How Many of The Three-Digit Numbers Are Divisible by 7? GMAT Problem Solving

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

1. 105
2. 106
3. 127
4. 128
5. 142

“How many of the three-digit numbers are divisible by 7?”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “Princeton Review GMAT Premium Prep”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

Solution and Explanation:

Approach Solution 1:

It is asked in the question to find out the number of three-digit numbers which are divisible by 7.
There is a general formula using which we can find the multiples of x in the given range.
Generally,
The number of multiples in the range = (last multiple of x in the range - first multiple of x in the range)/x + 1
For example, if it is asked to find out how many multiples of 5 are in the range -7,35, not inclusive.
Then the last multiple of 5 will be 30 and the first multiple of 5 will be 5.
Total number of multiples in the range = (30 - (-5))/5 + 1 = 35/5 + 1 = 8
Now coming to a given question,
The last 3 digit multiple of 7 will be = 994
The first three-digit multiple of 7 will be = 105
The number of three digit numbers which are multiples of 7 are = (994-105)/7 + 1
= 128

Correct Answer: D

Approach Solution 2:

It is asked in the question to find out the number of three-digit numbers which are divisible by 7.
Divide 1000 by 7 = 142.xxx
It means we have 142 (No rounding up) numbers that are multiples of 7 till 1000. For clarification if you round up, it means 143 * 7= 1001 and hence wrong
However, we asked to find out the range of three-digit numbers. 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 that are multiples of 7
Total 3-digit numbers = 142-14=128

Correct Answer: D

Approach Solution 3:

Hint: An arithmetic progression can be given by a, (a+d), (a+2d), (a+3d), ……

a, (a+d), (a+2d), (a+3d),….. where a = first term, d = common difference.

a,b,c are said to be in AP if the common difference between any two consecutive number of the series is same ie 

b−a=c−b⇒2b=a+cb−a=c−b⇒2b=a+c

Formula to consider for solving these questions

an=a+(n−1)dan=a+(n−1)d

Where d -> common difference

A -> first term

n-> term

an−>nthan−>nth term

Complete step-by-step answer:

We have AP starting from 105 because it is the first three-digit number divisible by 7.

AP will end at 994 because it is the last three-digit number divisible by 7.

Therefore, we have AP of the form 105, 112, 119…, 994

994 is the nthnth term of AP.

We need to find n here.

First term = a = 105, Common difference = d = 112 – 105 = 7

Using formula , to find nthnth term of arithmetic progression,

⇒994 = 105 + (n − 1) (7)⇒994 = 105 + 7n−7⇒896 = 7n⇒n = 128

Correct Answer: D

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