Units digit of 3, 33 , (33 )3 , ((33 )3 )3 , etc


By Hal 2001 (P3046) on Friday, April 27, 2001 - 10:24 am :

Hello NRich folks,

In the following question I don't have a clue of what it is asking for. Especially when it says 'unit place' (well I have an inkling but not certain).

Let a1 = 3, an+1 = (an )3 for n> =1
(Thus a2 = 33 , a3 (33 )3 and so on.)

What digit appears in the unit place of a7 ?

Hal.

PS: I forgot to mention, we can't use the calculator for this question.


By Kerwin Hui (Kwkh2) on Friday, April 27, 2001 - 02:03 pm :
Hal,

Unit place means the last digit of an integer.

Hint for this question: 34 º 1 mod 10.

Kerwin


By Hal 2001 (P3046) on Saturday, April 28, 2001 - 06:23 am :

Hi Kerwin.

Thanks for that.

I am sorry to say that I don't know what 1 mod 10 means? Can you be kind enough as to briefly explain what it means.

Thanks for your help.
Hal.


By Tom Hardcastle (P2477) on Saturday, April 28, 2001 - 12:29 pm :

A less sophisticated method of solving this problem is to consider what determines the value of the digit in the unit place for each member of the sequence. Try working out the unit place for n = 1, n = 2, n = 3 &c. if you still can't see the way it works.

Tom.


By Olof Sisask (P3033) on Saturday, April 28, 2001 - 02:36 pm :

1 mod 10 means that it's in the form 10n + 1 (where n is an integer).

/Olof


By Ali Korotana (P4195) on Saturday, April 28, 2001 - 03:38 pm :

All of the numbers in the series are some of the terms in the series 3n , where n increases. The unit place in every number 3n is 3,9,7 or 1.
If n/4 has no remainder, then 3n =1 mod 10
if n/4 has a remainder 1, then 3n =3 mod 10
if n/4 has a remainder 2, then 3n =9 mod 10
if n/4 has a remainder 3, then 3n =7 mod 10

If you can work out a value for n in the term a7 , then it will be quite easy to work out the unit place. This probably seems like quite a crude way way of working it out. The series described in the original thread is a special case of the above and I think you could work out a rule for that series. I don't really see why this is the case though, do all power series have a repeating set of unit places?


By Tom Hardcastle (P2477) on Saturday, April 28, 2001 - 04:52 pm :

Yes, all power series have a repeating set of unit places (although they may have a set of non-repeating unit places previous to this - in the same way that fractions give a repeating set following a non-repeating set).

The proof is simple:
Let an+1 = (an )k
Then un+1 is equal to the unit place of (un )k where un is the unit place of an .
Hence, if ua = ub then ua+1 = ub+1 .
Since un can only take 10 different values, for some a there exists b such that ua = ub .
Hence there is a repeating set of unit places.

Tom.


By Tom Hardcastle (P2477) on Saturday, April 28, 2001 - 05:10 pm :

The proof is simple, but that's not it. That's the proof for repetition in the series defined in the original question. The proof for power series should read
Let an+1 = an ×k.
Then un+1 is equal to the unit place of un ×k where un is the unit place of an .
Hence if ua = ub then ua+1 = ub+1 .
Since un can only take 10 different values, for some a there exists b such that ua = ub .
Hence there is a repeating set of unit places.

Both these proofs depend on the obvious property that the unit place of an+1 in the series is dependent only on the unit place of an .

Tom