Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


Math Forum
»
Discussions
»
sci.math.*
»
sci.math
Notice: We are no longer accepting new posts, but the forums will continue to be readable.
Topic:
Brainteaser n
Replies:
15
Last Post:
Aug 6, 2003 7:10 AM




Re: Brainteaser n
Posted:
Jan 13, 2003 2:22 PM


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*....*(10^1211)/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 reread (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



