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2004-02-09, 04:32
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#1 |
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6809 > 6502
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Aug 2003
101×103 Posts
2·3·7·233 Posts |
1,234,567,890
Here is one that should be slightly entertaining. I made it up out of my head, so you know who to blame. Post the last 2 numbers in this sequence (spoilerised and all).
 13,485,943,174,138 116 17 |
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2004-02-12, 14:25
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#2 |
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6809 > 6502
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Aug 2003
101×103 Posts
2×3×7×233 Posts |
No takers
 ?????????????????????
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2004-02-12, 18:05
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#3 |
Dec 2003
23 Posts |
Maybe I'm missing something, but as far as I can tell, you're not actually asking a question. Are you? You seem to be showing a sequence that is composed of three numbers:
13,485,943,174,138 116 17 the first number has 14 digits, the second one has three digits the last one has two digits. The you are asking what the last two numbers of that sequence are. Well, they are obviously 116 and 17. I, at least, fail to see the puzzle here. Maybe you could clarify what you're really looking for? |
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2004-02-12, 18:08
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#4 |
"Mike"
Aug 2002
25×257 Posts |
Maybe he meant the next two numbers in the sequence?
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2004-02-12, 18:26
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#5 |
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6809 > 6502
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Aug 2003
101×103 Posts
100110001110102 Posts |
Yes, the numbers posted are in a sequence. Based upon the numbers posted, what are the last 2 in the sequence?
The sequence will end. The last 2 numbers will show that you know the relationship between them. |
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2004-02-13, 08:16
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#6 |
Sep 2002
22×3×5 Posts |
9, 4
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2004-02-13, 14:36
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#7 |
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6809 > 6502
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Aug 2003
101×103 Posts
2·3·7·233 Posts |
Can you prove that that number is the end to all such sequences
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