Question

a scientist has discovered an organism that produces five offspring exactly one hour after its own birth, and then goes on to live for one week without producing any additional offspring. each replicated organism also replicates at the same rate. at hour one, there is one organism. at hour two, there are five more organisms. how many total organisms are there at hour seven?

○ 2,801

● 19,531

○ 19,607

○ 97,655

Explanation:

Step1: Identify the pattern

The number of organisms at each hour follows a geometric series. At hour 1: \(a_1 = 1\). At hour 2: \(a_2 = 1 + 5 = 6\) (original + 5 offspring). Wait, actually, the growth is such that each organism produces 5 offspring at the next hour. So the recurrence relation is \(a_n = a_{n - 1}+5\times a_{n - 1}=6\times a_{n - 1}\)? Wait, no. Wait, at hour 1: 1 organism. At hour 2: that 1 organism produces 5, so total \(1 + 5 = 6\). At hour 3: each of the 6 organisms (wait, no, the original organism lives for a week, so at hour 3, the original organism is still alive, and the 5 offspring from hour 2 each produce 5? Wait, no, the problem says "produces five offspring exactly one hour after its own birth, and then goes on to live for one week without producing any additional offspring". So each organism, at 1 hour after its birth, produces 5, then no more. So the timeline:

- Hour 1: 1 organism (born at hour 0, maybe? Wait, the problem says "At hour one, there is one organism. At hour two, there are five more organisms." So hour 1: 1. Hour 2: 1 (original) + 5 (offspring) = 6.

- Hour 3: The original organism is still alive (lives a week), and each of the 5 offspring from hour 2 will produce 5 at hour 3? Wait, no. Wait, the offspring are born at hour 2. So at hour 3, which is 1 hour after their birth (hour 2), each of the 5 offspring from hour 2 will produce 5. Also, the original organism (born at hour 1? Wait, maybe the initial organism is born at hour 1. Then at hour 2 (1 hour after birth), it produces 5. Then at hour 3, the initial organism is still alive, and the 5 offspring from hour 2 are now 1 hour old, so each produces 5. So hour 3: original (1) + 5 (from hour 2) + 5*5 (from the 5 offspring) = 1 + 5 + 25 = 31? Wait, no, wait the total at hour 2 is 6 (1 + 5). At hour 3, the 5 offspring from hour 2 each produce 5, so the new offspring are 5*5 = 25. So total at hour 3: 6 (hour 2) + 25 = 31? Wait, no, that's not right. Wait, the problem says "each replicated organism also replicates at the same rate". So the replication rate is: each organism, at 1 hour after its birth, produces 5, then no more. So the number of organisms at hour n is:

- Hour 1: 1 (organism A, born at hour 1? Or hour 0? The problem says "At hour one, there is one organism." So hour 1: 1.

- Hour 2: organism A produces 5, so total 1 + 5 = 6 (organisms A, B, C, D, E, F? Wait, no, 1 original + 5 offspring: 6 total.

- Hour 3: Organism A is still alive (lives a week), and each of the 5 offspring (B-F) from hour 2 are now 1 hour old (born at hour 2), so each produces 5. So the new offspring from B-F: 5*5 = 25. So total at hour 3: 6 (hour 2) + 25 = 31? Wait, no, that's 6 + 25 = 31? Wait, but 6*5 + 1? No, maybe the recurrence is \(a_n = a_{n - 1}+5\times a_{n - 1}\)? Wait, no. Wait, at hour 1: 1. Hour 2: 1 + 5*1 = 6 (since the 1 organism produces 5). Hour 3: 6 + 5*5 = 6 + 25 = 31? No, 5*5 is 25, but where does the 6 come from? Wait, no, maybe the recurrence is \(a_n = 1 + 5 + 5^2 + 5^3 + ... + 5^{n - 1}\)? Wait, no. Wait, let's list the hours:

- Hour 1: 1 = \(5^0\)

- Hour 2: 1 + 5 = \(5^0 + 5^1\)

- Hour 3: 1 + 5 + 5^2 = \(5^0 + 5^1 + 5^2\)

- Wait, no, that would be if each hour, the new offspring are \(5^{n - 1}\). But hour 2: 1 (original) + 5 (offspring) = 6 = 1 + 5. Hour 3: original (1) + 5 (from hour 2) + 5*5 (from the 5 offspring) = 1 + 5 + 25 = 31 = 1 + 5 + 25. Hour 4: 1 + 5 + 25 + 125 = 156? Wait, no, that can't be. Wait, maybe the formula is a geometric series where the first term \(a = 1\), and the common ratio \(r = 6\)? Wait, hour 1: 1. Hour 2: 6. Hour 3: 6*6 = 36? No, that doesn't match. Wait, the problem says "each replicated organism also replicates at the same rate". So the total number of organisms at hour n is the sum of a geometric series. Let's re-express:

At hour 1: \(N_1 = 1\)

At hour 2: \(N_2 = N_1 + 5\times N_1 = 6\times N_1 = 6\)

