integers between

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greenplant: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Sat Apr 04, 2009 8:09 am

by greenplant » Tue Apr 07, 2009 1:27 pm

Q. How many integers are between 100 and 150, inclusive, cannot be evenly divided by 3 nor 5?
a. 35
b. 27
c. 25
d. 26
e. 28

Could anyone tell me if there is a simple way to deduce the answer instead of going through each number between 100 and 150

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pakaskwa: Master | Next Rank: 500 Posts; Posts: 107; Joined: Wed Mar 04, 2009 4:39 am; Location: Vancouver; Thanked: 9 times; GMAT Score:750

by pakaskwa » Tue Apr 07, 2009 3:35 pm

It's a question testing knowledge on arithmetic progression. And it's quite tricky because it's also testing knowledge on sets.

For any arithmetic progression, a1, a2, ..., an, if we know common difference d, first number a1, and last number an, we can calculate how many numbers there are in the sequence:
n=[(an-a1)/d]+1

If an integer a is divisible by 3, (a+3) is divisible by 3. In the original question, from 100-150, first number that's divisible by 3 is 102, so a1=102;
Last number divisible by 3 is 150, so an=150
Common difference d=3, so
n=[(150-102)/3]+1=17
There are 17 numbers from 100-150 that are divisible by 3.

Likewise, when a1=100, an=150, d=5,
n==[(150-100)/5]+1=11
There are 11 numbers from 100-150 that are divisible by 5.

There are 51 numbers from 100 to 150 (you can use the same formula to get 51 if you need to). So numbers that are NOT divisible by 3 OR 5 is
51-17-11=23

But here's the tricky part: there are numbers divisible by both 3 AND 5 (ie, divisible by 15), and they were deducted twice from our calculation above. Once as in set divisible by 3; once as in set divisible by 5. So we need to add the number back:
Using same formula, first number divisible by 15 is a1=105,
Last number divisible by 15 is an=150, so
n=[(150-105)/15]+1=4

23+4=27

Answer is B.

The last step can also be calculated by sets formula (refer to OG11 page 116-117). Assume that from 100-150, there are n numbers can be divisible by 3, m numbers can be divisible by 5. Total numbers should be:
|n OR m| = |n| + |m| - |n AND m|
which is
|n or m| = 17+11-4=24
Total numbers not divisible by 3 or 5 is:
51-24=27

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cramya: Legendary Member; Posts: 2467; Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members; Thanked: 331 times

by cramya » Tue Apr 07, 2009 3:43 pm

3 * 34 = 102
3*50 = 150

There are 50-34+1 i.e 17 multiples of 3 between 100 and 150 inclusive

5*21 = 105
5*30= 150

There are 30 - 21 +1 i.e 10 multiples of 5 between 100 and 150 inclusive

Total = 27 but u would have counted multiples of both 3 and 5 twice i.e 115,130,145(counted twice once as a multiple of 3 and another time as a multiple of 5) so subtract one instance of each of them

27 - 3 = 24 distinct integers

The total number of integers between 100 and 150 inclusive are 150 - 100 +1 = 51

51- 24 = 27(numbers between 100 and 150 neither divisible by 3 nor 5)

Regards,
CR

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dtweah: Master | Next Rank: 500 Posts; Posts: 487; Joined: Fri Mar 27, 2009 5:49 am; Thanked: 36 times

Re: integers between

by dtweah » Tue Apr 07, 2009 4:00 pm

greenplant wrote:Q. How many integers are between 100 and 150, inclusive, cannot be evenly divided by 3 nor 5?
a. 35
b. 27
c. 25
d. 26
e. 28

Could anyone tell me if there is a simple way to deduce the answer instead of going through each number between 100 and 150

In the case of 3, going thru each number might be very prone to error though it might be quick for some people. Instead, I would use the formula for the nth term of an arithmetic sequence to find how many numbers are divisible by 3 btw 100, and 150, count how many numbers are divisible by 5, sum these and subtract from the sum the number. of integers divisble by 15, since these would have been double counted. I would then take this sum from the number of integers between 100 and 150 inclusive. So the Algorithm might run like this

N-(T + F - B) where T, F and B mean "The number of Integers divisible by Three, Five and Both 3 and 5 or 15, respectively. N is the number of integers between 100 and 150 inclusive.

150 = 102 +(T-1)3 (102 is the first number divible by 3 after 100.)
51 =3T
T=17

Since 5 is generally easier to handle I would count, not forgeting to start with 100. F=11
For B: 105 120 135 150 so, B=4

So 17+11-4=24

Now N= 150 -100 +1=51

51-24=27

I choose B.

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