As promised, here is the answer to Monday’s The Terrible Threes math Brain Twister!

First off, we should translate “what is the greatest value of x where n is an integer”. What the question is asking is how many 3’s can you divide out of 61! + 60! + 59!, or in other words how many factors of 3 does ‘n’ have.

Next, we’ll want to factor. The first step is to remove a 59! from each of the three numbers, since that is the biggest number in each: 59! [(61)(60) + (60) + 1].

Notice how (60) + 1 = 61. Notice as well that we already have ‘60’ 61’s (that’s what 61×60 means). By adding in the additional 61, we now have 61^2. Since ‘61’ is a prime, it has no factors of ‘3’, so we can focus our attention on 59!.

There are a lot of ‘3’s in 59!. First off, there are **19** multiples of ‘3’ (19×3 = 57). Secondly, there are 6 multiples of 9, which means an extra 6 factors of 3 (remember 9 = 3^2 and we already counted one of those threes when we found the multiples of ‘3’).

Finally, there are two multiples of 27 (27 and 54), which give us 2 additional factors of 3, since 27 = 3^3.

Therefore, **19 + 6 + 2 = 27**.

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Hey Chris,

First of all: I hate your math problems. But then again, I can’t let go if I see a new one coming up every Monday

For this one, I didn’t really understand the part where it says:

“There are a lot of ‘3’s in 59!. First off, there are 19 multiples of ‘3’ (19×3 = 57). Secondly, there are 6 multiples of 9, which means an extra 6 factors of 3 (remember 9 = 3^2 and we already counted one of those threes when we found the multiples of ‘3’).

Finally, there are two multiples of 27 (27 and 54), which give us 2 additional factors of 3, since 27 = 3^3″

How did you come up with the multiples? Did you went through every number and checked whether it is divisible by 3 or a multiple of 3? I find that rather exhausting – isn’t there a quicker way? Oh, and how likely is it that such questions appear on the GRE if you’re aiming for 165+ points?

Thanks