Terry Tao's Magic Money Plant Puzzle: The Full Solution 2026

Terry Tao, the UCLA professor often called the Mozart of math, recently shared a deceptively simple puzzle about a magic money plant. The puzzle challenges players to figure out coin totals, derive an expression for any day, and work backward from a final amount. It's a brilliant exercise in exponential growth and linear algebra, and we've got the full solution.

What Is the Magic Money Plant Puzzle?

In an online game, you earn coins by harvesting from plants. There are two types: a regular plant produces one coin per day, while a mother plant produces one coin and two seeds per day. A seed grows into a regular plant the following day. You start on Day 1 with a mother plants, b regular plants, and c coins, with each quantity no more than nine.

The puzzle asks three questions: (1) How many coins do you have at the end of days 2 and 3? (2) Find an expression for coins at the end of Day X. (3) If you have 427 coins at the end of Day 10, what are a, b, and c?

Step-by-Step Solution for Days 2 and 3

Let's calculate the coin totals day by day. At the end of Day 1, you have a + b + c coins. Each mother plant has also produced two seeds, but these have not yet grown into regular plants.

On Day 2, those seeds become regular plants and each produces a coin, so there are 2a additional coins from them. The original mother plants produce another a coins, and the original regular plants produce another b coins. Adding the previous total, we get:

• Day 2: 4a + 2b + c
• Day 3: 8a + 3b + c

For Day 3, the reasoning is similar: the seeds from Day 2's mother plants become regular plants, adding more coins. The pattern begins to emerge.

General Expression for Day X

To find the number of coins at the end of Day X, we need to account for the growth of the plant population. Each mother plant produces two seeds per day, which become regular plants the next day. This leads to an exponential increase in the number of regular plants over time.

By carefully summing the contributions, we can derive that the total coins at the end of Day X is:

Coins = a * (2^X - X - 1) + b * X + c

This formula captures the exponential growth from mother plants (the 2^X term) and the linear growth from regular plants.

Solving for a, b, and c When Day 10 Has 427 Coins

We are given that at the end of Day 10, the total coins equal 427. Plugging X = 10 into our formula:

427 = a * (2¹⁰ - 10 - 1) + b * 10 + c = a * (1024 - 11) + 10b + c = 1013a + 10b + c

Since a, b, and c are each between 0 and 9 (inclusive), we need to find integers that satisfy this equation. The term 1013a is much larger than the others, so a must be 0 because if a ≥ 1, 1013a would exceed 427. Thus a = 0.

Then we have 10b + c = 427. Since b and c are at most 9, 10b + c can be at most 99. This is impossible, so our initial assumption must be wrong. Let's re-examine the formula.

Wait, the formula might be incorrect. Let's derive it properly.

Let's denote the number of mother plants at the start of day d as M_d and regular plants as R_d. Initially, M₁ = a, R₁ = b, and coins C₁ = c.

Each day, the coins increase by the number of mother plants plus the number of regular plants (since each produces one coin). Also, each mother plant produces two seeds, which become regular plants the next day. So the number of regular plants at the start of day d+1 is R_d + 2M_d. The number of mother plants remains constant: M_d+1 = M_d = a.

Thus, the coin increment on day d is M_d + R_d = a + R_d.

We can write a recurrence for R_d: R₁ = b, and R_d+1 = R_d + 2a.

Solving this recurrence: R_d = b + 2a(d-1).

Now, the total coins at the end of day X is C₁ + sum_{d=1}^{X} (a + R_d) = c + sum_{d=1}^{X} [a + b + 2a(d-1)] = c + X(a+b) + 2a sum_{d=1}^{X} (d-1).

The sum sum_{d=1}^{X} (d-1) = sum_{k=0}^{X-1} k = X(X-1)/2.

So total coins = c + X(a+b) + a X(X-1).

Let's check with X=2: c + 2(a+b) + 2a = c + 4a + 2b, which matches. For X=3: c + 3(a+b) + 6a = c + 9a + 3b? Wait, earlier we had 8a+3b+c. There's a discrepancy. Let's recalculate.

At the end of day 1, coins = a+b+c.

Day 2: mother plants produce a coins, regular plants produce b coins, and the seeds from day 1 (2a) become regular plants and produce 2a coins. So total added = a + b + 2a = 3a + b. Total = a+b+c + 3a+b = 4a+2b+c. Correct.

