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M01-18

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New post 16 Sep 2014, 00:15
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Question Stats:

69% (01:13) correct 31% (01:12) wrong based on 350 sessions

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New post 16 Sep 2014, 00:15
4
7
Official Solution:

How many of the three-digit numbers are divisible by 7?

A. 105
B. 106
C. 127
D. 128
E. 142


Approach One: Divide all of the three-digit numbers \(999 - 100 + 1 = 900\)(Don't forget to add 1 to get the number of all the 3-digit numbers) by 7, which is 128.57, and then round it off to 128.

Approach Two: Find the first and the last multiples of 7 WITHIN the range (which is between 105 and 994). Find the positive difference, divide by 7, and add 1:
\(\frac{994-105}{7}+1=128\)


Answer: D
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New post 21 Dec 2014, 08:54
Hi Bunuel,

Plz clarify :why should we add 1 to the above solution
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New post 21 Dec 2014, 08:57
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New post 05 Jan 2015, 09:10
Bunuel wrote:
Official Solution:

How many of the three-digit numbers are divisible by 7?

A. 105
B. 106
C. 127
D. 128
E. 142


Approach One: Divide all of the three-digit numbers \(999 - 100 + 1 = 900\)(Don't forget to add 1 to get the number of all the 3-digit numbers) by 7, which is 128.57, and then round it off to 128.

Approach Two: Find the first and the last multiples of 7 WITHIN the range (which is between 105 and 994). Find the positive difference, divide by 7, and add 1:
\(\frac{994-105}{7}+1=128\)


Answer: D


Hi Bunuel,

Although I solved this question using approach 2 and got it correct, I would prefer to use the Approach 1 as advised by you as it is a shorter method to solve the same.

But I have a doubt regd Approach 1:
1) In approach 1 should we ALWAYS first add 1 and then divide by 7 (or whatever number's multiple we have to find out)?
2) In approach 1 should we ALWAYS round off the resultant number to a smaller number as done in this case or does it change in certain cases?

Basically will approach one work the same way in all such type of questions no matter what range and what number's multiple we have to find out?
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New post 05 Jan 2015, 09:12
rohitsaxena00 wrote:
Bunuel wrote:
Official Solution:

How many of the three-digit numbers are divisible by 7?

A. 105
B. 106
C. 127
D. 128
E. 142


Approach One: Divide all of the three-digit numbers \(999 - 100 + 1 = 900\)(Don't forget to add 1 to get the number of all the 3-digit numbers) by 7, which is 128.57, and then round it off to 128.

Approach Two: Find the first and the last multiples of 7 WITHIN the range (which is between 105 and 994). Find the positive difference, divide by 7, and add 1:
\(\frac{994-105}{7}+1=128\)


Answer: D


Hi Bunuel,

Although I solved this question using approach 2 and got it correct, I would prefer to use the Approach 1 as advised by you as it is a shorter method to solve the same.

But I have a doubt regd Approach 1:
1) In approach 1 should we ALWAYS first add 1 and then divide by 7 (or whatever number's multiple we have to find out)?
2) In approach 1 should we ALWAYS round off the resultant number to a smaller number as done in this case or does it change in certain cases?

Basically will approach one work the same way in all such type of questions no matter what range and what number's multiple we have to find out?


Explained here: totally-basic-94862.html#p730075
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New post 05 Jan 2015, 09:19
Bunuel wrote:
rohitsaxena00 wrote:
Bunuel wrote:
Official Solution:

How many of the three-digit numbers are divisible by 7?

A. 105
B. 106
C. 127
D. 128
E. 142


Approach One: Divide all of the three-digit numbers \(999 - 100 + 1 = 900\)(Don't forget to add 1 to get the number of all the 3-digit numbers) by 7, which is 128.57, and then round it off to 128.

Approach Two: Find the first and the last multiples of 7 WITHIN the range (which is between 105 and 994). Find the positive difference, divide by 7, and add 1:
\(\frac{994-105}{7}+1=128\)


Answer: D


Hi Bunuel,

Although I solved this question using approach 2 and got it correct, I would prefer to use the Approach 1 as advised by you as it is a shorter method to solve the same.

