Question

the sum of all two digit numbers which,when divided by 4,yield unity as a remainder is

A. 1190 B.1197 C.1210 D.none

Answer 1

Answer:

hjjjk

Step-by-step explanation:

Answer 2

Answer: C) 1210

Step-by-step explanation:

The two digits number which leave reminder 1 when divided by 4 are

13, 17, 21 ,....................................97

therefore it is an AP where a = 13 , d= 4 , l= 97 whee a if first term d is common difference and l is last term

$l= a+(n-1)d\\\\\Rightarrow 97 = 13+ (n-1)4\\\\\Rightarrow 84= (n-1)4\\\\\Rightarrow 21= n-1\\\\\Rightarrow n= 22$

Now sum is given by

$S_n=\dfrac{n}{2} (a+l)\\\\\Rightarrow S_n= \dfrac{22}{2} (13+97) = 11\times 110= 1210$

Hence the sum of all two digits number which when divided by 4 yield unity as remainder is 1210