repeating 1s quadratic
This is probably one of the hardest q25s I've done - at least for me. Following the ms isnt too bad, but figuring out exactly what to do for this q in less than 5 minutes is wild.
I was wondering if anyone had any alternative methods to approaching this q/advice to solving repeating digits qs -> should I always focus on using limits as that seems to be a key theme here or....?
I was wondering if anyone had any alternative methods to approaching this q/advice to solving repeating digits qs -> should I always focus on using limits as that seems to be a key theme here or....?
(edited 1 month ago)
A quick hack, Id try simple, specific values so x = 1 or 11 or 111 then youd have
p + q + r = 11
121p + 11q + r = 1111
12321p + 111q + r = 111111
The right hand sides are a bit of a guess, one digit greater than x^2 (it could be more) but p~9 seems to work and you get the
q + r = 2
11q + r = 22
111q + r = 222
So q = 2, r = 0.
As this is one of the answers then ...
p + q + r = 11
121p + 11q + r = 1111
12321p + 111q + r = 111111
The right hand sides are a bit of a guess, one digit greater than x^2 (it could be more) but p~9 seems to work and you get the
q + r = 2
11q + r = 22
111q + r = 222
So q = 2, r = 0.
As this is one of the answers then ...
(edited 1 month ago)
Oh yeah thats a nice hack, thanks for that
I think I thought of smth else, since you know (px^2+qx+r) and 10(px^2+qx+r) +1 = 111...., is that a possible recipe for some recursive solution?
The ms mentioned this at the bottom and I was just thinking about it but cant really figure out a way to get p,q,r
edit: also just wondering with your method, I think you assume p,q,r are positive with your guesses, but if you were doing this in a real exam, and you dont know if they are +ve or -ve (other than p) would you just try p+q+r = 1,11,111... until something sticks? I think that might be faster rather than doing it the ms way..
I think I thought of smth else, since you know (px^2+qx+r) and 10(px^2+qx+r) +1 = 111...., is that a possible recipe for some recursive solution?
The ms mentioned this at the bottom and I was just thinking about it but cant really figure out a way to get p,q,r
edit: also just wondering with your method, I think you assume p,q,r are positive with your guesses, but if you were doing this in a real exam, and you dont know if they are +ve or -ve (other than p) would you just try p+q+r = 1,11,111... until something sticks? I think that might be faster rather than doing it the ms way..
(edited 1 month ago)
Original post by mosaurlodon
Oh yeah thats a nice hack, thanks for that
I think I thought of smth else, since you know (px^2+qx+r) and 10(px^2+qx+r) +1 = 111...., is that a possible recipe for some recursive solution?
The ms mentioned this at the bottom and I was just thinking about it but cant really figure out a way to get p,q,r
edit: also just wondering with your method, I think you assume p,q,r are positive with your guesses, but if you were doing this in a real exam, and you dont know if they are +ve or -ve (other than p) would you just try p+q+r = 1,11,111... until something sticks? I think that might be faster rather than doing it the ms way..
I think I thought of smth else, since you know (px^2+qx+r) and 10(px^2+qx+r) +1 = 111...., is that a possible recipe for some recursive solution?
The ms mentioned this at the bottom and I was just thinking about it but cant really figure out a way to get p,q,r
edit: also just wondering with your method, I think you assume p,q,r are positive with your guesses, but if you were doing this in a real exam, and you dont know if they are +ve or -ve (other than p) would you just try p+q+r = 1,11,111... until something sticks? I think that might be faster rather than doing it the ms way..
There are a few ways to get inspiration. One is to note that x^2 is sort of
123454321
and the left digits * p must give the left digits of 1111.. in the answer as the other two terms will only affect the right digits (q and r must be small). Then noting that
123456789*9 ~ 111111111
so you can guess p is 9 or 90 or ... which gives the above.
Or you could note that x^2 has an odd number of digits so if p is <= 9 then the right hand side could have an even number of 1s (odd+1) and so will be divisble by 11. The left hand side is
x(px + q) + r = 11...11
and both sides are divisble by 11 when x has an even number of digits so r must be divisble by 11 so 0 or 11 or ...
So there are a few ways of guessing the answer and if you have something to check, its easy to verify what works.
A "recursive" solution of p=9,90,900 is pretty much as above. You simply need to reason about the right few digits to get the q and r to account for the left shift by one or two digits depending on the number of zeros in p.
Note the above of subbing values for simple cases is a standard problem solving technique. The multiple choice answer doesnt require a proof (like bmo would and what the model solution is) but a bit of simple reasoning with simple cases pretty much gets you there. Its easy to waste time on the actual smc if you try to do full/proof solutions to all questions. Similarly, Ive not really assumed q,r are positive and the previous pretty much works with them (possibly) being negative.
(edited 1 month ago)
Another approach:
px^2 + qx + r is all 1s.
Dividing by x, px+q+r/x is approximately a power of 10.
So the dominant term px is approximately a power of 10. To get 11...1 to a power of 10, you multiply by 9 (and add 1). So p = 9 (or potentially 9 x 10^k for some k).
As mqb says, you're not expected to prove all the possible options, just find one that works. I would guess less than 5% of people who solve the question would do so by a process similar to the given answer (with the proof, etc). Even if you asked for a full enumeration of solutions with proof, I think nearly everyone would find a solution by trial and error, extend to all the solutions, and then finally prove "these are all the solutions".
px^2 + qx + r is all 1s.
Dividing by x, px+q+r/x is approximately a power of 10.
So the dominant term px is approximately a power of 10. To get 11...1 to a power of 10, you multiply by 9 (and add 1). So p = 9 (or potentially 9 x 10^k for some k).
As mqb says, you're not expected to prove all the possible options, just find one that works. I would guess less than 5% of people who solve the question would do so by a process similar to the given answer (with the proof, etc). Even if you asked for a full enumeration of solutions with proof, I think nearly everyone would find a solution by trial and error, extend to all the solutions, and then finally prove "these are all the solutions".
Original post by DFranklin
Even if you asked for a full enumeration of solutions with proof, I think nearly everyone would find a solution by trial and error, extend to all the solutions, and then finally prove "these are all the solutions".
Fully agree with this. For a reasonable number of bmo(1) questions, the usual way to approach them is to work through simple cases/simplfied problem to spot patterns about how the solution works, then think about what algebra/arguments are required. So problem solving it, which is all you usually need for the smc, is (often the first step but) quite different from the full algebraic proof (bmo/smc model solution).
For such questions, dont get "analysis paralysis", just get stuck into it by simplifying the problem, subbing numbers, guessing solutions, ... Even when the first approach doesnt work, you often learn something.
(edited 1 month ago)
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can someone please explain what principle domain is and why the answer is a not c?Maths
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