Bank Account

Let Bal be the balance in case of no withdrawal, Bal = $100 + \sum_{i=1}^{7} p_{i}$ = 139

Next, get all possible combinations of $q_{i}$ where sum is $\geq10$ . Of course, we want to have the minimum number of $q_{i}$ s.

(Without writing the algorithm used) the combination are as follows: + the loss for withdrawing

• $q_{1} + q_{2}$ $\Rightarrow$ (4+9) + (5+4) = 22
• $q_{1} + q_{3} + q_{4}$ $\Rightarrow$ (4+5+2) + (5+8+4) = 28
• $q_{1} + q_{3} + q_{5}$ $\Rightarrow$ (4+5+5) + (5+8+9) = 36
• $q_{1} + q_{3} + q_{6}$ $\Rightarrow$ (4+5+2) + (5+8+6) = 30
• $q_{1} + q_{3} + q_{7}$ $\Rightarrow$ (4+5+6) + (5+8+3) = 31
• $q_{1} + q_{4} + q_{5}$ $\Rightarrow$ (4+2+5) + (5+4+9) = 29
• $q_{1} + q_{4} + q_{6} + q_{7}$ $\Rightarrow$ (4+2+2+6) + (5+4+6+3) = 32
• $q_{1} + q_{5} + q_{6}$ $\Rightarrow$ (4+5+2) + (5+9+6) = 31

• $q_{1} + q_{7}$ $\Rightarrow$ (4+6) + (5+3) = 18 $\Leftarrow$ Min

• $q_{2} + q_{3}$ $\Rightarrow$ (9+5) + (4+8) = 26

• $q_{2} + q_{4}$ $\Rightarrow$ (9+2) + (4+4) = 19
• $q_{2} + q_{5}$ $\Rightarrow$ (9+5) + (4+9) = 27
• $q_{2} + q_{6}$ $\Rightarrow$ (9+2) + (4+6) = 21

• $q_{2} + q_{7}$ $\Rightarrow$ (9+3) + (4+6) = 22

• $q_{3} + q_{5}$ $\Rightarrow$ (5+5) + (8+9) = 27

• $q_{3} + q_{7}$ $\Rightarrow$ (5+6) + (8+3) =22

• $q_{4} + q_{6} + q_{7}$ $\Rightarrow$ (2+2+6) + (4+6+3) = 23

Note: There might be some other combinations but are not written because they are redundant. For example, $a_{1} + a_{2} + a_{3} \geq10$ and $a_{1} + a_{3} \geq10$ . The extra addend on the first expression is unnecessary because we are looking the minimum loss.

Now, $Bal - Min \implies 139 - 18 = 121$