Infinite sequence proof

Can anyone help me answer this please-
Prove that no number in the infinite sequence 10,110,210,310,410... can be written in the form a^n where a is an integer and n is an integer > or equal to 2

Reply 1

Theloniouss

Universities Forum Helper

21

What have you tried? Where are you stuck?

Reply 2

bleepsheep

OP

6

I don’t even know where to start tbh

Reply 3

Theloniouss

Universities Forum Helper

21

Original post by bleepsheep

I don’t even know where to start tbh

Can you find the nth term of the sequence?

OP

It’s 100n-90

Universities Forum Helper

21

Original post by bleepsheep

It’s 100n-90

Can you write that in the form a^n?

Reply 6

jebjeb27

6

say any number in the sequence can be expressed via the general expression 100m-90
this is equal to 10(10m-9)
which is equal to 5*2(10m-9)
whatever m is equal to, the term will have differing prime factors (5 and 2) within it.
Therefore, no term can be expressed via a^n as a cannot be 5 and 2 simultaneously.

Reply 7

bleepsheep

OP

6

Original post by Theloniouss

Can you write that in the form a^n?

I can’t find how to. I’m a bit lost now sorry

Reply 8

davros

16

Original post by jebjeb27

say any number in the sequence can be expressed via the general expression 100m-90
this is equal to 10(10m-9)
which is equal to 5*2(10m-9)
whatever m is equal to, the term will have differing prime factors (5 and 2) within it.
Therefore, no term can be expressed via a^n as a cannot be 5 and 2 simultaneously.

You're told that a is an integer, not that it's prime, so it could be 5 x 2 =10. However, it's not difficult to patch this up correctly.