Hi Michael. I love your videos. I just think you have wrong answer in the first problem. I think it is 277771. The problem with your answer is that your answer is even when you add 299 odd numbers and it is not divisible by 299 even when it is 299*(average of serie A + average of serie B) .
Maybe someone else already noted this, but the first problem is solvable using nothing beyond Algebra 2, where they cover arithmetic sequences. The sum of an arithmetic sequence is the first plus the last times n/2. All we have to do is figure out n, but that can be done using the formula for the nth number an = a1 + (n-1)d. (d is the common difference.) Whether you proceed the "hard" way (thinking of it as two sequences, so more calculating), or recognizing that it's really one sequence, a B+ Algebra 2 student shouldn't have any trouble.
2:58 Should be 298, right? 4 * 298 + 2 = 1194 and 2 * 298 + 33 = 629. I guess it's a Classic Michael Penn mistake 😂 12:11 Dropping chalk 12:31 Good Place To Stop
Ahhh I realized what the mistake was at 3:27 he says the last term is the 299th term in the sum which is correct but the first term has n=0 so there are 299 terms and n=298
When counting the solutions for the second problem, first realize that 2010^20 does not end in a 1, sk including it does not change anything. Next realize there are 2010 integers between 1 and 2010 inclusively, and we can group these integers into 10 groups of 201 based off of their last digit. Only the groups that end in 1, 3, 7, or 9 will satisfy the condition, so wr have 4*201=804 integers.
I think n=298, not 299, and the result is 277771, a slightly more "interesting-looking" number than 279600, as is often the case for such math contest problems.
The first one, I grouped it as (2+33) + (6+35) + ... + (1194+629) = 35 + 41 + ... + 1823 i.e. going up by 6's. So it's the sum from 0 to 298 of (6n+35). Pair up the numbers from each end: 35 + 1823 = 41 + 1817 = ... = 1858. There are 149 pairs of numbers plus the center number which is the 149th number, which is 929. So the answer is 149*1858 + 929 = 277771
Every even number in any positive integer power is even. Every number that ends with 5 in any positive integer power ends with 5. Other odd numbers in 4th power end with 1, then in 20th=4*5 power they also end with 1. In any ten consecutive integers there are 4 such numbers. 2010 contains 201 tens, so the answer is 201*4=804
The second one I did a bit differently We can ignore all the even numbers, and 5 obviously Every number ending in 1 raised to any power will end in 1 so there are 201 such numbers The final digit of powers of three follows 3,9,7,1, which ends in a 1 every 4th power and since 4 divides 20 our number could end in 3 so there are 201 such numbers The final digit of powers of 7 follows 7,9,3,1 which ends in a 1 every 4th power so another 201 The final digit of powers of 9 follows a 9,1,9,1 pattern which ends in a 1 every 4th power so another 201 Much less elegant than the solution you provide but so simple to calculate you can solve it in your head.
I also did it this way but I assume that Micheal's approach can be applied in more number theory problems. Also the first question could be solved using arithmetic series formulas
I thought about finishing the second problem a bit differently: subtracting numbers in 1,2,...,2009 that have divisors 2 and 5 and then add divisors of 10 which we have overcounted, i.e. 2009-[2009/2]-[2009/5]+[2009/10]=2009-1004-401+200=804.
You can derive last digit of n also without knowledge of Eulers theorem: if gcd(n,10) > 1, then gcd(n^20,10) > 1 thus n^20 != 1 (mod 10). So we are left with last digits 1,3,7,9: 1 = 1^20 (mod 10) 9^2 = (-1)^2 = 1 (mod 10) thus 9^20 = 1^10 = 1 (mod 10) 3^4 = 9^2 = (-1)^2 = 1 (mod 10) thus 3^20 = 1^5 = 1 (mod 10) 7^4 = (-3)^4 = 9^2 = (-1)^2 = 1 (mod 10) thus 7^20 = 1^5 = 1 (mod 10) In fact, this is proof that gcd(n,10) = 1 n^4 = 1 (mod 10) which is exactly what Eulers theorem says here.
