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?

BunuelFor 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

rohitsaxena00In 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.)