But I have a doubt regd Approach 1:
1) In approach 1 should we ALWAYS first add 1 and then divide by 7 (or whatever number's multiple we have to find out)?
2) In approach 1 should we ALWAYS round off the resultant number to a smaller number as done in this case or does it change in certain cases?

Basically will approach one work the same way in all such type of questions no matter what range and what number's multiple we have to find out?


Explained here: totally-basic-94862.html#p730075


Hi Bunuel,

I just saw the link, it explains the 2nd approach to solve this question and I have understood that.
I was looking for the answer in the context of approach 1.
Is there any link to explain that?
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New post 05 Jan 2015, 10:18
1
Bunuel wrote:
rohitsaxena00 wrote:
Bunuel wrote:
Official Solution:

How many of the three-digit numbers are divisible by 7?

A. 105
B. 106
C. 127
D. 128
E. 142


Approach One: Divide all of the three-digit numbers \(999 - 100 + 1 = 900\)(Don't forget to add 1 to get the number of all the 3-digit numbers) by 7, which is 128.57, and then round it off to 128.

Approach Two: Find the first and the last multiples of 7 WITHIN the range (which is between 105 and 994). Find the positive difference, divide by 7, and add 1:
\(\frac{994-105}{7}+1=128\)


Answer: D


Hi Bunuel,

Although I solved this question using approach 2 and got it correct, I would prefer to use the Approach 1 as advised by you as it is a shorter method to solve the same.

But I have a doubt regd Approach 1:
1) In approach 1 should we ALWAYS first add 1 and then divide by 7 (or whatever number's multiple we have to find out)?
2) In approach 1 should we ALWAYS round off the resultant number to a smaller number as done in this case or does it change in certain cases?

Basically will approach one work the same way in all such type of questions no matter what range and what number's multiple we have to find out?


Explained here: totally-basic-94862.html#p730075


Bunuel
For eg: find multiples of 5 btw 4 and 26?
Acc to approach 1 : 26-4+1 = 23, 23/5 = 4.6 = 4 (rounded) which is the wrong answer
So approach 1 fails here right?
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New post 13 Feb 2015, 06:39
1
Hello,

I solved the question via A.P(Arithmetic progression)

The minimum three digit number divisible by 7 is 105 and maximum number is 994.
Hence,a=105,d=7,l=994
l=a+(n-1)d;
994=105+(n-1)7
by solving,we get
n=128.

Hope it helps!

Thanks.
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New post 13 Feb 2015, 09:16
2
rohitsaxena00 wrote:
rohitsaxena00 wrote:
Bunuel wrote:
Official Solution:

How many of the three-digit numbers are divisible by 7?

A. 105
B. 106
C. 127
D. 128
E. 142


Approach One: Divide all of the three-digit numbers \(999 - 100 + 1 = 900\)(Don't forget to add 1 to get the number of all the 3-digit numbers) by 7, which is 128.57, and then round it off to 128.

Approach Two: Find the first and the last multiples of 7 WITHIN the range (which is between 105 and 994). Find the positive difference, divide by 7, and add 1:
\(\frac{994-105}{7}+1=128\)


Answer: D


Hi Bunuel,

Although I solved this question using approach 2 and got it correct, I would prefer to use the Approach 1 as advised by you as it is a shorter method to solve the same.

But I have a doubt regd Approach 1:
1) In approach 1 should we ALWAYS first add 1 and then divide by 7 (or whatever number's multiple we have to find out)?
2) In approach 1 should we ALWAYS round off the resultant number to a smaller number as done in this case or does it change in certain cases?

Basically will approach one work the same way in all such type of questions no matter what range and what number's multiple we have to find out?



Bunuel
For eg: find multiples of 5 btw 4 and 26?
Acc to approach 1 : 26-4+1 = 23, 23/5 = 4.6 = 4 (rounded) which is the wrong answer
So approach 1 fails here right?


Hi, there is a fundamental difference in the two examples you have chosen which will determine whether method will will need to be rounded up or down.

In the above question you are asked to find all three digit number divisible by 7. Method 1 would have you follow this process \frac{(999-100+1)}{7}. Now, take the time and think about what you're actually doing when you calculate this, you've calculated the number of "steps" of magnitude 7 exist along the number line from 100 to 999 (inclusive).