Michael basically just included a derivation of the value of triangular numbers (sums of arithmetic sequences) within his work. There's not much to it.
Second problem seems like it should be 804, because any odd number not divisible by 5 to the 4th power ends in 1. 1^4=1, 3^4=81, 7^4=2401, 9^4=6561, and so on. All even numbers won't be odd when taken to the 4th power, and all numbers divisible by 5 will keep a 5 as they go up powers. n^20 is a power of n^4. 2009 is 201 sets of 10 digits, minus one even number. 201*4=804.
Hi Michael I love ur videos. However, I think there's a mistake from 3:00 to 3:12, because 299*4+2=1198 and 299*2+33=631. The actual number that satisfies the condition is 298, cuz if you check 298*4+2=1194 and 298*2+33=629. Bye, thanks for the content.
Computing the number of numbers in the second problem can be more easily done(just a bit more easily). Any number prime to 10 is of the form 10k+{1,3,7,9}. 0
The first problem can be decomposed into two sequences, 4n+2 + 2n+33, ranging from n=0 (2, 33) to n=298 (1194, 629). Taking the sum Sn = n/2 * (a(0) + a(298), do you not get 277,771? Alternatively, the sequences could be 4n-2 and 2n+31, from n=1 to n=299, and then the sum formula Sn = n/2 * (a(1) + a(299)) is equivalent and also gives 277,771
A possibly easier lower grade solution of the second problem (a bit work intensive tho): Notice that the ones digit in the multiplication of two numbers comes from the multiplication of their ones digits Now fill the table of multiplying all odd numbers from 1 to 9 by all odd numbers from 1 to 9 (neither number can be even because that results in an even digit and 1 isn't) Now notice that 1*1=1 3*7=21 9*9=81 Now since 1*...*1=1 And since 9*9=81 If we can write 9^20 as (9*9)^x it will end in 1 and it is possible for x=10 now since 3*7=21 We need to do some extra work... We need to trace on the 7s multiplication table starting at 7*1 and ending at 7*3 and counting how many multiplications did we do (or stopping after 9 multiplications hence 7 isn't a solution) 7*1=7 7*7=49 7*9=63 7*3=21 4 multiplications So if we can write 7^20 as (7*7*7*7)^x and we can at x=5 Usually we'll do the same process for 3 but since (9*9)^10 = 9^20 we only need to take the root of both sides to notice that (9*9)^5 = 3^20 hence 3^20 ends in 1 And the obvious one is 1^20 = 1 Now we have 4 candidates for the ones digit we can either do as Michael did or Use this little trick... Our range ends at 2009 We can either increase or decrease this number to a multiple of 10 while being careful to note any solutions added or ignored by changing our range In this case I'll choose 2010 since it creates no extra solutions Now divide it by 10 and multiply it by 4 (candidates for ones digit) Which gives 804 now remove the extra solutions created by increasing the range which is 0 so 804 is the final answer And that's a good place to stop
You take an algebraically intensive approach to the first problem (and arrive at the wrong answer). I have found with algebraic progressions that it is easier to explain to my students that you are looking for the average value of the series times the number of members of the series. The average value is the smallest + the largest divided by 2. The final result should be (2+1194 + 33+692)*299/2 = 277,771. Regarding the second question, the problem on the board is not quite the same problem as on the thumbnail, not that it makes a difference. Finally, we could make the second problem be "how many n are there such that the last two digits of n^20 are 01," and have the same result.
