Clever 4-term Sequences

Considering increasing and decreasing GP's same, we can write the four terms as a, ar, ar^2, ar^3 Clearly 'a' has to be a positive integer, but key fact is that 'r' need not to be. Let r = p/q where p and q are co-prime positive integers and also p > q as I am considering increasing GPs only. So the four terms of GP become: a, a(p/q), a(p/q)^2, a(p/q)^3 It can be easily inferred, for each term of GP to be a positive integer, that a is of the form k(q)^3. Now checking cases: (Ai) q = 1, p = 2 and a varies from 1 to 12 i.e. 12 sequences {1, 2, 4, 8}, {2, 4, 8, 16}, {3, 6, 12, 24}, {4, 8, 16, 32}, {5, 10, 20, 40}, {6, 12, 24, 48}, {7, 14, 28, 56}, {8, 16, 32, 64}, {9, 18, 36, 72}, {10, 20, 40, 80}, {11, 22, 44, 88}, {12, 24, 48, 96} (Aii) q = 1, p = 3 and a varies from 1 to 3 i.e. 3 sequences {1, 3, 9, 27}, {2, 6, 18, 54}, {3, 9, 27, 81} (Aiii) q = 1, p = 4 and a can be 1 only i.e. 1 sequence {1, 4, 16, 64} (B) q = 2, p = 3 and a = k(2)^3 = 8k where k varies from 1 to 3 i.e. 3 sequences {8, 12, 18, 27}, {16, 24, 36, 54}, {24, 36, 54, 81} (C) q = 3, p = 4 and a = k(3)^3 = 27k where k can be i only i.e. 1 sequence {27, 36, 48, 64} So total GPs possible are: 12 + 3 + 1 + 3 + 1 = 20 they are considering decreasing GP too and GP's whose common difference is 1.

soanswer is 20*2 + 100 = 140

Let the geometric sequence be a,ar,ar^2,ar^3. Notice that if r=1, then the sequence is just a,a,a,a. There are 100 possibilities for this case, namely a=1,2,3....100.

Furthermore, notice that if r>1, we can flip the sequence and write ar^3,ar^2,ar,a and obtain a sequence with ratio 1/r, which is less than 1. Thus all sequences with r<1 have a corresponding sequence where r>1, so we only need to find the number of sequences with r>1 and double that number.

If r is an integer, we have the cases r=2,3,4. If r>=5, ar^3>=125>100. For r=2, we have a=1,2,...12. For r=3, we have a=1,2,3. For r=4, we have a=1. Thus, when r is an integer, we have a total of 12+3+1=16 possibilities.

If r is not an integer, it must be a rational number, since all the terms of the sequence must be positive integers. We can write r=m/n. Notice that a must be a multiple of n^3, since ar^3 is an integer. Thus n can only be 2,3, or 4, since 5^3>100. If n=2, it becomes evident that m can only be 3, so r=3/2 and a=8,16,24. If n=3, m can only be 4 and a=27. Therefore, when r is rational but not an integer, we have 3+1=4 possibilities.

Thus, for r>1, we have a total of 16+4=20 possibilities. There must be 20 possibilities for r<1, then, so our final answer is 100+20+20=140.

What we first must substantiate is that is there exists a geometric sequence with a common ratio r, then there also exists another geometric sequence with common ratio 1/r.

Using this idea, lets first find all the geometric sequences with common ratio 1. They are:

(1, 1, 1, 1) (2, 2, 2, 2) (3, 3, 3, 3,) ..... And so on until we reach (100, 100, 100, 100), giving us a total of 100 sequences.

Since 1/1 (1/r) is also 1, we cannot add to our pre-existing sum.

Now let's consider the common ratio to be 2.

We have:

(1. 2, 4, 8) (2, 4, 8, 16) (3, 6, 12, 24) ..... Until we reach (12, 24, 48, 96) giving us a total of 12 sequences.

However, since 1/2 (1/r) yields a new common ratio, we get 12 more geometric, making the total 24.

Now let's find the number of geometric sequence with common ratio 3. They are:

(1, 3, 9, 27) (2, 6, 18, 54) (3, 9, 27, 81)

Giving us a total of 3 sequences. But since 1/3 (1/r) yields a new common ratio, we now have 6 sequences.

Let consider the number of sequences that have 4 as the common ratio. There's only one: (1, 4, 16, 64)

Since 1/4 yields yet another new common ratio, out total is now two.

Adding to our preexisting sums, we get 100 + 24 + 2 + 6 =132

However, we're still not done yet.

What if the common ratio was 2/3? Then we'd get:

(27, 18, 12, 8) (54, 36, 24, 16) (81, 54, 36, 24)

This yields a total of 3 sequences. Since (1/(2/3)) = 3/2, yielding yet another geometric sequence. so our total is 6.

But we still aren't done yet. What if the common ratio was 3/4?

we result in only one: (64, 48, 36, 27)

Since 1/(3/4) yields another sequence, we get two.

This brings our total now to: 132 + 6 + 2 = 140.

Regardless of the casework shown here, the key thing here is to notice that the sequences couldn't have exceeded a maximum common ratio of 4. since (4^3) would be the 4th term multiplied by the first term (and 5^3 exceeds 100). However, when fractions come, its important to note that the DENOMINATOR of the fraction is what triggers the common ratio (for instance, integer sequences with a common ratio of 3/4 would only require numbers that have 4^3 in them, since (3/4)^3 would be part of the last term). Using these ideas, we see that the common ratios possible are only:

1, 2, 3, 4, 1/2, 1/3, 1/4, 2/3, 3/2, 3/4, 4/3. (7/2 wouldn't work since the last term would be greater than 100).

