Counting attempt #1

Since $A_u$ is the sum of the positive integers whose squares divide $u$ , we have $A_u \;= \; \sum_{v^2|u}v$ and hence $A = \sum_{u=1}^{100}A_u \; = \; \sum_{u=1}^{100} \sum_{{1 \le v \le 10} \atop {v^2|u}}v \; = \; \sum_{v=1}^{10} \sum_{{1 \le u \le 100} \atop {v^2|u}}v \; = \; \sum_{v=1}^{10} v\big\lfloor \tfrac{100}{v^2}\big\rfloor$ noting that $\lfloor \tfrac{100}{v^2}\rfloor$ is the number of multiples of $v^2$ between $1$ and $100$ . Similarly $B \; = \; \sum_{u=1}^{100}B_u \; = \; \sum_{v=1}^4 v \big\lfloor\tfrac{100}{v^3}\big\rfloor$ The sum for $A$ is for $1 \le v \le 10$ , since that is the range of positive integers whose square is less than or equal to $100$ . Similarly the sum for $B$ is for $1 \le v \le 4$ since that is the range of positive integers whose cube is less than or equal to $100$ .

Thus we obtain $A = 280$ and $B = 137$ , and hence $A-B = \boxed{143}$ .