Question

What is the sum of all three-digit even numbers divisible by seventeen?
A. \[18846\]
B. \[18684\]
C. \[14688\]
D. \[16848\]

Answer 1

Hint: To solve this question we have to find the sum of all the three-digit numbers which are divisible by \[34\]. Because the numbers are even divisible by \[17\]. And \[17\] is the odd number so to make it an even number the three-digit we have to divide by \[34\]. use the first three-digit number \[100\] and then find the remainder and then add that remainder is added to \[100\] to make it completely divisible. And that number is the first three-digit number which gets completely divisible by\[34\]. Do the same step for the last three-digit number and make subtraction instead of addition. Then we will use the Arithmetic progression. Here we have the first and last term along with the common difference i.e. 34. By using this we can find the total sum.

Complete step-by-step answer:
\[100\] Is the first three digit number.
If \[100\] is divided by \[34\] then \[32\] is a remainder.
\[100\] is written as \[34 \times 2 + 32\]
To make the three-digit number we add \[2\] to the reminder and the number generates is \[102\]
\[102\] Is the smallest three-digit number that is divisible by \[34\].
\[999\] is the largest three-digit number.
If \[999\] is divided by \[34\] then \[13\] is a remainder.
\[999\] is written as \[34 \times 29 + 13\]
To make the three-digit number we subtract 13 from the reminder and the number generated is \[986\].
We make subtraction because if we add something then that number will go on the four-digit number.
\[986\] Is the largest three-digit number which is divisible by \[34\].
So, through these numbers, an arithmetic progression is formed.
First-term of (A.P.) is
\[a = 102\]
Common difference of (A.P.) is
\[d = 34\]
The last term of (A.P.) is
\[l = 986\]
Sum of the (A.P.) \[ = \] sum of all three digit number which are divisible by \[17\]
\[{n^{th}}\] term of an A.P
\[{a_n} = a + (n - 1)d\] …(i)
Sum of A.P
\[{s_n} = \dfrac{n}{2}(a + l)\] …(ii)
Here,
\[n\] is number of terms in an A.P
\[a\] is the first term of the A.P
\[l\] is the last term of the A.P
\[d\] is the common difference of the A.P
On putting all the values in equation (i)
\[986 = 102 + (n - 1)34\]
From here we are able to find the value of \[n\]
\[986 - 102 = (n - 1)34\]
On further solving
\[\dfrac{{884}}{{34}} = (n - 1)\]
On dividing the number.
\[26 = (n - 1)\]
On arranging
\[n = 27\]
Put all these values in equation (ii)
\[{s_n} = \dfrac{{27}}{2}(102 + 986)\]
On further solving
\[{s_n} = \dfrac{{27}}{2}(1088)\]
\[{s_n} = 27 \times 544\]
\[{s_n} = 14688\]
Sum of all the even three digit number which are divisible by \[17\] is
\[ \Rightarrow {s_n} = 14688\]

So, the correct answer is “Option C”.

Note: To solve this question using the first three-digit number \[100\] and then find the remainder and then add that remainder is added to \[100\] make a completely divisible. And that number is the first three-digit number which gets completely divisible by\[34\]. Do the same step for the last three-digit number and make subtraction instead of addition.