@jain4444 said:The total number of Integral solutions of uvw²x²y² =277200
Too many cases 
Confusion he confusion hai, solution ka pata nahi..
Confusion he confusion hai, solution ka pata nahi..@mailtoankit said:5 ?E(11) = 1050^10k mod 11 = 156^52 mod 10 = 650^6 mod 11 = 6^6 mod 11 = 216*216 mod 11 = 7*7 mod 11 = 49 mod 11 = 5
Yar i got rem(56^52/10)=6 from pattern method,
but m applying euler theorem and getting this
mod(10)=10(1-1/2)(1-1/5)=4,
so rem(56^4/10)=1 so
rem(56^4)^13/10=1
where m going wrong????
@jain4444 said:20 children r standing in a line outside a ticket window . 10 of them have one rupee coin each & 10 have two rupee coin each . the entry ticket is priced Rs1. if all arrangements of 2o children are equally likely , find probability that 10th child will be first to wait for change .
0 hi hona chahiye
@bs0409 said:if a and c are integer and 3how many equation of the form ax^2+12x+c=0 have real roots?1.362.393.444.48
44.
D >=0
144-4ac >=0
ac
taking values of a as 3,4,5,6,7
we get 13,10,8,7,6 values
total 44 values.
@pakkapagal said:Yar i got rem(56^52/10)=6 from pattern method,but m applying euler theorem and getting thismod(10)=10(1-1/2)(1-1/5)=4,so rem(56^4/10)=1 sorem(56^4)^13/10=1where m going wrong????
Why apply complicated theorem here?
Look, remainder when divided by 10 is nothing but the last digit....so rem of 56^52 (or 56^anything for that matter) with 10 is 6.
regards
scrabbler
Look, remainder when divided by 10 is nothing but the last digit....so rem of 56^52 (or 56^anything for that matter) with 10 is 6.
regards
scrabbler
@pakkapagal said:Yar i got rem(56^52/10)=6 from pattern method,but m applying euler theorem and getting thismod(10)=10(1-1/2)(1-1/5)=4,so rem(56^4/10)=1 sorem(56^4)^13/10=1where m going wrong????
56 and 10 are not co-prime
ohhhhh je to dekha hi nahi 
thnks 😃
@jain4444 said:The total number of Integral solutions of uvw²x²y² =277200
bhai options hain kya??
Too many cases are there..options hota toh eliminate kar ke slightly jaldi solve hota...
Is it a multiple of 12....??
@Logrhythm said:@scrabbler ye wala dekho fir zara...In how many ways 2310 be written as product of 3 different factors??
factors are :
11,7,5,3 and 2
so clubbing them and checking combinations
club any 3 and the rest two can be mulitiplied separately : no.of ways 5c3 =10
club any two , then club any two of the three left = 5c2*3c2=30
so I guess 40.
@pakkapagal said:Yar i got rem(56^52/10)=6 from pattern method,but m applying euler theorem and getting thismod(10)=10(1-1/2)(1-1/5)=4,so rem(56^4/10)=1 sorem(56^4)^13/10=1where m going wrong????
euler nahi lagega bhai...56 and 10 are not co-prime...
@mailtoankit said:euler nahi lagega bhai...56 and 10 are not co-prime...
ha bhai..dekha nhi tha achche se.
. @ScareCrow28 bhai ne btaya tab dekha :)
. @ScareCrow28 bhai ne btaya tab dekha :)Q>How many number of zeroes will be there at the end of 12! expressed in base 6?
options ;
4,5,6,7
options ;
4,5,6,7
@saurav205 said:Q>How many number of zeroes will be there at the end of 12! expressed in base 6?options ;4,5,6,7
5
[12/3] plus [4/3]
[12/3] plus [4/3]
Q>Find the remainder when 2222^5555 + 5555^2222 is divided by 7?
options :
1,3,0 and 5
options :
1,3,0 and 5
@saurav205 said:Q>How many number of zeroes will be there at the end of 12! expressed in base 6?options ;4,5,6,7
6 = 3*2
[12/3] = [4/3] = 1
hence, 4+1 = 5 zeroes...
@saurav205 said:Q>Find the remainder when 2222^5555 + 5555^2222 is divided by 7?options :1,3,0 and 5
0
@saurav205 said:Q>Find the remainder when 2222^5555 + 5555^2222 is divided by 7?options :1,3,0 and 5
0 hoga
P.Sedited
P.Sedited