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Find a_1, a_2, ... , a_10

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,685
Find \(\displaystyle a_1,\;a_2,\;a_3,\;\cdots\;, a_{10}\) given that:

i) They are in geometric progression, in this order.
ii) \(\displaystyle a_1,\;a_2,\;a_3,\;a_4,\;a_5,\;a_6\) have 4 digits and \(\displaystyle a_{10}\) have 5 digits.
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
[JUSTIFY]Clearly the geometric term $r$ has to be greater than 1. Let $1000 \leq a_6 < 10000$. Then we have the following constraints on $r$:

1. $a_1 \geq 1000$, that is, $a_6 \cdot r^{-5} \geq 1000 ~ ~ ~ \implies ~ ~ ~ r \leq \sqrt[5]{\frac{a_6}{1000}}$

2. $a_7 \geq 10000$, which implies $a_6 \cdot r > 10000 ~ ~ ~ \implies ~ ~ ~ r \geq \frac{10000}{a_6}$

3. $a_{10} < 10000$, so $a_6 \cdot r^4 < 100000 ~ ~ ~ \implies ~ ~ ~ r < \sqrt[4]{\frac{100000}{a_6}}$

So, given $1000 \leq a_6 < 10000$, any $r$ satisfying the condition below is a solution to the problem.
$$\frac{10000}{a_6} < r < \min{\left ( \sqrt[5]{\frac{a_6}{1000}}, \sqrt[4]{\frac{100000}{a_6}} \right )}$$
Note that this interval may be empty. I haven't done the calculations but it is obvious that for $a_6$ close to $1000$ there is no solution, and so on. Also the inequalities may be a bit wrong with respect to $\leq$ and $<$ because I'm lazy but I trust it will not be an issue.

Let's try it with $a_6 = 9183$ at random. Then we have:
$$\frac{10000}{a_6} \approx 1.09$$
$$\sqrt[5]{\frac{a_6}{1000}} \approx 1.56$$
$$\sqrt[4]{\frac{100000}{a_6}} \approx 1.82$$
Therefore our condition on $r$ is approximately:
$$1.09 < r < 1.56$$
Arbitrarily, let's pick $r = 1.2$. Then:
$$a_1 = a_6 \cdot r^{-5} \approx 3690$$
And the entire sequence follows, rounded up to integers:
$$a = \{ 3690, 4429, 5314, 6377, 7653, 9183, 11020, 13224, 15868, 19042 \}$$
$$\blacksquare$$

This approach assumes that $a_7$ has 5 digits, this wasn't explicitly specified in the problem and so this method does not capture all solutions (it does capture all solutions under the assumption that $a_7$ has 5 digits, though, as far as I can tell, and should be able to be tweaked to assume that $a_7$ has 4 digits but $a_8$ has 5 and so on).[/JUSTIFY]
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,685
Hi Bacterius,

Thanks for participating but I believe the problem is set for not allowing us to round the answer to the nearest integer.:eek:

While I managed to solve it by generating a sequence (that consists of only integers value of the first ten terms) that satisfied the aforementioned conditions, I'm not certain if that is the only pair of answer to this problem.
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
Hi Bacterius,

Thanks for participating but I believe the problem is set for not allowing us to round the answer to the nearest integer.:eek:

While I managed to solve it by generating a sequence (that consists of only integers value of the first ten terms) that satisfied the aforementioned conditions, I'm not certain if that is the only pair of answer to this problem.
Oh, right, I am stupid, haha. I got confused between digits and integers. Sorry. Well at least my solution works for reals with thresholds. I take it the geometric term is an integer too, then. Anyway, back to the drawing board (Tongueout)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
Bacterius's method looks like a good starting point. If the first term is $a$ and the common ratio is $r$, then we have the conditions $a\geqslant 1000$, $ar^5<10\,000$, $ar^9\geqslant 10\,000$. That tells us that $r^5<10$ and $r^9>10$, from which $10^{1/9}<r<10^{1/5}$, or $1.291<r<1.584$.

Next, if the numbers $ar^k\ (0\leqslant k\leqslant 9)$ are all integers, then $r$ must be rational, say $r=p/q$ with $p,q$ co-prime integers. If $ar^9 = \dfrac{ap^9}{q^9}$ is to be an integer then $a$ must be a multiple of $q^9$. But that forces $q$ to be $2$ (because $3^9 = 19683$, which already has five digits). Therefore $r = 3/2$ and $a$ must be a multiple of $2^9 = 512$. Since $a$ has four digits it must be at least $2*512 = 1024$. The next possibility would be $a=3*512=1536$. But then $ar^5 = 11664$, which is too big, since we are told that $a_6$ has four digits.

Therefore the only possibility is that $a=1024$ and $b=3/2$. The sequence is then
1024
1536
2304
3456
5184
7776
11664
17496
26244
39366​