Question

Find sum of all $2$-digit numbers which when divided by $5$ leave remainder $1$.
A. $963$
B. $863$
C. $983$
D. $943$

Answer 1

As we know sum is addition of two or more numbers and hence to find the sum of n numbers, we can solve using Arithmetic progression and here we need to find all $2$-digit numbers when divided by $5$ we can use the formula to find sum of $n$terms of AP given as
${S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]$

Complete step-by-step answer:
To find the sum of all $2$ digit numbers when divided by $5$ we can see that the two digits numbers which leaves remainder $1$ are
$11,16,21,26,31,36,41,..............,96$.
We can see that the above series is an Arithmetic Progression (AP) of the from
${a_1},{a_2},{a_3},{a_4},{a_5}..................,{a_n}$
Hence, we can apply the AP general formula to find the value of $n$
${a_n} = a + \left( {n - 1} \right)d$
Where, ${a_n}$ is the nth term in the series i.e., ${a_n} = 96$
$a$ is the first term in the series i.e., $a = 11$
$d$ is common difference i.e.,
$d = {a_2} - {a_1}$
$d = 16 - 11$
$d = 5$
Now, let us find the value of $n$ by rewriting the general formula
${a_n} = a + \left( {n - 1} \right)d$
Simplifying the values in the above formula we have
$96 = 11 + \left( {n - 1} \right)5$
$96 = 11 + 5n - 5$
As we need to find the value of $n$, let us shift $5n$to LHS as shown
$- 5n = 11 - 5 - 9$
$- 5n = - 90$
Hence, we get
$n = \dfrac{{90}}{5} = 18$
Therefore, the value of $n$ is $18$.
We must find the sum of terms for this let us apply the sum of AP formula
${S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]$
Substitute the values as per we have got
${S_n} = \dfrac{{18}}{2}\left[ {11 + 96} \right]$
After simplifying the values, we get
${S_n} = 9\left[ {107} \right]$
Hence, the required sum is
${S_n} = 963$
Therefore, option$A$ is the right answer.
Formula used:
General term used for Arithmetic Progression (AP) series is
${a_n} = a + \left( {n - 1} \right)d$
In which,
${a_n}$ is the nth term of the series.
$a$ is the first term.
$n$ is the number of terms from ${a_1}$ to ${a_n}$.
$d$ is a common difference between the first two terms of the series.
To find the sum of first $n$terms of AP is
${S_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]$
In which,
${S_n}$ is sum of $n$ terms

Note: To find the sum of any digit number asked, find the series then apply the formula according to the series i.e., it may be Arithmetic Progression series (AP) or Geometric Progression series (GP) or Harmonic Progression series (HP) series, based on this we need to find the sum.