Question

Let τ(m,n)=[T(m,n)T(n,m)], where [x] means x rounded down. What is τ(6⋅10^9,50⋅10^9)?

Answer 1

To calculate τ(6⋅10^9,50⋅10^9), we need to evaluate T(6⋅10^9,50⋅10^9) and T(50⋅10^9,6⋅10^9) separately, rounding down each result. T(6⋅10^9,50⋅10^9) is 4, and T(50⋅10^9,6⋅10^9) is 0. Therefore, τ(6⋅10^9,50⋅10^9) is [4 0].

Step-by-step explanation:

To find τ(6⋅10^9,50⋅10^9), we need to evaluate T(6⋅10^9,50⋅10^9) and T(50⋅10^9,6⋅10^9) separately, and then round down each result.

Let's start with T(6⋅10^9,50⋅10^9). In this case, we have M = 6⋅10^9 and t = 50⋅10^9.

Since t₂ represents the doubling time, we can calculate n = t/t₂ = 50⋅10^9/6⋅10^9 = 25/3.

Rounding down gives us n = [25/3] = 8. Therefore, T(6⋅10^9,50⋅10^9) = 4.

Now let's find T(50⋅10^9,6⋅10^9).

In this case, we have M = 50⋅10^9 and t = 6⋅10^9.

Calculating n = t/t₂ = 6⋅10^9/50⋅10^9 = 3/25.

Rounding down gives us n = [3/25] = 0.

Therefore, T(50⋅10^9,6⋅10^9) = 0.

Finally, we can calculate τ(6⋅10^9,50⋅10^9) as [4 0].