I solved the first one a little differently. Each pair can broken into (2+33)+6n where n goes from 0 to 298. So we have 299*(2+33)+(6+12+18+....+6*298) Factoring out the 6 in the second part gives 299*35+6*(1+2+....298) we can then solve the second half by adding the first and last terms and multiplying by 298/2 299*35+6(299)*149 and factoring 6 into 2*3 and multiplying the 2 into 149 to get 298 299*35+299*(298*3) and then grouping (35+298*3)*299 for 277771
if you would define the first sequence as (4n-2) and the second as (31+2n) it would be for n=1 to 299 and not 0 to 298. (and of course and then you could also combine them to (6n+29))
The thumbnail one is fairly easy on this one. All you have to do is check 3^20, 7^20, and 9^20 to see what digit they end with. (You know 1^20 = 1). So you find that it's all the odd numbers that are not multiples of five. So, between 1 and 2009, you have 2010/2 - 2010/10 = 804 integers where n^20 ≡ 1 (mod 10).
Hey guys, could you tell me what’s wrong with my solution of the second problem. So, I firstly noticed that any number that ends with 1 or 9 gives a number which ends with 1 in any power. Exp. numbers like 11, 21, 29, 19 and so on. So, next I found how many numbers end with 1 or 9 in the range of [1;2009]. And I found out that there are 402 numbers that end with 1 or 9. So, then I thought that this is the answer. But after watching the solution, the answer actually turned out to be two times bigger. So, what’s wrong with my solution?
phi(10)=phi(5)*phi(2)=4*1=4 So for every number relatively prime to 10, ord_10(n)|4|20, so for every number relatively prime to 10, n^20 ends with 1 there are exactly 4 numbers relatively prime to 10 mod 10 {1,3,7,9} So #n=2009/10*4=804
The numbers n^20 [n=0..9] which end with 1 are 4, namely 1, 3, 7, 9. From 1 to 2009 there are 201 decades thus the total number n^20 [n=1..2009] ending with 1 are 201 x 4 = 804 The first problem is simply $\sum _{n=0}^{298} (6 n+35) = 277,771$
*i'm being wrong the next lines and tim got an explanation for it, first reply for those saying n should be evaluated at 298, the mistake is rather expressing 1194 and 629 as 4n+2 and 2n+33 respectively 2, 6, 10, ..., 1194 form an arithmetic progression with 2 being the first term and the ratio 4 applying an=a1+r(n-1) where 'an' is the last term (1194), a1 is the first term (2), r is the ratio (4) and n is the number of terms, you get that 1194=4n-2 doing the same for the other progression, which is 33, 35, 37, ..., 629, you get that 629=2n+31 both equations tell n should be evaluated at 299
You would be correct if he were indexing his sequences starting from n=1, but he was indexing from n=0 (consistent with the n(n+1) closed form rather than the n(n-1) closed form). So, you're mistaken.
The first problem can be solved using AP .here in India we learn that in grade 10. Btw Michael , I love your videos and the thing I learn is how to approach any problem which is helpful in my Olympiad preparation
I've frequently noticed comments from India (and nowhere else) focused on when exactly in school the technique required for a problem is covered. Why is that?
@@TJStellmach Our mentality is f*cked due to a sh*t-ton of competitive exams Parents still force their children ( not all of them) to pursue Engineering or Medical without even paying attention to their children's own personal interests I am also from India and also preparing for these exams I am just glad that I know the actual truth before running for these exams
@@TJStellmach Although the OP stated a different exam which is not related to Engineering or Medical but I just thought that sharing my thoughts on those exams would be better
Hey UA-cam: I forgot to switch the audio to mono on this one. Sorry about that...
I like the stereo effect. It matches with were you are in the room
ASMR video when? 😂
@@z4zuse it’s way too wide for that
waaah, my brain.
Hi Michael. I love your videos.
I just think you have wrong answer in the first problem. I think it is 277771.
The problem with your answer is that your answer is even when you add 299 odd numbers and it is not divisible by 299 even when it is 299*(average of serie A + average of serie B) .
