In the following problems, Θ = 20 + units digit of your day of birth. For example, if you
were born on April 1st, then Θ = 20 + 1 = 21. If you were born on March 30th, then Θ = 20
+ 0 = 20. You can solve any three of the seven given problems.
1. Alok and Bhanu play the following game on arithmetic expressions. Given the
expression
N = (Θ + A)/B + (Θ + C + D)/E
where A, B, C, D and E are variables representing digits (0 to 9), Alok would like to
maximize N while Bhanu would like to minimize it.
To solve this problem, we need to find the values for A, B, C, D, and E that would maximize or minimize the expression N, given the conditions of the problem.
Since we want to find the maximum value of N, we need to choose values for A, B, C, D, and E that would make each term in the expression as large as possible.
First, let's analyze the expression:
N = (Θ + A)/B + (Θ + C + D)/E
The first term, (Θ + A)/B, will be maximized when Θ and A are both at their maximum possible values and B is at its minimum value.
Since Θ is given as 20 + the units digit of your day of birth, we need to determine the units digit of your day of birth. Let's assume it is 5. In that case, Θ = 20 + 5 = 25.
Now, we need to maximize A and minimize B. A can take any value between 0 and 9, so let's choose the maximum value, which is 9. For B, since it needs to be at its minimum value, let's choose 1.
So, the first term becomes (25 + 9)/1 = 34.
Now, let's analyze the second term, (Θ + C + D)/E. Similarly, we want to maximize Θ, C, and D, while minimizing E.
Using the same assumption for Θ (25), and considering C and D can take any value between 0 and 9, let's choose the maximum values for both, which is 9.
Now, we need to minimize E. Again, E can take any value between 0 and 9, so let's choose the minimum value, which is 0.
So, the second term becomes (25 + 9 + 9)/0, which is undefined since we cannot divide by 0.
To summarize, by choosing the values A = 9, B = 1, C = 9, D = 9, and E = 0 (assuming Θ = 25), the expression N becomes:
N = (Θ + A)/B + (Θ + C + D)/E = (25 + 9)/1 + (25 + 9 + 9)/0
The second term is undefined, so the maximum value of N is (25 + 9)/1 = 34.
Therefore, Alok can maximize the value of N to 34.
To minimize N, Bhanu would need to do the opposite. They would try to minimize the values of A, C, and D, while maximizing B and E. The resulting expression would be different, but the process would be similar.