Stumped by IQ test: help from uberbrain needed

[spire]

While clearing out some rubbish, I came across a 10-question IQ test. I am a sucker for these and got distracted. To my dismay, I couldn't answer one of the questions.

Can anyone answer and explain?

What is the next number in this series?

1, 2, 3, 5, 7, 11, 13, 15,...

 

[Stig-OT-Dump]

Primes. Next is 17, 19, 23....

 

[Gixxerman]

Not sure if this is correct.
But if you take the numbers 1,3,7,13 the active sequence and ignore the 2,5,11,15 numbers.
The distance increases by 2 every iteration for the active sequence.
I.E. 1->3 = +2, 3->7 = +4, 7->13 = +6, so next would be +8 = 13+8 = 21.
I have seen these sequences before where only alternate numbers make up the sequence, and the other numbers are decoys.
That is the best I can do.

 

[Gixxerman]

Primes. Next is 17, 19, 23....

15 is not prime. It is divisable by 3 and 5.

 

[gaz]

I've had a quick go, but it's late, so i failed. It's been a few years since i did my maths a-level.

I can't seem to get it to fit under any of the 3 types of number sequence; arithmetic, gemetric or fibonacci.
I'll have to re-look at my calculations tomorrow.

 

[marinyork]

They are the so called cyclic numbers.

1 is coprime to itself and the coprime of numbers less than or equal to itself. 2 likewise. 3 is obviously coprime to 2 . 4 has 1 and 3 coprime to itself but 4 and 2 are not coprime so it is junked. 5 is coprime to 1,2,3 and 4 and 4 itself is coprime to 5 so that's all right. If you bother to look at the totient function you'll get all the other numbers including

15 and φ(15)= 8 are coprime.
16 and φ(16)=8 are not. 17 and 6 are. 18 and 6 aren't etc. Quite a few of them stick out like a sore thumb where the totient function is 1 less than n it is obviously coprime or where n is even and the totient function is also even it can be ruled out.

A much better way of doing it is in group theory and cyclic groups. The klein group is isomorphic to the dihedral group of order 4 which is not cyclic so this is not isomophic to the cyclic group of order 4*. So 4 is binned. If you read your group theory which people may or may not be interested in there are dihedral groups of order 2n for 4 onwards so all even numbers bigger than this are binned. If p is a prime number by another theory in abstract algebra then it is a cyclic group. To prove n is not cyclic it is merely enough to find a counter example of a group of order n that is not cyclic. This is time consuming for larger numbers but there are a dozen theorems to help out.

 

[slowmotion]

Can you fix a dripping tap as well...?

 

[Aperitif]

They are the so called cyclic numbers.

1 is coprime to itself and the coprime of numbers less than or equal to itself. 2 likewise. 3 is obviously coprime to 2 . 4 has 1 and 3 coprime to itself but 4 and 2 are not coprime so it is junked. 5 is coprime to 1,2,3 and 4 and 4 itself is coprime to 5 so that's all right. If you bother to look at the totient function you'll get all the other numbers including

15 and φ(15)= 8 are coprime.
16 and φ(16)=8 are not. 17 and 6 are. 18 and 6 aren't etc. Quite a few of them stick out like a sore thumb where the totient function is 1 less than n it is obviously coprime or where n is even and the totient function is also even it can be ruled out.

A much better way of doing it is in group theory and cyclic groups. The klein group is isomorphic to the dihedral group of order 4 which is not cyclic so this is not isomophic to the cyclic group of order 4*. So 4 is binned. If you read your group theory which people may or may not be interested in there are dihedral groups of order 2n for 4 onwards so all even numbers bigger than this are binned. If p is a prime number by another theory in abstract algebra then it is a cyclic group. To prove n is not cyclic it is merely enough to find a counter example of a group of order n that is not cyclic. This is time consuming for larger numbers but there are a dozen theorems to help out.

I agree entirely. I would go into more Klein theory but I haven't got the bottle.

 

[ohnovino]

The answer is 666. They are the possible values of x that solve the equation:

(x-1)(x-2)(x-3)(x-5)(x-7)(x-11)(x-13)(x-15)(x-666) = 0

I tried this technique in a maths exam once, and the teacher had to concede it was right

 

[vernon]

I agree entirely. I would go into more Klein theory but I haven't got the bottle.

Klein had a twisted mind.

 

[yello]

Sometimes such questions are easy if you know what you were looking for.

I did it myself; scanned the sequence for the usual suspects. I drew a blank but if I had had some other candidates up my sleeve I may well have been able to identify the sequence on sight.

Could I have worked it out anyway? Now there's a question! I suspect not but the bottom line is I couldn't be arsed anyway!

 

[steve52]

all this coprime stuff is new to me ,i saw prime numbers , i must really beat my teachers if i meet them in the afterlife

 

[rich p]

Marin, my old mucker, I am suitably impressed.

Totally and utterly baffled too but that isn't hard, so you get no extra credit for that. I also dozed off halfway through and I thank you for that.

 

[Norm]

Klein had a twisted mind.

But he had some great nachtmusik.

 

[rich p]

But he had some great nachtmusik.

I thought it was a bit small