Maybe someone else already noted this, but the first problem is solvable using nothing beyond Algebra 2, where they cover arithmetic sequences. The sum of an arithmetic sequence is the first plus the last times n/2. All we have to do is figure out n, but that can be done using the formula for the nth number an = a1 + (n-1)d. (d is the common difference.) Whether you proceed the "hard" way (thinking of it as two sequences, so more calculating), or recognizing that it's really one sequence, a B+ Algebra 2 student shouldn't have any trouble.
2:58 Should be 298, right? 4 * 298 + 2 = 1194 and 2 * 298 + 33 = 629. I guess it's a Classic Michael Penn mistake 😂
12:11 Dropping chalk
12:31 Good Place To Stop
Ahhh I realized what the mistake was at 3:27 he says the last term is the 299th term in the sum which is correct but the first term has n=0 so there are 299 terms and n=298
yeah, I heard him mention 4n-2 so I think he was probably just mixing up the slightly different equations with an index change
The correct answer to the first part should have been 277771.
It's 2021-07-31 today but Michael still has not realized his mistake, unfortunately.
@@Horinius that’s what I got also
When counting the solutions for the second problem, first realize that 2010^20 does not end in a 1, sk including it does not change anything. Next realize there are 2010 integers between 1 and 2010 inclusively, and we can group these integers into 10 groups of 201 based off of their last digit. Only the groups that end in 1, 3, 7, or 9 will satisfy the condition, so wr have 4*201=804 integers.
Right, I did it this way too
Did it the same way too.
Yup, that's what I did.
Yeah, when I saw the pattern I realized it must be 40% of 2010. It becomes really easy as soon as you see that 1, 3, 7, and 9 will work every time.
The correct answer to the first part should have been 277771.
It's 2021-07-31 today but Michael still has not realized his mistake, unfortunately.
Shouldn't you have evaluated that first one at n=298? Setting n=299 gives 2n+33=631, not 629 as desired.
You are correct. I was wondering why I got a different solution.
@@wiktorbachta8576 what answer you got
@@advaykumar9726 i got 277771
Me too
@@Relrax me too I thought mine was wrong
I think n=298, not 299, and the result is 277771, a slightly more "interesting-looking" number than 279600, as is often the case for such math contest problems.
Yes , i also think that
I calculated it as sum of two arithmetic sequences
Glad I did it correctly, because for a moment I rechecked all of my solution and still found it to be 277771 :)
Yup, i got that answer too and reviewed it at least 3 times.
@@javiernasser3574 me too!
I did the same as Jacek Soplica...
The first one, I grouped it as (2+33) + (6+35) + ... + (1194+629) = 35 + 41 + ... + 1823 i.e. going up by 6's.
So it's the sum from 0 to 298 of (6n+35).
Pair up the numbers from each end: 35 + 1823 = 41 + 1817 = ... = 1858. There are 149 pairs of numbers plus the center number which is the 149th number, which is 929.
So the answer is 149*1858 + 929 = 277771
I was wondering why he didn’t just do a Gaussian sum. Maybe he did but with different terms? Either way would’ve worked I think
i find it amusing that math youtubers doht check their answers with a calculator omfg. 277771 is very much the correct answer!
Rohayhu Michal Penn!! Greetings from Paraguay!!
Will be in Paraguay in April 2022 when the airports are available and the epidemic situation suits. I have my family there, greetings from Germany :)
Every even number in any positive integer power is even. Every number that ends with 5 in any positive integer power ends with 5. Other odd numbers in 4th power end with 1, then in 20th=4*5 power they also end with 1. In any ten consecutive integers there are 4 such numbers. 2010 contains 201 tens, so the answer is 201*4=804
Basically same method I used. :) GMTA.
