How many three-digit numbers are divisible by 7

Answer: 105 is the very first three-digit integer divisible by seven.

The second number is 105+7=112.

The third number is 112+7=119.

So, the numbers 105, 112, 119,…, and so on are divisible by seven.

As we all are aware that the biggest three-digit integer is 999. If we divide this with 7, the remainder is 5, so it is not divisible by seven, and as a result, it cannot be the last three-digit integer to be divisible by 7.

Since 999-5=994 

So, 994 is the highest three-digit number that is divisible by seven. The current sequence of events in the series is as follows.

105, 112, 119, …, 994

Letting 994 be the A.P.’s nth term.

X = 105 is the first term

Y = 7 is a common difference.

The nth term of series Xn = 994.

So, we need to find the number of terms (n).

We all know that

 Xn= X+(n−1)Y

994=105+(n−1)7

994-105= (n−1)7

(n−1) = (995-105)/7

n= 890/7

N is approximately equal to 128. So, there are 128 3-digit numbers that are divisible by7.