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Let $a_1,a_2,...,a_{11}\in \mathbb{N}$ with $a_1<a_2<...<a_{11}$. If $\frac{1}{a_1}, \frac{1}{a_2},...,\frac{1}{a_{11}}$ forms an arithmetic sequence, find the smallest possible value of $a_1$.

My answer is 2520, I get this by finding the LCM of 1, 2, ..., 11, which means the arithmetic sequence is 1/27720, 2/27720, ..., 11/27720, 11/27720=1/2520, so the answer is 2520. But the answer should be 2310, what's wrong with my solution? Thank you.

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    $\begingroup$ I don't know if it helps, but interestingly $2*3*5*7*11=2310$. The LCM of all primes less than or equal to 11. $\endgroup$ – dineshdileep Dec 15 '12 at 4:36
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3 and 4 are in the set $\{1, 2, \dotsc , 11\}$, which you took the LCM of, so 12 (3 $\times$ 4) is also a divisor of 27720.

We can take advantage of this knowledge and instead of constructing the arithmetic sequence $\frac{1}{27720}, \frac{2}{27720}, \dotsc , \frac{11}{27720}$, we can construct the arithmetic sequence $\frac{2}{27720}, \frac{3}{27720}, \dotsc , \frac{12}{27720}$.

$\frac{12}{27720} = \frac{1}{2310}$, and 2310 $<$ 2520. So 2310 is the correct answer.

Trying this with 13 instead of 12 would require taking the LCM of an entirely different set, and trying it with things like 14 (2 $\times$ 7), which are also divisors of 27720, would still require taking the LCM with 13 involved, because you can't form an arithmetic sequence like this by going from $\frac{12}{27720}$ to $\frac{14}{27720}$ (arithmetic sequences require the differences between successive terms to be constant).

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  • $\begingroup$ Very clear, precise. Thank you. $\endgroup$ – JSCB Dec 15 '12 at 6:53

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