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integer problem

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by Mclaughlin » Tue May 13, 2008 9:50 pm
What is the total number of integers between 100 and 200 that are divisible by 3?
(A) 33
(B) 32
(C) 31
(D) 30
(E) 29


Where do i begin with this and what are some tips on how to handle questions like this in the future. thanks
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by arorag » Tue May 13, 2008 9:57 pm
HInt: any no. is divisible by 3 if sum of its digits are divisble by 3
e.q. 102, 105
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by Mclaughlin » Tue May 13, 2008 10:05 pm
Thanks! great hint. But how do I do this fast without having to list out all the nubmers if I'd rather spend time on harder problems. (if I'm lucky enough to get them)
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by akshatsingh » Tue May 13, 2008 11:26 pm
200/3 gives 66 and a remainder 2.
hence 66 nos, from 1 to 200 are divisible by 3

100/3 gives 33 and a remainder 1.
hence 33 nos, from 1 to 100 are divisible by 3

So from 100 to 200 there will be 66 - 33 = 33 nos divisible by 3.

My answer is A
Aks
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by shan_shine007 » Wed May 14, 2008 1:27 am
For any no. to be divisible by 3, the sum of the digits of the no. must be divisible by 3.
eg. take 102
sum of digits = 1+0+2 = 3 i.e, divisible by 3
the next no. divisible by 3 is 105 i.e, 3 more than d previous no.

so between 100 & 200 there are 99 integers
hence d no. of integers divisible by 3 between 100 & 200 is
99/3 = 33
ans a
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by Mclaughlin » Wed May 14, 2008 6:57 am
Thanks both of you! but I have a stupid question. How do you know that there are 99 intergers between 100 to 200. I thought it was 100 intergers.
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by bia » Wed May 14, 2008 7:12 am
There are 101 integers between 100 and 200. You can use the formula:
((max-min)/distance between two consecutive integer)) +1

We can solve above question much more easily by this formula:

max number divisible to 3 = 198
min number divisible to 3 = 102

So total number between 100 and 200 that is divisible to 3 = (198-102)/3 + 1 = 33
Bia
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by VP_Tatiana » Wed May 14, 2008 2:58 pm
Hi guys,

When a problem asks how many numbers are between two numbers, we want to exclude the endpoints. (Unless it says "inclusive.") So, in this problem we just want to look at the numbers 101-199. You can see that there are there exactly the same amount of numbers as between 1 and 99; there are 99 numbers.

If are examining the numbers between 100 and 200 inclusive, then we also count 100 and 200, and then we have 101 numbers.

In this case, since the endpoints are not divisible by 3, people were able to arrive at the right answer whether they considered the endpoints or not. However, counting the endpoints would have resulted in a wrong answer if the question were "how many numbers divisible by 2 are between 100 and 200?"

Best wishes,

Tatiana
Tatiana Becker | GMAT Instructor | Veritas Prep
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by bia » Wed May 14, 2008 9:27 pm
There are ((200-100)/2)+1 = 51 numbers divisible by 2 which are between 100 and 200

Is is right?
Bia
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by shan_shine007 » Wed May 14, 2008 10:53 pm
to find the no. of integers between 100 & 200 ... dont take into account the end pts (100 & 200) ...

So, no, of integers between 100 & 200 = (199- 101)+1= 99
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by Paramvir » Wed May 16, 2012 8:39 pm
Hello Guys,

I was going through this problem and reading the comments of various users. Can someone tell me if we can use the below formula for all such problems or not ? I mean if we have to find number of integers divisible by 2, 5, 7 etc...

((max-min)/distance between two consecutive integer)) +1

Or is it only valid for this particular problem ?

Thank You
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by Thrills4ever » Thu May 17, 2012 7:41 pm
These types of problems can be answered simply with a series of logical steps. Whenever you're dealing with sums or total numbers here is a simple formula.

Sum = Average * Number --> you can manipulate this formula to get whatever you want. In this case we want the number of integers divisible by 3 from 100-200.

Well let's take this question a step further, so you can see how this sequence works. Let's say we were asked this question instead:

"What is the sum of the numbers between 100-200 that are divisible by three"?

Remember Sum = Average * Total number.

Let's first find the average:

1) To find the average of any range of numbers, the first thing you must do is to begin by including only those numbers that the problem constrains you to. So we are told that we only want numbers divisible by 3. This means our "new" range of numbers becomes 102-198 because 100, 101, 199 and 200 are all numbers NOT divisible by 3. Now that we have the new range, we take the high number plus the low number and divide by 2

High number: 198
Low number 102
High + Low = 198+102 = 300
300/2 = 150 --> this is our AVERAGE.

Let's find the Total numbers between 100-200 that are divisible by three.

2) Again, we only include the smallest and largest numbers in the range that the problem states, in this case only those numbers that are divisible by 3. Thus our range is still 102-198. Now subtract these two numbers, divide by the constraint and ADD ONE (ALWAYS ADD ONE)

High number: 198
Low number: 102
High - Low = 198-102 = 96

Take 96 and divide by constraint (only numbers divisible by 3) and you get 96/3 = 32

ADD ONE to the new number to get 32 + 1 = 33

33 is thus the total amount of numbers between 100-200 that are divisible by 3

To find the sum, use the formula: Sum = Average*Number

Sum = 150*33 = 4950

This type of technique can be used to solve any type of problem, if it asks for the sum of only even integers from 1 -100 or the number of integers between 1 - 500 that are divisible by 4 or anything.

Hope this helps!
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by kullayappayenugula » Fri May 18, 2012 6:14 am
hi ,

We cn apply a formula of progressions.

since we have to find the divisible of 3 between 100 and 200

first find the first and last multiples of three between the two numbers

we find that the first multiple is 102 and the last multiple is 198


no apply the formula for finding the value of nth term in AP series.

an= a + (n-1) d

where an = nth term in series,
a is the first term in the series
d = difference between the terms in the series
n = the term we want to find

since we know last term is 198 => an= 198 , a=102, d=3

198=102+(n-1)3
=> 96=(n-1)3
=> 32=n-1
=> n=33

Therefore the answer is 33.

=>
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