@iLoveTorres said:bro isko summation wala logic laga ke bhi toh solve kar sakte hai.. you will get (summation)n^2*(n+1)^3 = [n*(n+1)*(2n+1)]/6*[((n+1)*(n+2))/2]^2
n^2 and (n+1)^3 is in a product.
@iLoveTorres said:bro isko summation wala logic laga ke bhi toh solve kar sakte hai.. you will get (summation)n^2*(n+1)^3 = [n*(n+1)*(2n+1)]/6*[((n+1)*(n+2))/2]^2
Summation (ab) is not (Summation (a))*(Summation(b))...
2 * 3 + 3 * 4 is not the same as 5 * 7 for example
regards
scrabbler
2 * 3 + 3 * 4 is not the same as 5 * 7 for example
regards
scrabbler
@scrabbler said:Summation (ab) is not (Summation (a))*(Summation(b))...2 * 3 + 3 * 4 is not the same as 5 * 7 for exampleregardsscrabbler
then how would you do a summation for n^5?
@iLoveTorres said:then how would you do a summation for n^5?
Exactly! That is why this approach is not profitable here!
@iLoveTorres said:then how would you do a summation for n^5?
BTW the formula is: = (1/6)n^6 + (1/2)n^5 + (5/12)n^4 - (1/12)n^2
@scrabbler said:Consider with 9 and 4.All terms divisible by 4. So of the form 4kAlso when divided by 9, the terms leave -1, 0, 0, -1, 0, 0....which in 99 terms adds to -33 equiv +3 with 9. So 9m+3 form as well.1st number satisfying both = 12.regardsscrabbler
Bhai , will thefourth term in the series be 4^2*5^3 ?
if yes ,then that divided by 9 will leave a remainder 2 , not -1..
Pl clarify..
if yes ,then that divided by 9 will leave a remainder 2 , not -1..
Pl clarify..
@iLoveTorres said:then how would you do a summation for n^5?
I wouldn't 😃
Why trouble with out-of syllabus stuff
enjoy maadi...
regards
scrabbler
Why trouble with out-of syllabus stuff
enjoy maadi...regards
scrabbler
@Logrhythm said:9*4all terms div by 4. -> 4x8,0,0,8,0,0,... -> 33*8 = 264 %9 = 3 -> 9y+3so 12...
bhai 4^2. 5^3 % 9 will be 2 i guess
@rnishant231 said:Bhai , will third fourth term in the series be 4^2*5^3 ?if yes ,then that divided by 9 will leave a remainder 2 , not -1..Pl clarify..
Main shortcut karne ki koshish mein galat hua I guess....when solving had got some longer thing which summed to -33....I felt ki shayad sab -1 hi the...rechecking...
Edit: Can't find the paper where I did it. :banghead:
Edit2: Haan so I checked the first 9 steps and got -1, 0, 0, -7, 0, 0, -4, 0, 0, after which repeat...so -12 = -3 and 11 such cycles so - 33....then no idea how, I typed it as -1 in all 3 cases 😞 Since answer same aaya I didn't realise
Sorry...
regards
scrabbler
Edit: Can't find the paper where I did it. :banghead:
Edit2: Haan so I checked the first 9 steps and got -1, 0, 0, -7, 0, 0, -4, 0, 0, after which repeat...so -12 = -3 and 11 such cycles so - 33....then no idea how, I typed it as -1 in all 3 cases 😞 Since answer same aaya I didn't realise
Sorry...regards
scrabbler
@techgeek2050 said:bhai 4^2. 5^3 % 9 will be 2 i guess
oh yes....ye toh primes ke sath wale mein kuch alag aa raha hai....recheck karta hun...
@scrabbler said:Main shortcut karne ki koshish mein galat hua I guess....when solving had got some longer thing which summed to -33....I felt ki shayad sab -1 hi the...rechecking...Edit: Can't find the paper where I did it. regardsscrabbler
@Logrhythm said:oh yes....ye toh primes ke sath wale mein kuch alag aa raha hai....recheck karta hun...
I am happy finally @scrabbler bhai ne paper use kar liya.. orally lgta hai out of fashion ho gaya hai..
Bhai logo lagta hai mein first time apne approach mein correct tha :P
@techgeek2050 said:bhai 4^2. 5^3 % 9 will be 2 i guess
write 4^2*5^3 in general terms (3a+1)^2 * (3a+2)^3 ....
(3a+1)^2 * (3a+2)^3mod 9 = (6a+1)*(8)*33
(6a+1)*(-1)(-3) mod 9
18a + 3 mod 9 =>3 mod 9
= 3*4 = 12
@scrabbler kar dia "tough part"...
P.S:i was waiting for quite some time to write this... lol.. :D
(3a+1)^2 * (3a+2)^3mod 9 = (6a+1)*(8)*33
(6a+1)*(-1)(-3) mod 9
18a + 3 mod 9 =>3 mod 9
= 3*4 = 12
@scrabbler kar dia "tough part"...
P.S:i was waiting for quite some time to write this... lol.. :D
@iLoveTorres said:I am happy finally @scrabbler bhai ne paper use kar liya.. orally lgta hai out of fashion ho gaya hai..Bhai logo lagta hai mein first time apne approach mein correct tha
Konsa approach? Summation wala? That's not gonna work bro! :(
@Asfakul said:Bhai why 60/6 ?
actually if all three x y z are different
suppose
2 4 5
this could be arranged as x y z in 3! ie 6 ways
now in 60 solutions as all values are different each solution is considered 6 times so i divided it by 6 to get number of solutions which are different absent arrangement.
@iLoveTorres said:Arre people told me "itna bhi eco-friendly mat bano" so I decided to try writing. Waise bhi WAT mein kaam aayega...handwriting improve karni hai :DI am happy finally @scrabbler bhai ne paper use kar liya.. orally lgta hai out of fashion ho gaya hai..
(And writing mein, I make more mistakes! Why!!!!) :(
regards
scrabbler
@scrabbler said:Arre people told me "itna bhi eco-friendly mat bano" so I decided to try writing. Waise bhi WAT mein kaam aayega...handwriting improve karni hai (And writing mein, I make more mistakes! Why!!!!) regardsscrabbler
@gautam22 said:Sir for m=0 there r 3 values of y=+2 , -2 , 0....yeah for -1 there r 2 values
Gautam, for m=0, there are four roots: 0,-2,2,4
Team BV - Vineet
@iLoveTorres said:I am happy finally @scrabbler bhai ne paper use kar liya.. orally lgta hai out of fashion ho gaya hai..Bhai logo lagta hai mein first time apne approach mein correct tha
nahi bhai wo toh still won't work...best toh @ScareCrow28 bhai wala hi lagg raha hai approach....wohi karna hai isme sahi hoga...
otherwise ye ganda sa pattern bann raha hai...
8,0,0,2,0,0,5,0,0 and then it repeats
11(8+2+5)%9 = 3
but aese karna wahiyaat hai....wohi karo crow bhai wala...
@ChirpiBird said:write 4^2*5^3 in general terms (3a+1)^2 * (3a+2)^3 ....(3a+1)^2 * (3a+2)^3mod 9 = (6a+1)*(8)*33(6a+1)*(-1)(-3) mod 918a + 3 mod 9 =>3 mod 9= 3*4 = 12 @scrabbler kar dia "tough part"... P.S:i was waiting for quite some time to write this... lol..
do you mind explaining this part
@The_Loser said:LCM and HCF of two numbers are 11! and 7! respectively. How many pairs of such numbers are possible?
u have to just put 2^4*3^2*5*11
4c0 +4c1 +4c2/2 = 1+ 4+ 3 = 8