Hoped you enjoyed! :D

Denote n is the common ratio When n=1 there are 100 sequences. When n=2, we have these following sequences: (1; 2; 4; 8), (2; 4; 8; 16), ..., (12; 24; 48; 96). Then you reverse these sequences, we have: (96; 48; 24; 12), ..., (16; 8; 4; 2), (8; 4; 2; 1). When n=3, we have these sequences: (1; 3; 9; 27), (2; 6; 18; 54), (3; 9; 27; 81) and their reverses. When n=4, we have these sequences: (1; 4; 16; 64) and its reverse. When n=\frac{3}{2}, we have these sequences: (8; 12; 18; 27), (16; 24; 36; 54), (24; 36; 54; 81) and their reverses. When n=\frac{4}{3}, we have the sequence: (27; 36; 48; 64) and its reverse. In total, we have 140 sequences.

The first case is when r=1 making all the terms the same (100 possiblities, one for each integer from 1 to 100). The second case is r=1/2 (which is essentially the same as r=2). Because the final term can be expressed as ar^3, the initial term must be divisible by 2^3 or 8 (12 possiblities, one for each multiple of eight from 8 to 96). For r=1/3 (or r=3) and r=2/3 (or r=3/2), there are 3 possiblities for each (multiples of 27 from 27 to 81). For r=1/4 (or r=4) and r=3/4 (or r=4/3), there is one possiblity (64). So 100+2(12+3+3+1+1)=140.

Let t is the common ratio If t=1, there are 100 of them. (1; 1; 1; 1) (2; 2; 2; 2) ... (100; 100; 100; 100)

If t=2, we have (1; 2; 4; 8) (2; 4; 8; 16) (3; 6; 12; 24) ... (12; 24; 48; 96) and their reverses.

If t=3, we have: (1; 3; 9; 27) (2; 6; 18; 54) (3; 9; 27; 81) and their reverses.

If t=4, we have (1; 4; 16; 64) and its reverse.

If t=\frac{3}{2}, we have (8; 12; 18; 27) (16; 24; 36; 54) (24; 36; 54; 81) and their reverses.

If t=\frac{4}{3}, we have (27; 36; 48; 64) and its reverse.

140 so far ... there could well be others. So we will have 140 sequences.

Denote n is the common ratio When n=1, there are 100 sequences: (1; 1; 1; 1) (2; 2; 2; 2) ... (100; 100; 100; 100) If n=2, these sequences are (1; 2; 4; 8) (2; 4; 8; 16) (3; 6; 12; 24) ... (12; 24; 48; 96) and their reverses. If n=3, we have: (1; 3; 9; 27) (2; 6; 18; 54) (3; 9; 27; 81) and their reverses. If n=4, the sequence is (1; 4; 16; 64) and its reverse. If n=\frac{3}{2}, we have (8; 12; 18; 27) (16; 24; 36; 54) (24; 36; 54; 81) and their reverses. If n=\frac{4}{3}, we obtaim (27; 36; 48; 64) and its reverse. In conclusion, there are 140 sequences.

Let the terms be $a, ar, ar^2, ar^3$ . We will consider the different values of $r$ . We will only consider values of $r \geq 1$ , since a sequence with $r < 1$ is just the reverse of a sequence with $r >1$ . When $r = 1$ , the sequence is $a,a,a,a$ , so we have 100 possibilities. We now consider the cases where $r > 1$ .

$r$ is the ratio of 2 integers. Let $r = \frac{p}{q}$ , where $\gcd(p,q) = 1$ and $p>q$ . Then our sequence is $a, \frac{ap}{q}, \frac{ap^2}{q^2}, \frac{ap^3}{q^3}$ . Since $p,q$ are coprime, we get that $q^3 | a$ , so we must have that $a = kq^3$ for some integer $k \geq 1,$ . This gives us sequences of the form $(kq^3, kq^2 p, kqp^2, kp^3)$ . Since each of these terms is less than 100, we must have $kp^3 < 100$ , or that $p \leq 4$ . Hence, we have $1 \leq q < p \leq 4$ .

For $q = 1$ , the ratio is simply $p$ an integer. When $p = 2$ , the sequence is $a,2a,4a,8a$ . So we need $0 < a$ and $8a \leq 100$ , so there are 12 values of $a$ that work.
When $p = 3$ , the sequence is $a,3a,9a,27a$ , so there will be 3 values of $a$ that work.
When $r = 4$ , the sequence is $a,4a,16a,64a$ , so there is one value of $a$ which works.
This gives a total of $12 + 3 + 1 = 16$ sequences.

For $q=2$ , we must have $p = 3$ , since $\gcd(p,q) = 1$ . This gives us the sequence $8k, 12k, 18k, 27k$ . For all terms to be less than 100, we have 3 choices for the value of $k$ .

When $q = 3$ , we must have $p=4$ . This gives the sequence $27a, 36a, 48a, 64a$ , so there is only 1 possible value of $a$ .

Finally, combining the 4 sequences when $r$ is not an integer with the 16 sequences when $r$ is an integer gives 20 sequences with $r > 1$ . There will be an equal number of sequences with $r < 1$ , for a total of 40 sequences. Combined with the 100 sequences when $r = 1$ , there are $140$ geometric sequences with all terms positive integers from 1 to 100.

The first case is when r=1 making all the terms the same (100 possiblities, one for each integer from 1 to 100). The second case is r=1/2 (which is essentially the same as r=2). Because the final term can be expressed as ar^3, the initial term must be divisible by 2^3 or 8 (12 possiblities, one for each multiple of eight from 8 to 96). For r=1/3 (or r=3) and r=2/3 (or r=3/2), there are 3 possiblities for each (multiples of 27 from 27 to 81). For r=1/4 (or r=4) and r=3/4 (or r=4/3), there is one possiblity (64). So 100+2(12+3+3+1+1)=140.