Question

Find the sum of all two digit numbers greater than \[50\] which when divided by \[7\], leave a remainder of \[4\].

Answer 1

Hint: In the above given question, we are given two digit numbers greater than \[50\] which when divided by \[7\] , leave a remainder of \[4\] . We have to find the sum of all such two digit numbers. In order to approach the solution, first we have to find all these two digit numbers.

Complete step by step answer:
Given that, all the two digits numbers greater than \[50\] which when divided by \[7\] , leave a remainder of \[4\]. We have to find their sum. Now, the first two digit number greater than \[50\] which when divided by \[7\] , leaves a remainder of \[4\] is \[53\] as,
\[ \Rightarrow 53 = 7 \times 7 + 4\]
And the greatest two digit number which when divided by \[7\] , leaves a remainder of \[4\] is \[95\] as,
\[ \Rightarrow 95 = 7 \times 13 + 4\]
Therefore, these numbers are \[7 \times 7 + 4\] , \[7 \times 8 + 4\] , \[7 \times 9 + 4\] , \[7 \times 10 + 4\] , \[7 \times 11 + 4\] , \[7 \times 12 + 4\] and \[7 \times 13 + 4\] .
Hence, there are total seven such numbers, and their sum \[{S_n}\] can be calculated as,
\[\Rightarrow {S_n} = \left( {7 \times 7 + 4} \right) + \left( {7 \times 8 + 4} \right) + \left( {7 \times 9 + 4} \right) + \\
  \left( {7 \times 10 + 4} \right) + \left( {7 \times 11 + 4} \right) + \left( {7 \times 12 + 4} \right) + \left( {7 \times 13 + 4} \right) \]

Taking some parts as common, we can write the above equation as,
\[ \Rightarrow {S_n} = 7\left( {7 + 8 + 9 + 10 + 11 + 12 + 13} \right) + \left( {7 \times 4} \right)\]
That gives us,
\[ \Rightarrow {S_n} = 7\left( {4 + 7 + 8 + 9 + 10 + 11 + 12 + 13} \right)\]
Adding the numbers inside the bracket, we get the equation,
\[ \Rightarrow {S_n} = 7\left( {20 + 20 + 20 + 14} \right)\]
That gives us,
\[ \Rightarrow {S_n} = 7 \times 74\]
That is,
\[ \therefore {S_n} = 518\]
That is the required sum of the seven obtained numbers which satisfy the given condition.

Therefore, the sum of all two digit numbers greater than \[50\] which when divided by \[7\],leave a remainder of \[4\] is \[518\].

Note: Since, the first two digit number required was obtained to be \[53\] , therefore the rest of the numbers which have two digits, i.e. they are less than \[100\] can be obtained directly by adding \[7\] to the first two digit number up to \[ < 100\].That gives us, \[53,53 + 7,53 + 7 + 7,... < 100\]. Therefore, there are seven numbers obtained as, \[53,60,67,74,81,88,95\] and their sum is \[518\] .