The average of (1,2, ... n) is (n+1)/2. Easy to calculate that remove 1 it increases by1/2 and remove n it decreases by1/2. So the original average is between 40.25 and 41.25, which means it is either 40.5 or 41. Therefore n=80 or 81. n=80 does not work because of divisibility. n=81 and 61 is the only answer.
@@goyaldev09 thanks! It's a good problem! I think what makes it interesting is there are a few ways to attempt this problem, a lot of which lead to dead ends! Might use it too for some of my mock interviews 🤓😂
Hi J pi, I was wondering if you could explain why you decide to find bounds for particular questions. I see it's quite an important technique in olympiads but I can never spot when to do it or why it may be useful. Thanks!
@@OliverGoodman-todd great question! It wasn't my original thought, and that's the case with lots of such Olympiad problems: you don't get the answer straight away. After having played about with it, I thought if I was able to narrow down the values of n, then I can just test case by case and that got me to think of inequalities.
@@ayushrudra8600 to be honest I did that too. But then I realised that sometimes in BMO, the most obvious technique isn't always the way to solve it. But that being said, with most BMO problems I end up trying quite a few different routes before (if I'm lucky) arriving at a solution
Sir, can you help me on a math problem? 33998.74 = 30000(1 + 0.042/n)^3n I tried using the Lambert W function but it didn't work, and I didn't want to approximate using Newton-Raphson method.
@@deadkiller4129 from where did this problem originate? Generally when you combine polynomial and exponent functions, the way to solve them is via Lambert W
@@JPiMaths it was from an interest problem to calculate the number of times interest is compounded per year. n=3 is a valid solution but I couldn't get it using Lambert W function
@@deadkiller4129 compound interest problems are a lot lower level than the Lambert W function, so I imagine they were just expecting you to plug in values of n and solve that way.
@@JPiMaths I know, but I thought it might be interesting to try actually solving it algebraically, as theoretically there shouldn't be any problems with the maths. I was in for a sleepless night that day🤣
I forgot the meaning of (40)3/4, lol, such bad notation, but they used to use it back in the day, in grade school and such, lol...I'm guessing 40 + 3/4...anyway, lol, a brute force way of doing this is simply to substitute n=2, then n=3, etc, for n, and solve for x, etc...should give a different answere each time...hell, lol, why not just solve for x?...It should be a function of n, etc...the thing I'm not 100% sure about is what that silly (very bad) notation meant, lol, I think that's what it meant...
The average of (1,2, ... n) is (n+1)/2. Easy to calculate that remove 1 it increases by1/2 and remove n it decreases by1/2. So the original average is between 40.25 and 41.25, which means it is either 40.5 or 41. Therefore n=80 or 81. n=80 does not work because of divisibility. n=81 and 61 is the only answer.
@@wesleydeng71 bingo!
Beautiful… will use this in mock interviews😂😂
@@goyaldev09 thanks! It's a good problem! I think what makes it interesting is there are a few ways to attempt this problem, a lot of which lead to dead ends! Might use it too for some of my mock interviews 🤓😂
brilliant use of upper and lower bounds! never thought about using that to figure out n.
@@YunruiHu thank you very much!!
Hi J pi, I was wondering if you could explain why you decide to find bounds for particular questions. I see it's quite an important technique in olympiads but I can never spot when to do it or why it may be useful. Thanks!
@@OliverGoodman-todd great question! It wasn't my original thought, and that's the case with lots of such Olympiad problems: you don't get the answer straight away.
After having played about with it, I thought if I was able to narrow down the values of n, then I can just test case by case and that got me to think of inequalities.
You can avoid testing case n=80 by noticing that 3(n-1)/4 must be an integer ==> 4|(n-1) ==> n=81.
@@derwolf7810 nice yep, that works!
oh thats actually smart - i was just gonna solve the diophantine equation and hope it wasn't crazy annoying as opposed to solving the inequality
@@ayushrudra8600 to be honest I did that too. But then I realised that sometimes in BMO, the most obvious technique isn't always the way to solve it. But that being said, with most BMO problems I end up trying quite a few different routes before (if I'm lucky) arriving at a solution
Sir, can you help me on a math problem?
33998.74 = 30000(1 + 0.042/n)^3n
I tried using the Lambert W function but it didn't work, and I didn't want to approximate using Newton-Raphson method.
@@deadkiller4129 from where did this problem originate? Generally when you combine polynomial and exponent functions, the way to solve them is via Lambert W
@@JPiMaths it was from an interest problem to calculate the number of times interest is compounded per year. n=3 is a valid solution but I couldn't get it using Lambert W function
@@deadkiller4129 compound interest problems are a lot lower level than the Lambert W function, so I imagine they were just expecting you to plug in values of n and solve that way.
@@JPiMaths I know, but I thought it might be interesting to try actually solving it algebraically, as theoretically there shouldn't be any problems with the maths. I was in for a sleepless night that day🤣
I forgot the meaning of (40)3/4, lol, such bad notation, but they used to use it back in the day, in grade school and such, lol...I'm guessing 40 + 3/4...anyway, lol, a brute force way of doing this is simply to substitute n=2, then n=3, etc, for n, and solve for x, etc...should give a different answere each time...hell, lol, why not just solve for x?...It should be a function of n, etc...the thing I'm not 100% sure about is what that silly (very bad) notation meant, lol, I think that's what it meant...
@@archangecamilien1879 haha yes, I much prefer improper fractions to mixed fractions 😂