In the first problem just combine the two APs and calculate as sum of a new AP with general term (6n + 35)
You've spelled Paraguay wrong on the board :)
The second one I did a bit differently
We can ignore all the even numbers, and 5 obviously
Every number ending in 1 raised to any power will end in 1 so there are 201 such numbers
The final digit of powers of three follows 3,9,7,1, which ends in a 1 every 4th power and since 4 divides 20 our number could end in 3 so there are 201 such numbers
The final digit of powers of 7 follows 7,9,3,1 which ends in a 1 every 4th power so another 201
The final digit of powers of 9 follows a 9,1,9,1 pattern which ends in a 1 every 4th power so another 201
Much less elegant than the solution you provide but so simple to calculate you can solve it in your head.
I did it the sae way
Very possible that Michael was conformable with his approach. It happens in math.
@@wisdomokoro8898 Oh definitely, he had a better set of tools to solve the problem. Its like I used a rock to crack the nut and he used a nutcracker
I also did it this way but I assume that Micheal's approach can be applied in more number theory problems. Also the first question could be solved using arithmetic series formulas
The correct answer to the first part should have been 277771.
It's 2021-07-31 today but Michael still has not realized his mistake, unfortunately.
I thought about finishing the second problem a bit differently: subtracting numbers in 1,2,...,2009 that have divisors 2 and 5 and then add divisors of 10 which we have overcounted, i.e. 2009-[2009/2]-[2009/5]+[2009/10]=2009-1004-401+200=804.
You can derive last digit of n also without knowledge of Eulers theorem:
if gcd(n,10) > 1, then gcd(n^20,10) > 1 thus n^20 != 1 (mod 10). So we are left with last digits 1,3,7,9:
1 = 1^20 (mod 10)
9^2 = (-1)^2 = 1 (mod 10) thus 9^20 = 1^10 = 1 (mod 10)
3^4 = 9^2 = (-1)^2 = 1 (mod 10) thus 3^20 = 1^5 = 1 (mod 10)
7^4 = (-3)^4 = 9^2 = (-1)^2 = 1 (mod 10) thus 7^20 = 1^5 = 1 (mod 10)
In fact, this is proof that gcd(n,10) = 1 n^4 = 1 (mod 10) which is exactly what Eulers theorem says here.
I would have gone ahead and computed the two triangular sums on the first problem.
Thank you, professor you’re always enlightening.
the second number problem can be formulated idiomatically:
*Count[Range@2009, n_ /; Last@IntegerDigits[n^20] == 1]*
We could have done this using arithmatic progression . (2 + 6 + 10... 1194 ) + (33 + 35 +37 ....... 629) .
This is how I did it!
Michael basically just included a derivation of the value of triangular numbers (sums of arithmetic sequences) within his work. There's not much to it.
Second problem seems like it should be 804, because any odd number not divisible by 5 to the 4th power ends in 1. 1^4=1, 3^4=81, 7^4=2401, 9^4=6561, and so on. All even numbers won't be odd when taken to the 4th power, and all numbers divisible by 5 will keep a 5 as they go up powers. n^20 is a power of n^4. 2009 is 201 sets of 10 digits, minus one even number. 201*4=804.
He's written PARAGAUY as well...That's ok Micheal we love you the same
U r phenomenal Michael. Bravo.
Hi Michael I love ur videos. However, I think there's a mistake from 3:00 to 3:12, because 299*4+2=1198 and 299*2+33=631. The actual number that satisfies the condition is 298, cuz if you check 298*4+2=1194 and 298*2+33=629. Bye, thanks for the content.
tembo, saludos desde Paraguay 😋🇵🇾
Great video btw
Computing the number of numbers in the second problem can be more easily done(just a bit more easily). Any number prime to 10 is of the form 10k+{1,3,7,9}. 0
The first problem can be decomposed into two sequences, 4n+2 + 2n+33, ranging from n=0 (2, 33) to n=298 (1194, 629). Taking the sum Sn = n/2 * (a(0) + a(298), do you not get 277,771?
