Mei's Savings Plan: Which Academic Year Does ¥500,000 Arrive?
Mei enters Kochi University in April 2026. On 1 April 2026 she opens a savings account with a deposit of ¥50,000. The bank credits interest at 1.2% per year on 31 March, and on 1 April of each following year she pays in ¥120,000 earned from a part-time job and family support. She never withdraws money. Her plan is to have at least ¥500,000 in the account for a field course and a laptop, and she wants to know, before she signs the rental contract, exactly which academic year the balance first reaches that figure. Her father warns her that if the bank instead applies the interest after the April deposit, the answer could move by a whole year. Mei also wonders whether a smaller number of larger deposits would get her there sooner.
1. Let B₀ = 50,000 be the balance on 1 April 2026, and Bₙ the balance just after the deposit on 1 April of the n-th following year. Write and solve the recurrence, and find the first n with Bₙ ≥ 500,000.
Read the worked answer
Each year the balance is multiplied by 1.012 on 31 March, then increased by 120,000 on 1 April:
Bₙ₊₁ = 1.012 Bₙ + 120000, B₀ = 50000.
Fixed point: c = 1.012c + 120000 gives c = -120000/0.012 = -10,000,000, so Bₙ = -10,000,000 + (50,000 + 10,000,000)(1.012)ⁿ = 10,050,000(1.012)ⁿ - 10,000,000.
Require 10,050,000(1.012)ⁿ ≥ 10,500,000, i.e. (1.012)ⁿ ≥ 1.0447761. Taking logs, n ≥ ln(1.0447761)/ln(1.012) = 0.0437936/0.0119287 = 3.671, so n = 4.
Check: B₃ = 10,050,000(1.036434) - 10,000,000 = 416,160 < 500,000, while B₄ = 10,050,000(1.048871) - 10,000,000 = 541,153 ≥ 500,000. The target is first reached with the deposit on 1 April 2030, during her fifth year of study - not during the four years she assumed.
2. Her father's variant: the bank credits interest after the April deposit, so every deposit also earns that year's interest. Does the date change?
Read the worked answer
Now Bₙ₊₁ = 1.012(Bₙ + 120000), B₀ = 50000. The constant added each year is effectively 120000 × 1.012 = 121,440, so Bₙ = 10,170,000(1.012)ⁿ - 10,120,000.
Require (1.012)ⁿ ≥ 10,620,000/10,170,000 = 1.0442478, giving n ≥ ln(1.0442478)/ln(1.012) = 0.0432995/0.0119287 = 3.630, so again n = 4.
The number of academic years does not move; the balance at n = 4 is 10,170,000(1.048871) - 10,120,000 = 546,918, about ¥5,800 more than before. Judgement: a shift of one year in the interest date is worth less than 5% of one deposit here, so the plan is robust to that detail, and the father's 'whole year' warning is wrong. What does move the date is the size of the deposit, not the interest.
3. How large would each annual deposit have to be for the target to be met with only three deposits (by 1 April 2029)? Comment on the cost of each year saved.
Read the worked answer
With a constant deposit D and the original interest timing, B₃ = 1.012³ × 50000 + D(1.012³ - 1)/0.012 = 51,821.7 + 3.03617 D.
Setting B₃ ≥ 500,000 gives D ≥ (500,000 - 51,821.7)/3.03617 = 448,178.3/3.03617 ≈ 147,600.
So the deposit must rise from ¥120,000 to about ¥147,600, an increase of roughly ¥27,600 per year (23%). Note how little of the target comes from interest: even at n = 3 the accumulated interest on all deposits is only a few thousand yen, so shortening the plan by one year is paid for almost entirely by the extra deposits. Judgement: if the goal is fixed in date as well as amount, increasing D is the only realistic lever; changing the interest timing cannot substitute for it.