GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 14 Oct 2019, 08:50

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# M01-18

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 58316

### Show Tags

16 Sep 2014, 00:15
2
12
00:00

Difficulty:

35% (medium)

Question Stats:

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

### HideShow timer Statistics

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

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

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 58316

### Show Tags

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:
$$\frac{994-105}{7}+1=128$$

_________________
Intern
Joined: 10 Jul 2014
Posts: 12

### Show Tags

21 Dec 2014, 08:54
Hi Bunuel,

Plz clarify :why should we add 1 to the above solution
Math Expert
Joined: 02 Sep 2009
Posts: 58316

### Show Tags

21 Dec 2014, 08:57
1
amargb wrote:
Hi Bunuel,

Plz clarify :why should we add 1 to the above solution

Check this: totally-basic-94862.html#p730075
_________________
Intern
Joined: 14 Sep 2012
Posts: 20

### Show Tags

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:
$$\frac{994-105}{7}+1=128$$

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?
Math Expert
Joined: 02 Sep 2009
Posts: 58316

### Show Tags

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:
$$\frac{994-105}{7}+1=128$$

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
_________________
Intern
Joined: 14 Sep 2012
Posts: 20

### Show Tags

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:
$$\frac{994-105}{7}+1=128$$

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?
Intern
Joined: 14 Sep 2012
Posts: 20

### Show Tags

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:
$$\frac{994-105}{7}+1=128$$

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?
Intern
Joined: 08 Apr 2014
Posts: 15
Location: United States
Concentration: Strategy, Technology
GPA: 4

### Show Tags

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.
Intern
Joined: 09 May 2013
Posts: 22
GMAT 1: 730 Q49 V40
GPA: 3.44

### Show Tags

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:
$$\frac{994-105}{7}+1=128$$

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.)
Intern
Joined: 05 Dec 2013
Posts: 26

### Show Tags

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 ?
Math Expert
Joined: 02 Sep 2009
Posts: 58316

### Show Tags

17 Apr 2015, 04:12
Randude wrote:
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 ?

210 = 7*30, so the next number divisible by 7 is 210 + 7 = 217.
_________________
Intern
Joined: 23 Apr 2016
Posts: 14

### Show Tags

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?
Math Expert
Joined: 02 Sep 2009
Posts: 58316

### Show Tags

09 Jul 2016, 09:16
expertesp wrote:
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?

GMAT Club's questions are mostly quite difficulty. One should not attempt them if the fundamentals are not strong enough.

As for your question: the number of 3-digit integers is 900, not 999.
_________________
Intern
Joined: 23 Apr 2016
Posts: 14

### Show Tags

09 Jul 2016, 11:01

Anyone else?

Posted from my mobile device
Intern
Joined: 23 Apr 2016
Posts: 14

### Show Tags

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
Math Expert
Joined: 02 Sep 2009
Posts: 58316

### Show Tags

09 Jul 2016, 23:10
expertesp wrote:

Anyone else?

Posted from my mobile device

This was an advice. Was not intended as an insult in any way. If you attempt hard questions with weak fundamentals: 1. the chances that you understand the question and the theory behind it is slim; 2. you will waste the question; 3. you will shake your confidence.
_________________
SVP
Joined: 26 Mar 2013
Posts: 2341

### Show Tags

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

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.
Intern
Joined: 23 Apr 2016
Posts: 14

### Show Tags

10 Jul 2016, 04:52
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!

Posted from my mobile device
Current Student
Joined: 29 Sep 2016
Posts: 17
Location: United States

### Show Tags

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.
Re: M01-18   [#permalink] 24 Mar 2017, 14:18

Go to page    1   2    Next  [ 24 posts ]

Display posts from previous: Sort by

# M01-18

Moderators: chetan2u, Bunuel