Alternatively, the sequences could be 4n-2 and 2n+31, from n=1 to n=299, and then the sum formula Sn = n/2 * (a(1) + a(299)) is equivalent and also gives 277,771
Yes thats what i got too
Same answer for me ! I'm going to check if I didn't make a mistake.
Checked ! We are right
A possibly easier lower grade solution of the second problem (a bit work intensive tho):
Notice that the ones digit in the multiplication of two numbers comes from the multiplication of their ones digits
Now fill the table of multiplying all odd numbers from 1 to 9 by all odd numbers from 1 to 9 (neither number can be even because that results in an even digit and 1 isn't)
Now notice that
1*1=1
3*7=21
9*9=81
Now since 1*...*1=1
And since 9*9=81
If we can write 9^20 as (9*9)^x it will end in 1 and it is possible for x=10
now since 3*7=21
We need to do some extra work...
We need to trace on the 7s multiplication table starting at 7*1 and ending at 7*3 and counting how many multiplications did we do (or stopping after 9 multiplications hence 7 isn't a solution)
7*1=7
7*7=49
7*9=63
7*3=21
4 multiplications
So if we can write 7^20 as (7*7*7*7)^x and we can at x=5
Usually we'll do the same process for 3 but since (9*9)^10 = 9^20
we only need to take the root of both sides to notice that (9*9)^5 = 3^20
hence 3^20 ends in 1
And the obvious one is 1^20 = 1
Now we have 4 candidates for the ones digit we can either do as Michael did or
Use this little trick...
Our range ends at 2009
We can either increase or decrease this number to a multiple of 10 while being careful to note any solutions added or ignored by changing our range
In this case I'll choose 2010 since it creates no extra solutions
Now divide it by 10 and multiply it by 4 (candidates for ones digit)
Which gives 804 now remove the extra solutions created by increasing the range which is 0 so 804 is the final answer
And that's a good place to stop
I think u can use arithmetic progression to evaluate this may be
You take an algebraically intensive approach to the first problem (and arrive at the wrong answer). I have found with algebraic progressions that it is easier to explain to my students that you are looking for the average value of the series times the number of members of the series. The average value is the smallest + the largest divided by 2. The final result should be (2+1194 + 33+692)*299/2 = 277,771.
Regarding the second question, the problem on the board is not quite the same problem as on the thumbnail, not that it makes a difference.
Finally, we could make the second problem be "how many n are there such that the last two digits of n^20 are 01," and have the same result.
the first number problem is easy to code:
*total = Total@Table[i, {i, 33, 629, 2}] + Total@Table[i, {i, 2, 1194, 4}];*
*sum = Sum[i, {i, 33, 629, 2}] + Sum[i, {i, 2, 1194, 4}];*
*total == sum == 277771*
I solved the first one a little differently. Each pair can broken into (2+33)+6n where n goes from 0 to 298.
So we have 299*(2+33)+(6+12+18+....+6*298)
Factoring out the 6 in the second part gives
299*35+6*(1+2+....298)
we can then solve the second half by adding the first and last terms and multiplying by 298/2
299*35+6(299)*149 and factoring 6 into 2*3 and multiplying the 2 into 149 to get 298
299*35+299*(298*3) and then grouping (35+298*3)*299 for 277771
if you would define the first sequence as (4n-2) and the second as (31+2n) it would be for n=1 to 299 and not 0 to 298.
(and of course and then you could also combine them to (6n+29))
Use sum of arithmetic progression formula, it's easier
The thumbnail one is fairly easy on this one. All you have to do is check 3^20, 7^20, and 9^20 to see what digit they end with. (You know 1^20 = 1). So you find that it's all the odd numbers that are not multiples of five.
So, between 1 and 2009, you have 2010/2 - 2010/10 = 804 integers where n^20 ≡ 1 (mod 10).
Two colors of chalk? Looks like a variant of bprp.
In the 1st problem, the value of n should be 298, thus changing the final answer.
