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Numbers game

OK, while I am waiting, here is another maths quiz:

Which two numbers go in tenth place in this sequence?

1) 1
2) 2
3) 9
4) 8
5) 625
6) 12 and 18
7) 117649
8) An infinite set of numbers starting 24, 40, 56...
9) An infinite set of numbers starting 36, 225, 441...
10) ?? and ??

Answers will be revealed in good time.

Comments

( 6 comments — Leave a comment )
pgmcc
May. 23rd, 2007 06:16 am (UTC)
If I recall correctly your last number quiz coincided with an airport wait. I to that one wildly wrong.

I hope your terminal waiting is not too tedious. Not long now. You should be boarding soon.
secritcrush
May. 23rd, 2007 06:35 am (UTC)
2 & 16
smellingbottle
May. 23rd, 2007 07:41 am (UTC)
I am cowering.
bohemiancoast
May. 23rd, 2007 08:10 am (UTC)
80, 1250

I loved, loved, loved it. I got it by doing the obvious google, and then turned up the most marvellous website ever. That didn't quite give the answer, but it does have a sequence which is 'the smallest' for each n, whereas you have given all the results for each n. However, it is not hard to work out the other answer.

For primes there is only ever one answer, which is p^(p-1). For any number where it has any square factor there will be infinite answers I think.
pwilkinson
May. 23rd, 2007 05:11 pm (UTC)
the most marvellous website ever

The Online Encyclopedia of Integer Sequences, by any chance? A marvellous site indeed.
nickbarnes
May. 23rd, 2007 09:02 am (UTC)
Hmm. 117649 is the meatiest number there, and is 7^6. And 625 is 5^4. So it looks like f(n) = n^(n-1), and in fact it is that for n = 1, 2, 3, 5, 7. OK, so that's it for prime n, and for n=1 for some reason.
Let's look at the other entries. f(4)=8? 8 = 2^3. So it's k^(n-1), for some value of k. Except for n=6, 8, 9.
I notice that f(n) is a set of numbers, all of which are divisible by n. Let's suppose that's important. Let's look at the cases where f(n) is not the empty set. For n=6, we have 12 and 18. What do these numbers have in common apart from divisibility by 6?
=== Spoiler ===


Factors?
12: {1, 2, 3, 4, 6, 12}
18: {1, 2, 3, 6, 9, 18}
Hmm.
6p: {1, 2, 3, 6, p, 2p, 3p, 6p}
Hmmmmmmmm.

625: {1, 5, 25, 125, 625}
p^(p-1): 1, p, p^2, ... p^(p-1)
24: {1, 2, 3, 4, 6, 8, 12, 24}
8p: {1, 2, 4, 8, p, 2p, 4p, 8p}
Observe that prod_i{p_i^k_i} has prod_i{k_i+1} factors. So to find numbers with N factors, consider the factorization of N itself. Numbers with, say, 9 factors must either be p^8 or p^2.q^2. So the only such numbers divisible by 9 itself are 3^8 and 9p^2.
Similarly, numbers with 8 factors are p^7, p^3.q, and pqr. To be divisible by 8 itself: 2^7 and 8p.
Numbers with pq factors are r^(p-1).s^(q-1). To be divisible by pq itself, this leaves p^(p-1).q^(q-1) and q^(p-1).p^(q-1). For n=6, this is 12 and 18. For n=10, this is 80 and 1250.
( 6 comments — Leave a comment )

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