Questions about the world of GMAT Math from other sources and general math related questions.
JackH825
Course Students
 
Posts: 7
Joined: Fri Jan 31, 2020 3:05 pm
 

Divisibility

by JackH825 Thu May 28, 2020 10:06 am

How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3 ?

15

16

17

18

19

What's the best way to approach the above?
Thanks!
Sage Pearce-Higgins
Forum Guests
 
Posts: 1336
Joined: Thu Apr 03, 2014 4:04 am
 

Re: Divisibility

by Sage Pearce-Higgins Fri May 29, 2020 4:27 am

Please observe the forum guidelines: to make this as useful as possible to others, please include the source of the problem and the correct answer.

When faced with a pattern, or counting cases, or unusual cases, I strongly recommend that you pick some examples. What's the first number that would give a remainder of 1 when divided by 3? If you haven't thought about that already, then you're probably making things too complicated. The first number is 1, then 4, then 7, and you can probably see a pattern. Actually, the answers show you that the number of cases is small enough that you could probably count them on your fingers or write them all out in less than a minute. The risk of missing a case means that I would probably take that crude approach.

However, more theoretically, every multiple of 3 is followed by a number that gives a remainder of 1 when divided by 3. So how many multiples of 3 are there up to 50? Simply divide by 3: 50/3 = 16. We need to make sure that the last number is still included, so 48 + 1 = 49, that's fine. Also, don't forget that we also need to add 1 as another case, making 17 in total.