@erm said:What is the remainder if the sum 2012^2013+2013^2012 is divided by 2012.2013?
=((2012)^2012 / 2013) +) (2013)^2011 / 2012)
= ((-1)^2012 + (1) ^2011)
=2
=((2012)^2012 / 2013) +) (2013)^2011 / 2012)
= ((-1)^2012 + (1) ^2011)
=2
x^2 + y^2 = 3479, x,y E N
How many values (x,y) can take?
@erm said:What is the remainder if the sum 2012^2013+2013^2012 is divided by 2012.2013?
2012 and 2013 are co prime
reminder by 2012 = 1
reminder by 2013 = -1
so reminder will be in the form of 2012a+1 or 2013b-1
2012a+1 = 2013b-1
2013b - 2012a = 2
as a and b both are integers
so
a = b = 2
so reminder = 2012*2+1 or 2013*2-1 = 4025
@culdip said:2012 and 2013 are co primereminder by 2012 = 1reminder by 2013 = -1so reminder will be in the form of 2012a+1 or 2013b-12012a+1 = 2013b-12013b - 2012a = 2 as a and b both are integerssoa = b = 2so reminder = 2012*2+1 or 2013*2-1 = 4025
fir @nramachandran ke sol me kya problem hai....
just abv ur post..
just abv ur post..
@erm said:@culdip why u took same values for a and b
a and b are positive integers and no other value is possible, where a
2013b - 2012a = 2
a + 2012(a - b) = 2
only a = b = 2 satisfies.
@nramachandran said:x^2 + y^2 = 3479, x,y E N How many values (x,y) can take?
All perfect squares are of form 4k or 4k + 1
=> Sum of two perfect squares will be of form 4n or 4n + 1 or 4n + 2 but never of form 4n + 3
But 3479 is of form 4n + 3, so no solution
@Dexian said:fir @nramachandran ke sol me kya problem hai....just abv ur post..
In case of (a*b)/(x*y) we can not break it like (a/x)*(b/y) and find individual remainders and then multiply to get the answer
It should be:- (remainder when a is divided by x*y)*(remainder when b is divided by x*y)
@nramachandran said:x^2 + y^2 = 3479, x,y E N How many values (x,y) can take?
zero.
Sum can never be 3479
:neutral:
@fireatwill said:a girl has to climb 12 steps .she climbs in either a single step or 2 steps simultaneously .in how many ways can she do it ?
Can do either
6 "2"s or
5 "2"s and 2 "1"s or
4 "2"s and 4 "1"s or
3 "2"s and 6 "1"s or
2 "2"s and 8 "1"s or
1 "2" and 10 "1"s or
12 "1"s
So 6C6 + 7C5 + 8C4 + 9C3 + 10C2 + 11C1 + 12C0
= 1 + 21 + 70 + 84 + 45 + 11 + 1
= 233?
Might be some calculation error. Please check!
regards
scrabbler
6 "2"s or
5 "2"s and 2 "1"s or
4 "2"s and 4 "1"s or
3 "2"s and 6 "1"s or
2 "2"s and 8 "1"s or
1 "2" and 10 "1"s or
12 "1"s
So 6C6 + 7C5 + 8C4 + 9C3 + 10C2 + 11C1 + 12C0
= 1 + 21 + 70 + 84 + 45 + 11 + 1
= 233?
Might be some calculation error. Please check!
regards
scrabbler
@chillfactor (remainder when a is divided by x*y)*(remainder when b is divided by x*y) yahi kiya hai.......... fir kahan galat ho raha hai...
=(2012^2013+2013^2012)/(2012*2013)
=(2012^2013)/2012*2013+(2013^2012)/2012*2013
=((2012)^2012 / 2013) +) (2013)^2011 / 2012)
= ((-1)^2012 + (1) ^2011)
=2
=(2012^2013+2013^2012)/(2012*2013)
=(2012^2013)/2012*2013+(2013^2012)/2012*2013
=((2012)^2012 / 2013) +) (2013)^2011 / 2012)
= ((-1)^2012 + (1) ^2011)
=2
@nramachandran said:x^2 + y^2 = 3479, x,y E N How many values (x,y) can take?
an square can end with 0 1 4 5 6 9
so we can have
a2^2 + b5^2 = 3479
a0^2 + b3^2 = 3479
a0^2 + b7^2 = 3479
a8^2 + b5^2 = 3479
possible value of a2 = 2, 12 22, 32, 42, 52,
possible values of a0 = 0, 10, 20, 30, 40, 50
also by keeping in mint that last 2 digit of 3479-a2^2 or 3479-a0^2, 3479-a8^2 should also be a square
3479-a0^2 = last 2 digits 79 so can not be square
3479- a2^2 = last 2 digits => 75, 35, 95, 55, 35, 75, so can not be square
3479- a8^2 = last 2 digits = 15, 55, 95, 35, 75 so can not be squar
so no value exists.
@jaspunit said:find remainder =?(2010^2011 + 2011^2012 +2012^2013)/(2010.2011.2012)
doesn't look like a competitive exam question.
Anyways, do you have options?
:neutral:
@Dexian said:@chillfactor(remainder when a is divided by x*y)*(remainder when b is divided by x*y) yahi kiya hai.......... fir kahan galat ho raha hai...=(2012^2013+2013^2012)/(2012*2013)=(2012^2013)/2012*2013+(2013^2012)/2012*2013=((2012)^2012 / 2013) +) (2013)^2011 / 2012)= ((-1)^2012 + (1) ^2011)=2
If remainder is 'r' when x is divided by y, then remainder will be 'nr' when nx is divided by ny
Here 2012^2012 when divided by 2013 remainder will be 1
=> 2012^2013 when divided by 2012*2013 remainder will be 2012
Same way for the other part
Hence remainder will be 2013 + 2012 = 4025
Q. Two of the sides of a scalene triangle are 14 and 16. How many different integral values third side can take?
a) 20
b) 23
c) 25
d) 27
a) 20
b) 23
c) 25
d) 27
@jaspunit said:tan6 tan 42 tan 66 tan78 = ?
can be written as
sin6*sin42*sin66*sin78/cos6*cos42*cos66*cos78
Numerator=1/16
Denominator=-1/16
FV:-1
@vijay_chandola
@vijay_chandola said:Q.Two of the sides of a scalene triangle are 14 and 16. How many different integral values third side can take?a) 20b) 23c) 25d) 27
d??
@vijay_chandola said:Q.Two of the sides of a scalene triangle are 14 and 16. How many different integral values third side can take?a) 20b) 23c) 25d) 27
D
@vijay_chandola said:Q.Two of the sides of a scalene triangle are 14 and 16. How many different integral values third side can take?a) 20b) 23c) 25d) 27
since scalene, the 3rd side, let it be x, so x2 ... so x can take 3,4... 29 except 14,16 and hence 25 values...