# How many numbers are there between 1 and 1000 which when divided by 7 leave the remainder 4?

Answer

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**Hint:**For solving this question you should know about the arithmetic progression A.P. and if we make the series for it then firstly we will find the first term and then we will find next same terms with a common difference of 7 because the question is asking for the same.

**Complete step by step answer:**

According to our question it is asked to calculate how many numbers there are between 1 and 1000 which when divided by 7 leave the remainder 4.

So, as we know that if any statement is given which shows about a type of a number, how it will be implemented, then we first make the first term which can be generated by the given information and then we find the next terms as a series wise or as an arithmetic progression wise.

So, according to the given statement in question we will evaluate that if we divide a number by 7 then it leaves a remainder 4.

So, we can write, if a number is divided by 7 and the remainder 4, then it must be in the form of \[7k+4\].

Let, \[{{T}_{k}}=7k+4\] from 1 to 1000

First term is ‘4’

And the common difference is ‘7’

And last term is \[\left( l \right)=998\]

\[\begin{align}

& \because l=a+\left( n-1 \right)d \\

& 998=4+\left( n-1 \right)7 \\

& 998=7n-3 \\

& 7n=1001 \\

& n=143 \\

\end{align}\]

So, the total terms are 143.

**Note:**While solving these types of problems you have to mind that you must make your first term with an exact remainder and common difference. And the term must be satisfied with the statement given in the question.

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