Molly Mae Posted September 30, 2011 Report Share Posted September 30, 2011 (edited) So pi never repeats, eh? Assume that an arbitrary number extends into infinity without ever repeating in whole. Tell me: what is the number of digits needed to guarantee that any string of 5 numbers is repeated? Edited September 30, 2011 by Molly Mae Quote Link to comment Share on other sites More sharing options...

0 Guest Posted September 30, 2011 Report Share Posted September 30, 2011 100,000 combinations, so by 100,001st string must be a repeat somewhere...string starts at 100,001 ends at 100,005 Quote Link to comment Share on other sites More sharing options...

0 Guest Posted September 30, 2011 Report Share Posted September 30, 2011 Think I'm missing overlaps, but someone else can expand if I did. Quote Link to comment Share on other sites More sharing options...

0 Molly Mae Posted September 30, 2011 Author Report Share Posted September 30, 2011 100,000 combinations, so by 100,001st string must be a repeat somewhere...string starts at 100,001 ends at 100,005 Good enough for me. Now: What if no number can ever appear twice in a row? Quote Link to comment Share on other sites More sharing options...

0 Guest Posted October 1, 2011 Report Share Posted October 1, 2011 I see two interpretations for Molly Mae's additional restriction "Now: What if no number can ever appear twice in a row?" Assuming that you mean that the irrational number never has two of the same digit in a row, then For the first question, Maurice determined that for each of the 5 digits, there existed 10 possibilities, hence repeating must occur when all these possibilities are completed which is 10^5=100,000 and the next sequence must repeat a previous sequence. now if we add the limitations that no number can ever appear twice in a row, this changes the problem to 10 possibilities for the first digit but only 9 possibilities for each subsequent digit so answer is 10 * 9^4=65,610 before a repetition must occur. On the other hand, if the intent is to allow the irrational number to have same digits adjacent but find a repeating sequence that does not allow adjacent numbers we must add the number of combinations that do allow adjacent numbers onto the number derived in original question this would be 100,000 + (100,000-65,610) See above spoiler for explanation for 65,610. Answer is: 134,390 Quote Link to comment Share on other sites More sharing options...

0 Guest Posted October 2, 2011 Report Share Posted October 2, 2011 I'm wondering why all the respondents have been so willing to accept Maurice's answer of 100,000. Maurice himself points out that he hasn't considered overlaps, and that makes a difference. For example, if we started with even the simplest and most obvious way to list the possibilities (00000, 00001, 00002, etc.), even before we finished writing the second combination, we'd get the repeating sequence 00000. I recognise that the OP didn't ask for the minimum number of digits. But surely that is implied, isn't it? Otherwise, it's just answering the trivial question "How many 5-digit numbers are there?" (including leading zeroes, if you want to be pedantic ) I'm sure the intent of the problem was to be more complex (and clever) than this. On the other hand, if we do strictly accept the wording of the problem (despite what I said in my other spoiler), then I will be the first to submit 100,000 as the answer to the variation as well! After all, if we don't care about it being a minimum, then the answer that works for the general case also works by default for the more restricted case as well. Quote Link to comment Share on other sites More sharing options...

0 Molly Mae Posted October 3, 2011 Author Report Share Posted October 3, 2011 (edited) I'm wondering why all the respondents have been so willing to accept Maurice's answer of 100,000. Maurice himself points out that he hasn't considered overlaps, and that makes a difference. For example, if we started with even the simplest and most obvious way to list the possibilities (00000, 00001, 00002, etc.), even before we finished writing the second combination, we'd get the repeating sequence 00000. I recognise that the OP didn't ask for the minimum number of digits. But surely that is implied, isn't it? Otherwise, it's just answering the trivial question "How many 5-digit numbers are there?" (including leading zeroes, if you want to be pedantic ) I'm sure the intent of the problem was to be more complex (and clever) than this. Right, I wasn't asking for the minimum, else it would have strictly been 6. Which is also not clever. My wording was simple and straightforward for two reasons: If pi extends into infinity, can we ever anticipate a string of 5 numbers to repeat? 10 numbers? 12,462 numbers? EDIT: (The second reason is that I was bored on a Friday...which happens often where I am.) On the other hand, if we do strictly accept the wording of the problem (despite what I said in my other spoiler), then I will be the first to submit 100,000 as the answer to the variation as well! After all, if we don't care about it being a minimum, then the answer that works for the general case also works by default for the more restricted case as well. Edited October 3, 2011 by Molly Mae Quote Link to comment Share on other sites More sharing options...

## Question

## Molly Mae

So pi never repeats, eh?

Assume that an arbitrary number extends into infinity without ever repeating in whole.

Tell me: what is the number of digits needed to guarantee that any string of 5 numbers is repeated?

Edited by Molly Mae## Link to comment

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