Question

Find the sum of all two digit numbers which when divisible by 4, yields 1 as remainder.

Answer 1

Hint: Here, using given conditions, write the numbers. It can be seen that the sequence of numbers forms A.P. with first term 13, common difference 4 and last term 97. Find the number of terms using formula. At the last apply formula to find the sum of terms, using first term, last term and the number of terms.

Complete step-by-step answer:
 If the number is divided by 4 and yield remainder 1, then the number should be of the form 4k + 1.
Smallest two- digit number that gives remainder 1 when divided by 4 is 13 (when k = 3) first term of A.P is 13.
Largest two- digit number that gives remainder 1 when divided by 4 is 97 (when k = 24) last term of A.P. is 97.
Series: 13, 17, 21,..., 97
Now, we have
First term = 13, common difference = 17 – 13 = 4 and last term = 97
$[L = a + (n – 1)\times d]$
$97 = a + (n − 1)\times d$
$\Rightarrow 97 = 13 + (n − 1)\times 4 $
$\Rightarrow 84 = (n − 1)\times 4$
$\Rightarrow (n − 1) = 21 $
$\Rightarrow n = 22 $
Sum of series = $\dfrac{n}{2}$ ​[first term + last term]
Sum of series = $\dfrac{{22}}{2}$ ​[13 + 97] = $11\times (110) = 1210$
So, the correct answer is “1210”.

Note: The students must note that we take the first term from 13 as 12 is the first two digit number that is divisible by 4 and hence adding 1 to it 13 becomes first number of the series , do not confuse and start the series with 9 or 5 as these are single digit numbers.