How many of the three-digit numbers...

[hokie4life]

---

Members don't see this ad.

How many of the three-digit numbers containing no
digits other than 2, 3 or 4 are divisible by 3?
A. 2
B. 3
C. 4
D. 6
E. 9

answer = 9
I can't follow kaplan's explanation and plus it seems like it would take so long...help please ​

 

[Decan]

I'm not sure what Kaplan says...but the rule to seeing if something is divisible by 3 or is: add up all the numbers and if the sum is divisible by 3, the number is divisible by 3. For example, lets look at 93...9+3=12. 12 is divisible by 3 so 93 would be divisible by 3 as well (this was a very simple example, but you get the point).

I dont know if that helps or not...

 

[hokie4life]

thanks for replying! and, okay i def. understand this 3 mutliple rule but how do you apply it to get the answer faster? you still gotta list out everything that you think is a probable number and then check it with this rule right? that's the fastest way i can think of doing it still but am worried it's not fast enough. what do you think?

 

[Streetwolf]

222, 333, and 444 definitely work since they repeat the same digit 3 times.

From there you know that any number with each of those digits would work. So that's 6 more (234, 243, 324, 342, 423, 432; also just 3! = 6).

Then you have to sit and think for a second. You need the numbers with 2 of one and 1 of another. If you have two 3s and a 2 or a 4, this won't work. If you have two 2s and a 3 or a 4, this won't work either. And if you have two 4s and a 2 or a 3, this won't work.

So 6 + 3 = 9.