BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

arithmetic

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 123
Joined: Mon Feb 07, 2011 12:11 pm
Followed by:1 members

arithmetic

by rupsk » Fri Jul 22, 2011 3:23 pm
How many positive integers between 200 and 300 (both inclusive) are not divisible by 2, 3 or 5?

A. 3

B. 16

C. 75

D. 24

E. 26

Is there any other way then below solution?

I had taken below approach
as it should not be divisible by 2 and 5 last digit can not be 2,4,6,8,0,5.
first digit will be 2.
and last digit will be 1,3,7,9
so if we take 1 as last digit then 2+1=3 now if it does not have to be divided by 3 the second digit should be such that there sum will not be multiple of 3. so it can have value 1,2,4,5,7,8
similarly found for 3,7 and 9 and got the total as 26 nos.
Top
Source: — Problem Solving |
User avatar
GMAT Instructor
Posts: 1031
Joined: Thu Jul 03, 2008 1:23 pm
Location: Malibu, CA
Thanked: 716 times
Followed by:255 members
GMAT Score:750

by Brian@VeritasPrep » Fri Jul 22, 2011 5:05 pm
Wow - really nice method, rupsk! And great question. I did it similarly to you but in a slightly different way so I might as well post.

Like you, I knew that the evens were all out, and those ending in 5 were all out, but that some even numbers and some multiples of 5 would also be multiples of 3. Since the 3s were the toughest, I did the 5s and 2s first:

1) There are 101 total numbers in this set, so I'll start with that

2) 51 of them are even (using the property of inclusive sets just like in step 1)

3) I've already counted the multiples of 5 that end in 0, so I only need the 10 that end in 5. So I'm already up to 61 numbers that don't count.

4) NOW it gets pretty interesting. But I've already eliminated the evens and those that end in 5, so I'm looking for the ODD multiples of 3 that don't end in 5. Well, I want to find a pattern so I'll start with:

201 --> and note that, by adding 6, I'll hit all the odd multiples of 3, so
206
213
219
225 ---BUT this one doesn't count...
231
237
243
249 ---> and here I notice a pattern. I go for 4 terms and then skip one because then I'm at a units digit of 5 when I get to:
255 (which doesn't count)

The pattern (which you could easily continue just writing out if you want but if this were a larger set you might want to just use the pattern) is:

2 sets of numbers that count in the first ten
2 set that count in the second ten
NONE in the third
start again

So I know I'll have: 2 + 2 + none (that's 225) + 2 + 2 + none (that's 255) + 2 + 2 + none (that's 285) + 2 (and that's it...that gets us to the 290s). So there are 14 multiples of 3 I need to eliminate.

So my totals are:

51 evens
10 that end in 5
14 that are odd multiples of 3 that don't end in t (using the pattern above)

75 numbers I can't use out of 101, so there are 26 numbers I CAN use, and the answer is E.


I think that's pretty similar to what you did, but maybe a different take if that clicks for people. With multiples/divisibility it's often convenient to look for patterns, so I've trained my mind to work that way and it's really helped...
Brian Galvin
GMAT Instructor
Chief Academic Officer
Veritas Prep

Looking for GMAT practice questions? Try out the Veritas Prep Question Bank. Learn More.
Top
Master | Next Rank: 500 Posts
Posts: 111
Joined: Tue Dec 30, 2008 1:25 pm
Location: USA
Thanked: 28 times
GMAT Score:770

by goalevan » Fri Jul 22, 2011 5:06 pm
You had almost the same approach as I did:

Out of the possible unit digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 we can knock out 0, 2, 4, 5, 6, and 8 because they are divisible by 5 or by 2.

This leaves unit digits 1, 3, 7, 9, which would give 4*10 = 40 integers without considering multiples of 3 that end with these digits.

The first digit will always be 2, since we knocked out 300 on the first step.

So, for 200-299, we want to find how many combinations of 2+x+y are divisible by 3, where x is a digit 1-10 and y is 1, 3, 7, or 9.

For units digit 1, 2+x+1 = 3+x is to a multiple of 3 for x = 0, 3, 6, 9
For units digit 3, 2+x+3 = 5+x is to a multiple of 3 for x = 1, 4, 7
For units digit 7, 2+x+7 = 9+x is to a multiple of 3 for x = 0, 3, 6, 9
For units digit 9, 2+x+9 = 11+x is to a multiple of 3 for x = 1, 4, 7

4 + 4 + 3 + 3 = 14 integers between 200 and 299 with unit digits of 1, 3, 7, or 9 are multiples of 3, so removing these integers we have 40 - 14 = 26.
Top
Post new topic Post Reply
• Page 1 of 1