Find the number of 6 -term strictly increasing geometric progressions, such that all terms are positive integers less than 1000.

Answer 1

The smallest first term is 1
so term(6) = r^5
then r^5 < 1000
r < 3.98

so r could be 3 or 2

if r = 3 , then a possible sequence is
1 3 9 27 81 243 ******
if r = 2, we could have
1 2 4 8 16 32 ****

or the first term could be 2
then term(6) = 2r^5
2r^5 < 1000
r^5 < 500
r < 3.4 , r could be 3 or 2

for r = 3 another possible sequence is
2 6 18 54 162 486 ****
for r = 2 a possible sequence is
2 4 8 16 32 64 ****

could a = 3 ?
then term(6) = 3r^5 < 1000
r^5 < 1000/3
r < 3.19 , so r could be 3 or 2
if r = 3 , possible sequence:
3 9 27 81 243 729 ****
if r = 2 , possible sequence :
3 6 12 24 48 96 ****

how about a = 4 ?
4r^5 < 1000
r^5 < 250
r < 3.01 , let r = 3
possible sequence:
4 12 36 108 324 972 ****
let r = 2, then we could have
4 8 16 32 64 128 ****

how about a = 5
5r^5 < 1000
r^5 < 200
r< 2.88 , so r = 2
sequence is
5 10 20 40 80 160 ****

how about a = 10 ?
10r^5 < 1000
r^5 < 100
r < 2.5 , so let r = 2
sequence:
10 20 40 80 160 320 ****

how about a = 50 ? NO!
50r^5 < 1000
r^5 < 20
r < 1.8 , so we have to use r = 1, but at r = 1, the terms would stay the same

So, as long as r > 1 , we can have a possible sequence
but the smallest value of r we can use is 2
a(2^5) = 1000
a = 31.25

if a = 1, we have 2
if a = 2 we have 2
if a = 3, we have 2
if a = 4 , we have 2
if a = 5 we have 1
if a = 6, we have 1
....
if a = 30 , we have 1
if a = 31, we have 1

check: if a = 31
31a^5 < 1000
r^5 < 1000/31
r < 2.0032 , could use r = 2
so the sequence would be
31 62 124 248 496 992

if a = 32
32r^5 < 1000
r^5 < 1000/32
r < 1.999 , which would mean we have to use r = 1, which of course would NOT be an increasing sequence

So the last value of a we could use is a = 31

so I count 35 such sequences
( I marked the first few with ****)

Answer 2

so what is the final answer?

Answer 3

39

Answer 4

39 is correct. thanks.

Answer 5

Why are there 39? I too used the exact same method Reiny used and can't figure out how people keep ending up with 39...

Answer 6

I also counted 35, I got 31 for r=2 and 4 for r=3, im not sure where you guys got another 4 possibilities :/

Answer 7

use r=3/2, nowhere does it say r is an integer.
64,96,144,216,324,486