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# M01-18

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Current Student
Joined: 29 Sep 2016
Posts: 17
Location: United States

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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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Joined: 03 Mar 2018
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02 Apr 2018, 09:06
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?

I have the same doubt as rohit. Approach 2 works all the time, there is no doubt on its credibility. But approach 1 is much more quicker than approach 2(since we dont need to find 1st and last terms). If the answer choices are not close we can safely use Approach 1(if 127 is not given as an option). Due to the glitches in approach 1, as explained by kyles in this thread, we cannot always use Approach 1.

chetan2u / Bunuel
Please corroborate my opinion regarding approach 1
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Joined: 08 Aug 2018
Posts: 6
GMAT 1: 210 Q1 V1

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03 Sep 2018, 13:16
Quick solution:

2) Only (C) and (D) are very close together 127 128
3) Think like the test-maker, he probably wants you to confuse one for the other
4) Think about it, if you were normally doing this problem, the real issue is that you miscount, by excluding the last integer (i.e. the common mistake would be to exclude the last integer when you use normal subtraction vs. inclusive subtraction (e.g. how many digits between 2 and 5 inclusive is 5 - 2 + 1 = 4 (Correct) vs. 5 - 2 = 3 (Incorrect and Trap answer choice) Therefore, the right answer must be one higher than the trap answer
5) Go back to step 2 look at 127 and 128, the test maker is trying to trick us into choosing 127 (the one lower - miscounted) therefore, 128 should be the right answer.
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Joined: 10 May 2018
Posts: 7

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04 Oct 2018, 09:30
sarimzahid2395 wrote:
Quick solution:

2) Only (C) and (D) are very close together 127 128
3) Think like the test-maker, he probably wants you to confuse one for the other
4) Think about it, if you were normally doing this problem, the real issue is that you miscount, by excluding the last integer (i.e. the common mistake would be to exclude the last integer when you use normal subtraction vs. inclusive subtraction (e.g. how many digits between 2 and 5 inclusive is 5 - 2 + 1 = 4 (Correct) vs. 5 - 2 = 3 (Incorrect and Trap answer choice) Therefore, the right answer must be one higher than the trap answer
5) Go back to step 2 look at 127 and 128, the test maker is trying to trick us into choosing 127 (the one lower - miscounted) therefore, 128 should be the right answer.

I solved using approach 2(presented by Bunuel) in 1:31 mins but I'm looking for a quicker solution. What you say makes sense to me but I have a question - answer choices A and B are close together as well - 105, 106 resp. What process did you use in eliminating these choices?
Re: M01-18   [#permalink] 04 Oct 2018, 09:30

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# M01-18

Moderators: chetan2u, Bunuel