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FP3 (MEI) help!

Hey, how would I prove that, for an integer x written as a_na_n-1...a₂a₁a₀, x=a_n+a_n-1+...+a₂+a₁+a₀ (mod 9) ?

There are no worked answers in the back of the FP3 textbook so it's difficult to help myself! T_T

Thanks,
Ed

Reply 1

Ed4434

Another shameless self-bump.

Reply 2

ghostwalker

One way would be to start by splitting your given number thus:

a_na_{n-1}...a_2a_1a_0=a_n10^n+a_{n-1}10^{n-1}+...+a_210^2+a_110^1+a_0

And consider (work out) what

10^n

etc. is equal to in mod 9

PS: Not always a good idea to "bump" unless your thread is dropping off the first page of the forum; at first glance people might think someone had already replied to it.