| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Prove that for any x, x being a positive integer
there exist a multiple of x such that when this multiple
is expressed in decimal notation, will starts with the
digits 1234567890.
Eg. kx = 1234567890......... , k being a positive integer
Eddie Leung.
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 552.1 | I switched k and x in this... | MODEL::YARBROUGH | Fri Aug 01 1986 13:13 | 7 | |
The initial sequence is irrelevant. Consider any initial sequence
S. Now think about the K different numbers
s, ss, sss, ... [k]s
Either one of them leaves a remainder = 0 when divided by K [qed]
or two of them leave the same nonzero remainder (by the pigeonhole
principle). Subtract the smaller of these two from the larger and you
get the number s...s0...0, which leaves remainder = 0 [qed].
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| 552.2 | squeezing all the integers into a small space | MODEL::YARBROUGH | Fri Aug 01 1986 20:06 | 3 | |
A corollary of .-1 is that the set of divisors of the set of integers
consisting simply of a sequence of 1's followed by a sequence of 0's
contains ALL the integers [>0].
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| 552.3 | easy... | JON::MORONEY | Madman | Fri Aug 01 1986 21:31 | 8 |
Let N be the number of decimal digits in x>0. Therefore, 10^N-1 will be >= x. Then let A1=1234567890*10^N+10^N-1 and A2=1234567890*10^N. Both A1 and A2 will be of the form 1234567890nn...nn. A1-A2 is equal to 10^N-1, and is >= x. Let z=A2 mod x. Z will follow 0 < z < x. Then if y=x-z, A1+y is a multiple of x. But y <= x <= A1-A2, so A1+y will be <= A2 so it will start with the sequence 1234567890. -Mike | |||||