The danger with method 1 is this may over count, or under count the number of divisible numbers. For example let's look at the given problem. The first three-digit number divisible by 7 is 105, but we started counting at 100. In effect, this gives us an "extra" 5 integers being counted. On the high limit, 994 is the largest three-digit number divisible by 7; but method 1 has us counting an additional 5 integers to 999. This means we've accounted for an additional 10 integers. Because 10 > 7 we end up with an answer that is too high and we must truncate. (For those interested

Now, the opposite is true of the example proposed by rohitsaxena00

In this example we are looking for integers divisible by 5 but the range provided is from 4 to 26. This means we only "over count" by 2 integers. 2 < 5 so we will have to round up.

This all becomes much more clear when viewed on the number line:

Let's look at the total number of steps of magnitude 5 between 4 and 26:
4-9-14-19-24-... (The steps are represented by the "-" so you can see there are 4 steps plus some extra [0.2 in this case])

Now let's play with the range so that we end up with a remainder greater than 5; assume a range from 1 to 28:
1-6-11-16-21-26-... (You can now see there are 5 steps with the same extra of 0.2)

Hopefully this helps shed some light on the questions. I recommend practicing solving these questions with method 2, it takes a little bit more time, but the GMAT will almost always offer answers to bait you into forgetting to add 1.

(As a side note, if you have to guess on a question like this I do not recommend choosing E and I would always lean to B or D as the common error is forgetting to add 1.)
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New post 16 Apr 2015, 14:51
I see most of the folks have quickly calculated 105 and 994 to be divisible by 7. I understand that its doable by simply checking numbers one by one, but is there an alternate way to do this ?
For example -- Find the first number divisible by 7 between 211 and 300 ?
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New post 17 Apr 2015, 04:12
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New post 09 Jul 2016, 08:59
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Why should I not just divide 999 by 7? Why must it be 999-100+1?
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New post 09 Jul 2016, 09:16
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New post 09 Jul 2016, 11:01
This answer was very rude and not helpful.

Anyone else?

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New post 09 Jul 2016, 12:36
There is another Forum post that explains the same problem solution (just different questions) but also explains the why something is done.
http://gmatclub.com/tests-beta/view-8624268.html
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New post 09 Jul 2016, 23:10
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New post Updated on: 17 Apr 2018, 13:07
1
1
Bunuel wrote:
How many of the three-digit numbers are divisible by 7?

A. 105
B. 106
C. 127
D. 128
E. 142



We have alternative approach with same line of logic presented by "Bunuel":

divide 1000 by 7 = 142.xxx It means we have 142 (No rounding up) numbers are multiple of 7 till 1000. For clarification if you round up, it means 143 * 7= 1001 and hence wrong

However, we asked about THREE Digit numbers ONLY. It means we need to find multiple of 7 till 100'

divide 100 by 7 = 14.xxxx It means we have 14 (No rounding up) numbers are multiple of 7

Total 3-digit numbers = 142-14=128

Answer: D

Originally posted by Mo2men on 10 Jul 2016, 04:22.
Last edited by Mo2men on 17 Apr 2018, 13:07, edited 1 time in total.
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New post 10 Jul 2016, 04:52
The alternative approach is helpful!
I also went back to review the basics. I realised that when I focus only in verbal for a week Mt quant knowledge disappears slowly. It is not see enough rooted yet in my memory.

@bunel: I got this question as part of one of the tests. So it was chosen automatically. I did get some 700 right! But I have 3 weeks and a long way to go to become better and need to repeat repeat repeat now. And hopefully also remember the next day.

Thanks for the quick replies everyone!

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New post 24 Mar 2017, 14:18
Hi - I used a different approach. Would my approach work for other similar problems, or I just got the correct answer as a fluke?

I divided 100/7 to see that 7 went into every group of 100 14 times with a remainder of 2. I multiplied 14 by 9 (the sets of 100 in this question) to get 126. I then multiplied the remainder 2 by 9 as well to get 18 (which 2 7's fit into). 126+2=128.
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Re: M01-18   [#permalink] 24 Mar 2017, 14:18

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