A) 3221
B) 3121
C) 3203
D) 3303
Correct Answer: B
Solution :
\[{{S}_{A}}=\]sum of numbers between 100 & 200 which are divisible by 7. \[\Rightarrow \]\[{{S}_{A}}=105+112+....+196\] \[{{S}_{A}}=\frac{14}{2}[105+196]=2107\] \[{{S}_{B}}=\]Sum of numbers between 100 & 200 which are divisible by 13. \[{{S}_{B}}=104+117+....+195=\frac{8}{2}[104+195]=1196\]\[{{S}_{C}}=\]Sum of numbers between 100 & 200 which are divisible by both 7 & 13. \[{{S}_{C}}=182\] \[\Rightarrow \] \[H.C.F.\left( 91,n \right)>1={{S}_{A}}+{{S}_{B}}{{S}_{C}}=3121\]
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