Let A={n∈N∣n2≤n+10000} n2≤n+10000 n2−n≤10000 ⇒n(n−1)≤100×100 ⇒A={1,2,3,......,100} Now, B={3k+1∣k∈N} B={4,7,10,13,...} and C={2k∣k∈N} C={2,4,6,8,...} So, B−C={7,13,19,......,97,...} So, A∩(B−C)={7,13,19,......,97} This form an AP with common difference (d=6) ⇒97=7+(n−1)6 n=