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Re: Brainteaser n
Posted:
Jan 13, 2003 2:22 PM
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On Mon 13 Jan 2003 17:07:52 +1100 Bill Hart <wb_hart@maths.mq.edu.au> wrote
>Better make that something like
>
>
>1*11*111*....*(10^121-1)/9
>
>Bill.
You propose the product of the first 121 repunits.
REP(1)=1, REP(2)=11, REP(3)=111, etc.
You picked up the clue with all those obvious ..909090.. and ...900900...
prime factors, I assume, and the high powers of 3, 11 and 37 among the
factors.
But you are wrong. Any of the following show you why (and (c) suggests
how):
(a) Note that 1.11111... ^ 21 < 10 and 1.11111... ^ 22 > 10.
The number of decimal digits in the product you offered is therefore about:
121 * 122 / 2 - 21 / 22 * 120 = ~7266 digits
But I had written:
>> X, an integer with just over 6100 decimal digits
:-)
By trivial algebra,
REP(2n)=REP(n)*(10^0+10^n)
REP(3n)=REP(n)*(10^0+10^n+10^2n) etc.
Bear this in mind for the next few sections.
(b) Now in my decomposition of X into prime factors, REP(19)^5 and
REP(23)^4 were shown as factors.
If you think about this a little, you will realize that had the factors of
X included REP(114=19*6) and REP(115=23*5), (at least) an extra REP(19) and
REP(23) would have been expected among the prime factors - right?
Once you think you have understood this, read (c), and then re-read (b) and
see why "(at least)" was used above.
(c) REP(2)^60 is a factor (and no other REP(2)s).
Does that suggest that REP(120) was present? ;-)
Nope. REP(2)^2 is a factor of REP(2nREP(2)) (I leave it for you to see
why), so there are 5 "extra" REP(2)s in the primal decomposition.
The other REP(2)^55 account for terms up to REP(110) or REP(111).
If we were to include multiplicands up to REP(121), there would have been 4
more REP(2)s, and REP(2)^64 would have been a factor.
(d) As can be seen from trivial algebra, REP(6) is divisible by 7.
So REP(6n) is divisible by 7, and REP(7*6n) is divisible by 7^2. Right?
Now 7^20 appears in the primal decomposition, 18 from REP(6n) and 2 extra
from the two REP(7*6n).
There would have been 2 further "7"s (from REP(114), REP(120) had we gone
up to REP(121).
(e) 11^2 = 121 is not a repunit. 111 is a repunit.
But I had written:
>> The expression is elegant in its economy.
Probably, you missed the 37 hint too ("Provide a short (37"). :-)
Learn from the pastry!
--
Bourbaki
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