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- #1

- Feb 14, 2012

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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.

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- #1

- Feb 14, 2012

- 3,802

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.

- Jan 26, 2012

- 644

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

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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- Feb 14, 2012

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

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.

- Jan 26, 2012

- 644

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 boardBacterius,

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

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.

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- Feb 7, 2012

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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

1536

2304

3456

5184

7776

11664

17496

26244

39366