tutorinseoul Posted March 15, 2017 Report Share Posted March 15, 2017 Hi! Here is a possible Section A question for 2017. Please do attempt and share your thoughts! Thank you. Reply Link to post Share on other sites More sharing options...
TheNintendoChip Posted March 16, 2017 Report Share Posted March 16, 2017 (edited) Perhaps there's an easier way to do this, but this is what I did: We can represent any number in base 10 by multiplying powers of 10 by numbers from 0-9 and adding them together. For example, 247=2(10)^2+4(10)^1+7(10)^0. Then we can write the number that is 2n 1s as the sum from k=0 to 2n-1 of 10^k. Notice that we have the upper bound as 2n-1 and not 2n because the amount of summands will actually be Upper bound - Lower bound +1=2n-1-0+1=2n, and the amount of summands must be equal to the amount of digits. We can write the number that is n 2s as 2 times the sum from k=0 to n-1 of 10^k. Thus we have reduced the problem to showing that [sum from k=0 to 2n-1 of 10^k]-2[sum from k=0 to n-1 of 10^k]. Using the formulas for finite geometric series (sum from k=0 to n-1 of r^k is [r^k-1]/[r-1]), we can reduce this to [10^(2n)-1]/[10-1]-2[10^n-1]/[10-1]. Factor out the 1/[10-1] and distribute the -2 to get: 1/9[10^2n-2*10^n+1]. You should recognize that [10^2n-2*10^n+1] is a perfect square, namely [10^n-1]^2. Thus we have 1/9[10^n-1]^2, which is [[10^n-1]/3]^2, which is a perfect square. I believe we might have to show that [10^n-1]/3 is an integer though. To do so, recall that a^n-b^n=[a-b][sum from k=0 to n-1 of a^k*b^(n-1-k)]. If we let a=10 and b=1 then we have that 10^n-1^n=[10-1][sum from k=0 to n-1 of 10^k]. This is evidently divisible by 3 and thus is an integer as it is the product of two integers. The 2nd term is in fact n 1s. So we have actually also shown that it is exactly equal to n 3's squared (this can also be seen by the two examples given). I did SL though and I've never encountered this before. Let me know if any of that was confusing. It's kind of hard to write out sums by hand Edited March 16, 2017 by TheNintendoChip Reply Link to post Share on other sites More sharing options...
Vioh Posted March 16, 2017 Report Share Posted March 16, 2017 (edited) @TheNintendoChip Your solution is correct, but it seems unnecessarily complicated. I found another solution that doesn't require us to use geometric series (solution attached as picture below). @tutorinseoul Correct me if I'm wrong, but are you sure that this is the type of problem that is likely to be in an IB exam? To me, this seems more like an Olympiad riddle rather than a problem for math SL. Edited March 16, 2017 by Vioh Small fix to the equations 1 Reply Link to post Share on other sites More sharing options...
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