Day 3: mother plants produce a coins, regular plants now include original b plus 2a from day 1 seeds, so they produce b+2a coins. Also, the mother plants from day 2 produce 2a seeds that become regular plants on day 3, but those don't produce coins until day 4. So total added on day 3 = a + (b+2a) = 3a+b. Total = 4a+2b+c + 3a+b = 7a+3b+c. But earlier we had 8a+3b+c. Which is correct?

Let's re-read the problem: "a mother plant produces a single coin and two seeds per day. (A seed will grow into a regular plant the following day.)" So on day 1, mother plants produce a coins and 2a seeds. Those seeds grow into regular plants on day 2. So on day 2, we have the original a mother plants, original b regular plants, and 2a new regular plants. They all produce coins: a from mothers, b from original regulars, and 2a from new regulars. So total coins produced on day 2 = a + b + 2a = 3a + b. Plus the c coins from day 1? Actually, the coins accumulate. At the end of day 1, we have a+b+c coins. At the end of day 2, we add the coins produced on day 2, which is 3a+b, so total = a+b+c + 3a+b = 4a+2b+c. That matches.

Now day 3: On day 2, the mother plants produced another 2a seeds, which grow into regular plants on day 3. So on day 3, we have a mother plants, and regular plants = original b + 2a (from day 1 seeds) + 2a (from day 2 seeds) = b+4a. They produce coins: a from mothers, and b+4a from regulars, total = a + b + 4a = 5a + b. Add to previous total: 4a+2b+c + 5a+b = 9a+3b+c. But earlier I had 8a+3b+c. So which is right? Let's list explicitly.

Day 1 start: M=a, R=b, C=c.

Day 1 production: coins = a + b. Seeds produced = 2a. End of day 1: coins = c + a + b. Seeds = 2a (will become R on day 2).

Day 2 start: M=a, R = b + 2a. Coins = c+a+b.

Day 2 production: coins = a + (b+2a) = 3a+b. Seeds = 2a. End of day 2: coins = c+a+b + 3a+b = c+4a+2b. Seeds = 2a (will become R on day 3).

Day 3 start: M=a, R = (b+2a) + 2a = b+4a. Coins = c+4a+2b.

Day 3 production: coins = a + (b+4a) = 5a+b. End of day 3: coins = c+4a+2b + 5a+b = c+9a+3b.

So the correct expression for day 3 is 9a+3b+c, not 8a+3b+c. The source article had 8a+3b+c, which might be a typo. We'll proceed with the correct derivation.

Thus, the general formula for total coins at the end of day X is:

Coins = c + X(a+b) + a X(X-1)

Let's verify for X=1: c + 1(a+b) + a*0 = a+b+c, correct. X=2: c+2a+2b+2a = c+4a+2b, correct. X=3: c+3a+3b+6a = c+9a+3b, correct.

Now, for X=10, total coins = c + 10(a+b) + a*10*9 = c + 10a+10b + 90a = c + 100a + 10b.

We are given that at the end of day 10, coins = 427. So:

100a + 10b + c = 427

Since a, b, c are each between 0 and 9 inclusive, we can solve this digit by digit. The equation is essentially the three-digit number 'abc' in base 10 equals 427. So a=4, b=2, c=7. Check: 100*4 + 10*2 + 7 = 400+20+7=427. Perfect.

Thus, the initial numbers are: a=4 mother plants, b=2 regular plants, and c=7 coins. The background to this story is set out in Terence Tao: The Mozart of Math and His Magic Puzzle. A related thread runs through A Pocket Full of Slagford Review: Retro Puzzle Adventure. Earlier coverage in Einstein's 90-Year Math Puzzle Solved by Spanish Researcher shows how the argument developed.

Why This Puzzle Matters

This puzzle is a beautiful illustration of how exponential growth can be hidden in simple rules. It also shows the power of mathematical modeling: by setting up recurrences and solving them, we can predict outcomes far into the future. Terry Tao's ability to craft such an elegant problem speaks to his genius. A closer look at the same subject appears in Tao of Glass Review: Stage Magic Meets Minimalist Master. Anyone following the wider picture can start with Tao of Glass Review: McDermott's Stage Magic Meditates on Glass. The same theme surfaces in Tao of Glass Review: McDermott’s Stage Magic Meets Glass’s Music, which takes a different angle.

For those interested in more puzzles and mathematical challenges, exploring similar problems can sharpen your problem-solving skills. Whether you're a student or a lifelong learner, engaging with these puzzles is a rewarding way to keep your mind sharp. Context on how this played out before sits in Solve the 3x3 Grid Puzzle: Can You Match a Math Genius?. A companion piece, Solve the 3x3 Grid Puzzle: 216 Ways Explained for Math Fans, approaches the same subject from another angle.

FAQ