Prime Factor Form.....like you've never seen it.

Who doesn't enjoy a good maths puzzle?  The normal barrier to such enjoyment is finding a problem of appropriate difficulty for yourself or those you teach.  Too easy and it becomes pointless and too hard and you fail to make any progress which isn't fun for anyone.

And then I stumbled across this problem...the advice is to have a go first and I thoroughly endorse that advice.  My solution, which goes around the houses, is below.  At the end I realized there was a far quicker way but that's why you should do it yourself first - you'll end up with a far better understanding of what is going on.

This is a great activity if structured well for your class.

Don't cheat - have a go first before looking at my solution.

Original problem  (Click the title!)

Given that there is a pattern in the sequence:

22 x 34, 23 x 34, 22 x 35, 24 x 34, 22 x 34 x 5,  23 x 35, …

Write the next six terms in the sequence, all in prime-factored form. Having accomplished that, write the 200th term in prime-factored form, and describe a method that will provide the prime factorization of the nth term of the sequence.

My Solution

So after realizing I couldn't solve it by simply looking at it and doing some mental arithmetic I decided to make a start and "do something"...

I changed each prime factor form back into its original number.

2² x 3⁴ = 324

2³ x 3⁴ = 648

2² x 3⁵ = 972

2⁴ x 3⁴ = 1296

2² x 3⁴ x 5 = 1620

2³ x 5³ = 1944 

I felt like I was making progress and could sense there must be a pattern in the powers of 2, 3 and 5 but couldn't really see it.....

The rule of 324

Further investigation showed that the sequence was increasing by 324 each time.  

This made it easy to find the next 6 numbers in the sequence.  2268, 2592, 2916, 3240, 3565 and 3888.

But how to do this in prime factor form? How could I figure out the prime factor form for the 12th, 20th or 100th term? 

You may have already spotted the quicker way...but this is the route I took...

When in doubt.....create a table :)

And then I came to a bit of a standstill.  I knew I could keep adding 324 and also knew there must be a pattern in the powers in the table...but it wasn't that clear to me what it was....

So, I went back to the original question and re-pondered this bit....

Having accomplished that (finding the next 6 numbers in the sequence), write the 200th term in prime-factored form, 

The 200th term?....Currently all I could do was keep adding on 324 loads of times which would work but wasn't mathematically pleasing at all.  But this is a linear sequence so the nth term is straight forward as the pattern is 324, 648, 972, 1296, 1620 etc giving an nth term of 324n.

So, the 200th term becomes easy;    324 x 200 = 64 800.

But I was still left with a nagging sensation that I am missing something obvious and feel I'm nowhere near answering this...

and describe a method that will provide the prime factorization of the nth term of the sequence.

Well, I know that the nth term is 324n.....and then it dawned on me (apologies it was obvious to you all along!).  I wonder what 324 is in prime factor form.

Well, it is 2² x 3⁴

Referring back to my table it all became so clear!  

The nth term of this sequence can be worked out by taking the prime factor form of 324n.

Or, in other words  2² x 3⁴ x n

So, I knew that the 11th term was 3565 so thought I'd check the formula worked...

2² x 3⁴ x n  becomes  2² x 3⁴ x 11 as it is the 11th term.  As 11 is prime I now see why this is the only number in the first 12 terms of this sequence to have an 11 when written in prime factor form.

If you want to check that it works for the 6th term you will see that 2² x 3⁴ x 6 works as 6 is obviously 2 x 3 in prime factor form leading to 2³ x 3⁵

Have fun!