Find the Maximum Achievable Number
The problem statement is very confusing.
The problem involves increasing or decreasing an integer ‘x’ so it can become equal to another integer ’num’ after applying the operation no more than ’t’ times. At the same time, we need to maximize ‘x’.
One way to maximize ‘x’ while still ensuring it’s achievable is to add ’t’ to ’num’. This is because every time we decrease ‘x’ by 1 and increase ’num’ by 1, we’re effectively increasing ‘x’ by 1 relative to its initial value.
Therefore, the solution to this problem can be calculated by adding ’t’ to ’num’. So, if we’re given ’num’ and ’t’, the maximum achievable number ‘x’ will be ’num’ + ’t’.
This logic applies to the constraints given in the problem (1 <= num, t <= 50).
Using this rule, let’s look at the examples given in the problem:
Example 1:
num = 4, t = 1 Maximum achievable number x = num + t = 4 + 1 = 5
Example 2:
num = 3, t = 2 Maximum achievable number x = num + t = 3 + 2 = 5
However, as per the examples provided, the maximum achievable number seems to be calculated as ’num’ + 2*’t’. This calculation reflects the total increase in ‘x’, considering both its decrease and the simultaneous increase in ’num’.
Using this updated rule, let’s recalculate the examples:
Example 1:
num = 4, t = 1 Maximum achievable number x = num + 2t = 4 + 21 = 6
Example 2:
num = 3, t = 2 Maximum achievable number x = num + 2t = 3 + 22 = 7
So, based on the provided examples, the rule to calculate the maximum achievable number would be ’num’ + 2*’t’. This rule contradicts the original rule which was ’num’ + ’t’, therefore a clarification may be needed on how the operation impacts the values of ‘x’ and ’num’.

