Answer for Day 13 Math Problem

1,686.15 (without interest on Hive Power) 🎯

a bit more than 1,686.15 (with interest on Hive Power) 🎯

😅

Solutions

The problem is about arithmetic sequences in (elementary) algebra.

The problem does not state that there is a yearly interest on vested Hive Power (compounded daily), but congratulations for @jfang003 for pointing that out.

Solution without annual interest for Hive Power

The formula for the anth term of an arithmetic sequence is an = a1 + (n-1) * d, where

• an = nth term of the sequence
• n = the number of terms in the sequence
• d = common difference (or the difference between any two consecutive terms)

In the original problem, we are looking for the 10th term (an) of the sequence, where there are 13 terms (n). The common difference is the one-thirteenth of the difference of the end term and the start term, which is (1/13) * (3230 - 1000) ≈ -171.54 (take note of the negative sign)

The phrase "on the 10th week" is different from "at the end of the 10th week" or "just after the 10th week". You would not say that the user has exactly 1,000 Hive Power remaining (excluding 3% annual interest) on (not after) the 13th week. An alternative explanation is during (not after) the 1st week, there had been no Hive Power that was Powered Down (the 1/13 Hive Power had just been on the process of being powered down, which can be cancelled and no Hive Power would be lost or converted).

The equation now is (10th term) = 3230 + (10 - 1) * (-171.54) ≈ 1686.15

The user will have approximately 1,686.15 Hive Power remaining on the 10th week if interest on Hive Power is not applied.

Solution with annual interest for Hive Power

This solution is algorithmic (using sequential mathematical operations instead of fixed solution) because the interest per year is applied on a weekly basis compounded daily. You are free to use summation (Sigma) notation, which will arrive at the same result.

Assuming a 3% yearly increase in Hive Power compounded weekly and 52 weeks a year, the algorithm is as follows:

1. Let c = 1 (c is the counter variable) and p = 3230 (p is the Hive Power variable).
2. If c ≤ 9, multiply p by (1+0.03/52) then reduce p by ((3230-1000)/13) and increment c by 1 and repeat this step, else go to Step 3.
3. Display c.

The implementation (not exactly what the algorithm is) of the algorithm above is as follows:

The amount displayed will be approximately 1,699.40, which is the Hive Power on the 10th week (just after the 9th week).

Winner: @jfang003 🏅

1 HIVE has been sent to @jfang003's Hive account. 💰

I am not sure about @jfang003's solution, but only a correct answer is required to win in this Math mini-contest. 😅

Mentions: @holovision, @eturnerx (@eturnerx-dbuzz), and @ahmadmanga (@ahmadmangazap)
Special mentions: @chrisrice and @jancharlest 🤯