How Many 3 Digit Numbers are Divisible by 7?

Three-Digit Numbers Divisible by 7

The smallest and largest three-digit number are 100 and 999, respectively. There are many numbers between 100 and 999 which are divisible by 7.

For knowing how many 3 digit numbers are divisible by 7, we will have to make an arithmetic sequence and use the formula to get the required answer. Let’s see how to do it. 

The lowest 3-digit number wholly divisible by 7 is 105.

Next number divisible by 7 = 105 + 7 = 112.

Next number = 112 + 7 = 119. 

Thus we have a series of 3-digit numbers divisible by 7 as:

105, 112, 119, … which forms an arithmetic sequence. 

We will have to find the last number of the above sequence, which is also the largest three-digit number divisible by 7.

The largest three-digit number = 999

When we divide 999 by 7, the remainder left will be 5.

So, 999 − 5 = 994 is the largest three-digit number divisible by 7, which will be the last number of our arithmetic sequence. 

So now the arithmetic sequence becomes 105, 112, 119, …, 994.

The total number of terms in this sequence will equal the total 3 digit numbers divisible by 7.

To find the total number of terms in this sequence, we will use the formula.

T(n)=a+(n-1) d

WhereT(n)=n^{t h} term of the sequence = 994.

a = first term = 105

d= difference = 112 – 105 = 7

n = number of terms in the sequence (To find)

Now, substitute the values in the formula – 

     994=105+(n-1) 7

\Rightarrow 994=105+7 n-7

\Rightarrow 7 n-7=994-105

\Rightarrow 7 n-7=889 \Rightarrow 7 n=889+7

\Rightarrow 7 n=896

\therefore n=\frac{896} {7}=128

So, it is clear that there are 128 terms in the series. That means there are 128 three-digit numbers which are divisible by 7.

Examples

1. Which is the first 3 digit number divisible by 7?

The lowest three-digit number is 100, and 105 is the smallest 3 digit number divisible by 7. We can use the divisibility rule of 7 to verify our answer. 

Last digit of the number 105 = 5

Multiply it by 2, and we get 2 × 5 = 10

Rest of the number = 10

Now on subtracting 10 from 10, we get zero.

Therefore, 105 is divisible by 7. 

 

2.  How many three-digit numbers leave 3 as a remainder in each case when divided by 7?

First three-digit number = 100.

If we divide 100 by 7, the remainder left is 2.

So, 101 is the first three-digit number leaving the remainder as 3 upon dividing by 7.

Next such number 101+7 = 108

Similarly 997  is the last three-digit number leaving the remainder as 3 upon dividing by 7.

So the arithmetic sequence is 101, 108, 115, … ,997

Here a = 101

d = 7

T(n)=997

n = ? (To find)

Formula is T(n)=a+(n-1) d

Therefore, 997 = 101 + (n – 1) × 7

Solving the equation, we get n = 129.

So, there are 129 such numbers.