could someone please help me here? I know it is a stupid question but could someone tell me the way to factorize y^4+2y^3-y^2-2y+1 into (y^2+y-1)(y^2+y-1). Im sorry if i am posting it in the wrong section. I get stuck with equations like this and do not know how to proceed if i need it in the form of product of 2 factors. thanks:)
what is the remainder when 128^500 is divided by 153?
seems tough: pls provide approach for this. i applied euler for reducing power ,, aftr thn digits remaining were not favourable to get it done in 1-2 mins:(
seems tough: pls provide approach for this. i applied euler for reducing power ,, aftr thn digits remaining were not favourable to get it done in 1-2 mins:(
ans 2 hai kya?
i think post 2528 will help. I think 67 is d rite answer.
could someone please help me here? I know it is a stupid question but could someone tell me the way to factorize y^4+2y^3-y^2-2y+1 into (y^2+y-1)(y^2+y-1). Im sorry if i am posting it in the wrong section. I get stuck with equations like this and do not know how to proceed if i need it in the form of product of 2 factors. thanks:)
Hi, Infact i too prefer to leave such type of equations for 2nd round :(
but still below is the approach i followed fr the above equation
y^4+2y^3-y^2-2y+1 put 1 , its not giving 1 , hence y-1 is not the factor = y^4 + y^3 + y^3 - y^2 - y - y + 1 (opened it to see what we can combine) We have 1 (0 power of y) and 4 power of y Hence I was looking for something that can be grouped as y^2 * y^2 after multiplication and 1*1 () So in two factors we have something like which gives 1*1 and y^2 * y^2 = y^4 (meaning that constant of term will be 1 and constant of y^2 in both factors will be 1 ) So I tried for last pair as - y^2 - y + 1 = y^4 + y^3 + y^3 - y - y^2 - y + 1 = y^4 + y^3 + y^3 - y - ( y^2 + y -1) y^3 - y (in this term if I try I will get y^2 and y of diffrnt sign) = (y^4 + y^3) + y^3 - y - ( y^2 + y -1) = y^2 (y^2 + y^1) + y^3 - y - ( y^2 + y -1) == y^2 (y^2 + y -1) + y^2 + y^3 - y - ( y^2 + y -1) == y^2 (y^2 + y -1) + (y^2 + y^3 - y ) - ( y^2 + y -1) == y^2 (y^2 + y -1) + y*(y^2 + y - 1) - ( y^2 + y -1) = (y^2 + y -1)(y^2 + y - 1)
Puys, Pls help to provide any simple approach other than mine
Hi, Infact i too prefer to leave such type of equations for 2nd round :(
but still below is the approach i followed fr the above equation
y^4+2y^3-y^2-2y+1 put 1 , its not giving 1 , hence y-1 is not the factor = y^4 + y^3 + y^3 - y^2 - y - y + 1 (opened it to see what we can combine) We have 1 (0 power of y) and 4 power of y Hence I was looking for something that can be grouped as y^2 * y^2 after multiplication and 1*1 () So in two factors we have something like which gives 1*1 and y^2 * y^2 = y^4 (meaning that constant of term will be 1 and constant of y^2 in both factors will be 1 ) So I tried for last pair as - y^2 - y + 1 = y^4 + y^3 + y^3 - y - y^2 - y + 1 = y^4 + y^3 + y^3 - y - ( y^2 + y -1) y^3 - y (in this term if I try I will get y^2 and y of diffrnt sign) = (y^4 + y^3) + y^3 - y - ( y^2 + y -1) = y^2 (y^2 + y^1) + y^3 - y - ( y^2 + y -1) == y^2 (y^2 + y -1) + y^2 + y^3 - y - ( y^2 + y -1) == y^2 (y^2 + y -1) + (y^2 + y^3 - y ) - ( y^2 + y -1) == y^2 (y^2 + y -1) + y*(y^2 + y - 1) - ( y^2 + y -1) = (y^2 + y -1)(y^2 + y - 1)
Puys, Pls help to provide any simple approach other than mine
Originally Posted by Srihari123 View Post Please tell the approach for these kind of problems Remainder for (2^164)/164 is ?? 2^164/2^2*41 2^162/41 (2^10*2^10................16 times)*2*2/41 (1025-1*1025-1*1024-1...................16times)*2*2/41 {41*25=1025} (-1*-1*-1*-1....................16times)*2*2 1*2*2 4*2^2 4*4 16
i really dont get it,,neither is the method of cancelling out some common factors correct,,,example....128/96..if the above method is correct then,,we can cancel out the common factors and the answer wud be 2(8/6 left out)..which is absolutely wrong!! plz help...
(2^164)/164 (2^164)/164 (for using euler method, numerator n denominator both should be co-primes. So (2^164)/164 =(2^164)/41*4= (2^164)/(41*2^2) So (2^162)/41 now we can use Euler method (but we will keep in mind the cancelled part later Now euler no of 41 is 40. 41*(1-1/41)=40 So 2^40/41 or 2^160/41 >> remainder is 1 So remaning part is 2*2=4 + the cancelled part 2*2 as actually number was (2^164)/164 Not (2^162)/41 >hence remainder is 2*2*2*2=16
ten students solved a total of 35 question in a maths olympiad each question was solved by exactly one student .there is at least one student who solved exactly one problem, at least one student who solved exactly two problem and at least one student who solved exactly three problems what is the minimum no.of students who has/have solved at least five problem? a)1 b)2 c)3 d)none
in a particular country all the members are expressed with the help of three alphabets a,b,c 15 is written as abc 6 is written as bc 60 is written as bcbc how would one write 17 in that country? a)abb b)bab c)baa d)aba
if there are 45 match played. A, B, C played 30, 23, and 18 respectively. Then, what is least no. of matches in which more than one of them played? please suggest how to solve this and other min. - max. problems.
I get stuck with equations like this and do not know how to proceed if i need it in the form of product of 2 factors. thanks:)
for this. i applied euler for reducing power ,, aftr thn digits remaining were not favourable to get it done in 1-2 mins:(
