Question

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

Answer 1

Hint: The given problem can be solved by using the concept of arithmetic progression (sequence of numbers in order such that the difference between the consecutive terms is constant).

Complete step by step solution:
Since \[11\] is the first number which when divided by of \[7\] leave the reminder 4, and \[998\] is the last number which when divided by of \[7\] leave the reminder 4.Then the arithmetic progression can be written as: \[11,18,25...998\] it can be solved by using the formula \[{T_n} = a + (n - 1)d\] in which the last term is \[998\]
To find the solution of the given problem we can use the formula of arithmetic progression is given by
\[{\text{Last term = first term + }}\left( {{\text{n - 1}}} \right){{ \times \; common \;difference}}\]
It can be written as
\[{T_n} = a + (n - 1)d\]
By taking the first term \[a = 11\]
To calculate the common difference \[d = 18 - 11 = 7\]
\[{T_n}\] is the last term in the progression and is given \[ = 998\]
On substitution we get
\[998 = 11 + (n - 1) \times 7\]
\[ \Rightarrow 998 - 11 = (n - 1) \times 7\]
On simplification we get
\[ \Rightarrow \dfrac{{998 - 11}}{7} = (n - 1)\]
\[ \Rightarrow \dfrac{{987}}{7} = (n - 1)\]
\[ \Rightarrow \dfrac{{987}}{7} + 1 = n\]
\[ \Rightarrow \dfrac{{987 + 7}}{7} = n\]
On simplification we get
\[ \Rightarrow \dfrac{{994}}{7} = n\]
\[ \Rightarrow n = 142\]
Therefore, there are \[142\] numbers are there between 1 and 1000 which when divided by 7 leave reminder 4
So, the correct answer is “142”.

Note: In the above problem \[1000\] is not a number which when divided by 7 leave reminder 4.so the last number in the progression is \[998\] if we divide it by 3, we get a reminder as 4 so we have taken \[998\] as a last number of a progression and similarly 11 as a first number of the progression.