1. combine every two term in the
sequence. Further question: Find the normal term of the sequence.
Good Place To Start At 12:11
Number of ns = number not divisible by 2 or 5.
What root of the mathematics is this?
Hey guys, could you tell me what’s wrong with my solution of the second problem. So, I firstly noticed that any number that ends with 1 or 9 gives a number which ends with 1 in any power. Exp. numbers like 11, 21, 29, 19 and so on. So, next I found how many numbers end with 1 or 9 in the range of [1;2009]. And I found out that there are 402 numbers that end with 1 or 9. So, then I thought that this is the answer. But after watching the solution, the answer actually turned out to be two times bigger. So, what’s wrong with my solution?
You missed the fact that any number that ends in 3 or 7 gives a result ending in 1 when raised to any power that's a multiple of 4.
phi(10)=phi(5)*phi(2)=4*1=4
So for every number relatively prime to 10, ord_10(n)|4|20, so for every number relatively prime to 10, n^20 ends with 1
there are exactly 4 numbers relatively prime to 10 mod 10 {1,3,7,9}
So #n=2009/10*4=804
It's crazy how much harder the USA math olympiad is compared to the Paraguay one, these questions would probably appear on the AMC lol
Paragauy
The numbers n^20 [n=0..9] which end with 1 are 4, namely 1, 3, 7, 9.
From 1 to 2009 there are 201 decades thus the total number n^20 [n=1..2009] ending with 1 are
201 x 4 = 804
The first problem is simply
$\sum _{n=0}^{298} (6 n+35) = 277,771$
Donde estan mis kps paraguayos 😈🇵🇾
n =298 not 299
There are 804 numbers using number theory
The sum gives me 277771
All odd numbers not ending with 5. So the answer is 804. No one would need to use Fermat's theorem or Euler's theorem to solve this problem.😃
Paraguayan math💪🏾🫶🏾
bruh the seconda question is too easy, every odd number that arent divisible by 5 are possible, the answer is 804
You should make more videos about 1 hard problem rather than 2 easy ones.
*i'm being wrong the next lines and tim got an explanation for it, first reply
for those saying n should be evaluated at 298, the mistake is rather expressing 1194 and 629 as 4n+2 and 2n+33 respectively
2, 6, 10, ..., 1194 form an arithmetic progression with 2 being the first term and the ratio 4
applying an=a1+r(n-1) where 'an' is the last term (1194), a1 is the first term (2), r is the ratio (4) and n is the number of terms, you get that 1194=4n-2
doing the same for the other progression, which is 33, 35, 37, ..., 629, you get that 629=2n+31
both equations tell n should be evaluated at 299
You would be correct if he were indexing his sequences starting from n=1, but he was indexing from n=0 (consistent with the n(n+1) closed form rather than the n(n-1) closed form). So, you're mistaken.
thank you, i rushed it and you're right
:(
The first problem can be solved using AP .here in India we learn that in grade 10.
Btw Michael , I love your videos and the thing I learn is how to approach any problem which is helpful in my Olympiad preparation
I've frequently noticed comments from India (and nowhere else) focused on when exactly in school the technique required for a problem is covered. Why is that?
Really, wow. Define frequently. How many are we talking about here per michaels video. Would also appreciate a link to these comments
@@TJStellmach Our mentality is f*cked due to a sh*t-ton of competitive exams
Parents still force their children ( not all of them) to pursue Engineering or Medical without even paying attention to their children's own personal interests
I am also from India and also preparing for these exams
I am just glad that I know the actual truth before running for these exams
@@TJStellmach Although the OP stated a different exam which is not related to Engineering or Medical but I just thought that sharing my thoughts on those exams would be better
@@shohamsen8986 Look in any comment section for channels like this one or similar (Mind Your Decisions, etc.)
i say 804 at first sight
Paraguay? That doesn't exist
(Latinoamerican joke)
Not funny
@@nicogh1spy 😭