every time so you would have to choose the first block, then the second, and so on. Try to see where you can go from there 😉

Note: Sorry this is so late I doubt you're active on this forum anymore.

@knowledgeabledog Thanks!

@rationaltiger Correct!

=1+11+111 times 68
=12+548
=560

24+3/4(24)(2)+(3/4^2)(24)(2)+(3/4^3)(24)(2)+.....
(The x2 is because the ball drops the same distance up and down)

This is an infinite arithmetic sequence.

Let S = 3/4(24)(2)+(3/4^2)(24)(2)+... = 36+(3/4)(36)+(3/4^2)(36)+....
So 4/3S=48+36+(3/4)(36)+....
So 4/3S-S=1/3S=48
So S=144
So S+24=168?