At hour 3: \(N_3 = N_2 + 5\times N_2 = 6\times N_2 = 6^2\times N_1 = 36\)? Wait, but that contradicts the earlier thought. Wait, maybe the original organism, after producing 5, continues to live but doesn't produce more, but the offspring do. Wait, no, the problem says "each replicated organism also replicates at the same rate". So every organism, when it's 1 hour old, produces 5, then lives on. So the number of organisms at each hour is:

- Hour 1: 1 (organism A, born at hour 1)

- Hour 2: A produces 5 (B, C, D, E, F), so total 1 + 5 = 6 (A, B, C, D, E, F)

- Hour 3: A is still alive (lives a week), B, C, D, E, F are now 1 hour old (born at hour 2), so each produces 5. So new organisms: 5*5 = 25. So total at hour 3: 6 + 25 = 31? Wait, but 6 + 25 = 31. Then hour 4: A is alive, B-F are alive, and the 25 new organisms from hour 3 are now 1 hour old, so each produces 5. So new organisms: 25*5 = 125. Total at hour 4: 31 + 125 = 156? No, that's not a geometric series with ratio 6. Wait, maybe the initial interpretation is wrong. Let's look at the answer choices. The options include 19531, which is \(6^5 + ...\)? Wait, let's compute the sum of a geometric series. Let's assume that at each hour n (starting from n=1), the number of organisms is \(N_n = 1 + 5 + 5^2 + 5^3 + ... + 5^{n - 1}\)? Wait, no. Wait, hour 1: 1 = \(5^0\)

Hour 2: 1 + 5 = \(5^0 + 5^1\)

Hour 3: 1 + 5 + 5^2 = \(5^0 + 5^1 + 5^2\)

Hour 4: \(5^0 + 5^1 + 5^2 + 5^3\)

...

Hour 7: Sum from k=0 to 5 of \(5^k\)? Wait, no, hour 1 is k=0, hour 2 is k=0 to 1, ..., hour 7 is k=0 to 6? Wait, no, let's check the hours:

Hour 1: 1 (k=0: \(5^0\))

Hour 2: 1 + 5 = \(5^0 + 5^1\) (k=0 to 1)

Hour 3: \(5^0 + 5^1 + 5^2\) (k=0 to 2)

...

Hour n: Sum from k=0 to n-1 of \(5^k\)

Wait, the sum of a geometric series is \(S_n = \frac{5^n - 1}{5 - 1}\) when the first term is 1 and ratio 5. Wait, no, the sum of \(1 + 5 + 5^2 + ... + 5^{m}\) is \(\frac{5^{m + 1} - 1}{4}\).

Wait, let's check hour 2: m=1 (terms 5^0 and 5^1), sum is \(\frac{5^2 - 1}{4} = \frac{25 - 1}{4} = 6\), which matches (1 + 5 = 6).

Hour 3: m=2 (terms 5^0, 5^1, 5^2), sum is \(\frac{5^3 - 1}{4} = \frac{125 - 1}{4} = 31\), which matches the earlier calculation.

Hour 4: m=3, sum \(\frac{5^4 - 1}{4} = \frac{625 - 1}{4} = 156\)

Hour 5: m=4, \(\frac{5^5 - 1}{4} = \frac{3125 - 1}{4} = 781\)

Hour 6: m=5, \(\frac{5^6 - 1}{4} = \frac{15625 - 1}{4} = 3906\)

Hour 7: m=6, \(\frac{5^7 - 1}{4} = \frac{78125 - 1}{4} = \frac{78124}{4} = 19531\). Ah! That matches the option 19,531. So the correct formula is the sum of a geometric series with first term 1, ratio 5, for 7 terms (from hour 1 to hour 7, which is 7 terms: k=0 to 6, since hour 1 is k=0, hour 7 is k=6). Wait, the number of terms from hour 1 to hour 7 is 7? Wait, hour 1: term 1 (k=0), hour 2: term 2 (k=0,1), ..., hour 7: terms k=0 to 6 (7 terms). So the sum is \(\frac{5^7 - 1}{5 - 1} = \frac{78125 - 1}{4} = 19531\).

Step2: Apply the geometric series formula

The sum of a geometric series is given by \(S_n = \frac{a(r^n - 1)}{r - 1}\), where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms.

Here, \(a = 1\) (hour 1: 1 organism), \(r = 5\) (each hour, the number of new organisms is 5 times the previous new organisms? Wait, no, the ratio is 5 because each organism produces 5, and the total number of organisms at each hour is the sum of the series. Wait, actually, the total number of organisms at hour \(n\) is the sum of \(1 + 5 + 5^2 + ... + 5^{n - 1}\). So for \(n = 7\) (hour 7), the number of terms is 7 (from \(5^0\) to \(5^6\)).

So \(S_7 = \frac{5^7 - 1}{5 - 1}\)

Compute \(5^7 = 78125\)

Then \(5^7 - 1 = 78124\)

Divide by 4: \(78124 \div 4 = 19531\)

Answer:

19,531