Decimal Expansion Buddies
2 7 1 1 = 0 . 0 0 3 6 9 0 0 3 6 9 . . .
3 6 9 1 = 0 . 0 0 2 7 1 0 0 2 7 1 . . .
If the decimal expansion of a 1 can be written as a string of 5 digits repeated over and over ad infinitum, where the rightmost digits of the string form the number b and any leading digits are 0 's, (as per the example), then a and b are considered decimal expansion buddies .
How many distinct pairs of decimal expansion buddies are there?
Note: ( a , b ) and ( b , a ) are considered to be the same pair of numbers, and therefore are only counted once.
Bonus: Can you generalize this for any period length?
The answer is 6.
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3 solutions
I think I've misunderstood "solely leading zeros", I thought there should be at least one leading zero and eliminated possibilities like: 9 1 = 0 . 1 1 1 1 1
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Good point. There would be 3 pairs for which this problem would arise, in which case there are only 3 buddy pairs left, namely ( 4 1 , 2 4 3 9 ) , ( 1 2 3 , 8 1 3 ) and ( 2 7 1 , 3 6 9 ) .
@Garrett Clarke I think that Kazem has a valid concern. Do you think that the phrasing of the question or the posted answer should be changed?
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I'm not very good at English, If solely means that there is at least one zero, then yeah, there are 3 such pairs.
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@Kazem Sepehrinia – O.k., great, (and your English is excellent as far as I'm concerned :)). I think that the "solely" could probably be dropped from the phrasing of the question, since "period of length 5 " is more definitive. My concern is more with the fact that, for example, 0 . 1 1 1 1 1 has period 1 and not period 5 .
With n = 6 we would have the pair ( 7 , 1 4 2 8 5 7 ) . The decimal expansion of both a 1 and b 1 would have a "valid" period of 6 , but one of them would have no leading zeroes. So I think that it would be better to leave out any mention of "solely leading zeroes" to avoid confusion. I've notified Garrett, so we'll see how he wants to proceed.
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@Brian Charlesworth – Right :) let's wait for Garrett! There is an interesting point here, 9 9 9 9 9 1 = 0 . 0 0 0 0 1 1 1 = 0 . 9 :))
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@Kazem Sepehrinia – Ahhhh I'm just seeing this now! Everyone is making very good points. I had great difficulty in trying to word this question while I was making it, honestly I'm looking for suggestions. @Kazem Sepehrinia , @Brian Charlesworth , @Bufang Liang : any thoughts on specific changes that can be made?
I want the problem to stay geared toward the intended solution, meaning that the answer is 6 because 99999 has 12 divisors. How should I go about changing it?
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@Garrett Clarke – I think that to preserve the elegance of the problem and the intended solution you might need to drop any mention of period 5 and just say something like:
"If the decimal expansion of a 1 is a string of 5 digits repeated over and over ad infinitum, where the rightmost digits of the string form the number b and any leading digits are 0 's, (as per the example), then a and b are decimal expansion buddies."
A bit awkward, but I think it stays true to your concept and keeps the present solution intact.
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@Brian Charlesworth – Perfect, thank you all for your help!
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@Garrett Clarke – Great. Hopefully Kazem and Bufang approve of the rephrasing as well. :)
@Brian Charlesworth – This was what I read when I tried this evening. 8 October 2015. I ought to sacrifice a trial as 0's was thought to be 00 only. Ambiguity was there.
@Garrett Clarke – How about "a repeating decimal with period of length 5 which is made up of the decimal expansion of b (could have leading 0's)"
@Kazem Sepehrinia – As long as one of them such as 1/ 99999 satisfies, the one follows shall not be doubted. In fact, the second equation is true. These are considered as one count.
I think it would be more appropriate to change the wording, as accounting for excess cases makes it harder to generalize and makes the problem not as elegant.
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@Bufang Liang – Yes, I agree. It's Garrett's question so it's his choice as to how to proceed, but I think that the wording will need to be revised.
same problem here....i think (41,2439) should also b ruled out....because zeros are different in both decimal expansion....which leaves just 2 pairs....
To confirm, yes, there does not exist an analytic approach to solving this for any period length, since it requires prime factorizing repunits. However, once the prime factorization is found, the rest of the problem is trivial.
Oh. inthought 1 is not included
The problem would have been cleaner if the trivial set {1,99999} was excepted. But then again, some people might get a clue from that number 99999 on how to solve this. I would hate to be the one to have to figure out how to best word this problem.
Yes, this was a tricky one to word. I had to sift through several possible rephrasing suggestions to Garrett before coming up with one that was unambiguous but did not give too much away. It's a bit awkward but at least it's serviceable. And I kind of like the 1 , 9 9 9 9 9 buddies because it brings up the curious fact that 1 1 = 0 . 9 . :)
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Exactly, I thought excluding the trivial case would help people get the idea to find the divisors of 99999 far to early in the problem. I wanted people to realize that on their own and find (1,99999) as an interesting example, not the other way around.
Let 1/a =10^{-n}b+10^{-2}b+...................... these terms are in GP Hence, 1/a=10^{-n}b/1-10^{-n} * 1/ab=10^{n}-1 *
ab=10^{n}-1
for n=5, ab=99999=3^{2} 41^{1} 271^{1} ==> no. of ways on expressing ab is (2+1)\times (1+1)\times (1+1) = \boxed {6}
We require that a 1 = b ∗ ( 1 0 5 1 + 1 0 1 0 1 + . . . . ) = 1 0 5 − 1 b
⟹ a b = 1 0 5 − 1 = 9 9 9 9 9 = 3 2 × 4 1 × 2 7 1 ,
which has ( 2 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 1 2 positive divisors, and thus 6 divisor pairs, i.e., 6 pairs of "buddies".
As for a generalization for period n buddies, we just need to find the number of divisor pairs whose product is 1 0 n − 1 . For example, for period 6 there are 3 2 buddies, since 9 9 9 9 9 9 = 3 3 × 7 × 1 1 × 1 3 × 3 7 .
A follow-up question could be: for which n are there the most buddies? For n = 1 8 there are 3 2 0 buddies, which is the most up to n = 2 0 . (This might be best determined by a computer program; there doesn't seem to be an analytic approach for this one.)