@amresh_maverick said:Remainder when 2^1990 is divided by 1990
2*5*199
2k
5p-1
2^10%199 = 199z+29
so 1024...
@amresh_maverick said:Remainder when 2^1990 is divided by 1990
2^1990 = 2^10 * 2^1980mod 1990 =1024.
@saurav205 said:Bhai approach pls..N i thought you were sleep..
So did I.
I can't read my handwriting, kuch to scribble kiya papge pe. Maine kuch to split kiya denominator ko as 199 * 10.
Then 2^1990/10 ka remainder = last digit = 4. So answer must end in 4 (isliye options manga tha)
Also 2^1990/199 = (2^199)^10/199. Now I think 2^199/199 should give 2 so this would remainder of 2^10/199 = 29. (Note to self: need to revise Fermat's little thm)
So we want a number of form 199k+29 and 10m+4 and I got 1024 as first such. Not sure whether that makes sense...
regards
scrabbler
I can't read my handwriting, kuch to scribble kiya papge pe. Maine kuch to split kiya denominator ko as 199 * 10.
Then 2^1990/10 ka remainder = last digit = 4. So answer must end in 4 (isliye options manga tha)
Also 2^1990/199 = (2^199)^10/199. Now I think 2^199/199 should give 2 so this would remainder of 2^10/199 = 29. (Note to self: need to revise Fermat's little thm)
So we want a number of form 199k+29 and 10m+4 and I got 1024 as first such. Not sure whether that makes sense...
regards
scrabbler
@scrabbler said:So did I.I can't read my handwriting, kuch to scribble kiya papge pe. Maine kuch to split kiya denominator ko as 199 * 10.Then 2^1990/10 ka remainder = last digit = 4. So answer must end in 4 (isliye options manga tha)Also 2^1990/199 = (2^199)^10/199. Now I think 2^199/199 should give 2 so this would remainder of 2^10/199 = 29. (Note to self: need to revise Fermat's little thm)So we want a number of form 199k+29 and 10m+4 and I got 1024 as first such. Not sure whether that makes sense...regardsscrabbler
Not much ...will check tomorrow morning now..
Too sleepy to think..
@Logrhythm said:2*5*1992k5p-12^10%199 = 199z+29 so 1024...
bhai kaise kiya fermat's little theorem ya chinese remainder theorem?
@bs0409 said:OA-2...........If x/y + y/z +z/x = 1Find the value of (x^3 + y^3 + z^3)/xyz?
ans bata do bhai
@tukka_king said:bhai kaise kiya fermat's little theorem ya chinese remainder theorem?
tukks bhai...first of all, welcome to the qa thread... :D
yaar mujhe concepts ke naam nahi pata hote...aese karke bata deta hun..
1990 = 2*5*199
2 se divisible hai so 2k form ka hoga remainder
(2^2)^995 when divided by 5 gives remainder as 4 or -1
euler's of 199 is 198
1990%198 = 10
so 2^10%199 = 29
now remainder is smallest number of the form -> 2k = 5p+4 = 199z+29
which is 1024... 😃
@Logrhythm said:tukks bhai...first of all, welcome to the qa thread... yaar mujhe concepts ke naam nahi pata hote...aese karke bata deta hun..1990 = 2*5*1992 se divisible hai so 2k form ka hoga remainder (2^2)^995 when divided by 5 gives remainder as 4 or -1euler's of 199 is 198 1990%198 = 10so 2^10%199 = 29now remainder is smallest number of the form -> 2k = 5p+4 = 199z+29which is 1024...
thanx bhai ab se fllow kartey jaunga ye thread ....thoda gyaan mil raha hai.....
@Logrhythm said:tukks bhai...first of all, welcome to the qa thread... yaar mujhe concepts ke naam nahi pata hote...aese karke bata deta hun..1990 = 2*5*1992 se divisible hai so 2k form ka hoga remainder (2^2)^995 when divided by 5 gives remainder as 4 or -1euler's of 199 is 198 1990%198 = 10so 2^10%199 = 29now remainder is smallest number of the form -> 2k = 5p+4 = 199z+29which is 1024...
thanx bhai ab se follow kartey jaunga ye thread .....thoda gyaan mil raha hai ....
@bs0409 said:OA-2...........If x/y + y/z +z/x = 1Find the value of (x^3 + y^3 + z^3)/xyz?
I have a doubt, the least value of x/y +y/z +z/x is 3, by AM>=GM. so how can it be 1?
@karl said:I have a doubt, the least value of x/y +y/z +z/x is 3, by AM>=GM. so how can it be 1?
That is if x, y, z all positive. Clearly can't be the case here...:( But aage I have no idea what to do. Algebra :P
regards
scrabbler
regards
scrabbler
@karl said:I have a doubt, the least value of x/y +y/z +z/x is 3, by AM>=GM. so how can it be 1?
it has 2 b 3 ..1 kahan se aaya??
@karl said:I have a doubt, the least value of x/y +y/z +z/x is 3, by AM>=GM. so how can it be 1?
wohi meine dekha tha...but fir shayad imaginary values lena ka koi funda vunda ho....meine kaha skip kar deta hun...waise hi kamzor hun thoda dimaag mein... 
@scrabbler - ye wala ques dekhna.....agar aap so na gaye ho toh..
@amresh_maverick said:I am also getting 3 Approach?. I took x=y=z=k as no other condition is given and got 3
if you take x=y=z then toh the question seems wrong sir... how can x/y+y/z+z/x=1 then? it has to be 3. 
.. explain thoda pliss...
.. explain thoda pliss... @koyal1990 said:if you take x=y=z then toh the question seems wrong sir... how can x/y+y/z+z/x=1 then? it has to be 3. .. explain thoda pliss...
han wohi toh baat chal rahi hai....ki x/y+y/z+z/x has a min val of 3 (when all x, y and z are +ve) so 1 nahi ho sakta...abb 1 hone ke liye negative shegative kuch lena padega...chalo so jao gn.. :)
@Logrhythm said:han wohi toh baat chal rahi hai....ki x/y+y/z+z/x has a min val of 3 (when all x, y and z are +ve) so 1 nahi ho sakta...abb 1 hone ke liye negative shegative kuch lena padega...chalo so jao gn..
sahi mein main bhi kya soch raha hun 
, sona chiaye
, sona chiaye @amresh_maverick said:I am also getting 3 Approach?. I took x=y=z=k as no other condition is given and got 3
bro if u take x=y=z=k ,then question becomes (x/y + y/z + z/x) = 3
but its given its equal to 1.
Not consistent with the question. I dont think thats a valid assumption... :/
@bs0409 said:OA-2...........If x/y + y/z +z/x = 1Find the value of (x^3 + y^3 + z^3)/xyz?
Integers, or real numbers, "x,y,z". Real numbers pe i am getting more than 1 solution.
@bs0409 said:If p=abc and Q=uv are three digits and two digits natural numbers respectively,such that u and v must be distinct integers.How many pairs of P and Q are there in total which gives the same result when we multiply abc with uv as the product of cba with vu?a.2 b.8 c.5 d.cbd
OA = 5 ?
as per Question -
abc*uv = cab*vu
=> [100a+10b+c] [10u+v] = [100c+10b+a] [10v+u]
=> 111 [au-cv] = 10 [v(b-a) + u(c-b)] --------- (1)
From (1), We have -
[au-cv] must be a factor of 10 i.e 0 & 10. ----- (2)
from (2)
for 0,
=> b = (av - cu)/(v - u) ------ (3)
for 10,
=> b = (111+av-cu)/(v-u) ------------- (4)
Now, since b -> [0,9]
=> b
for 0,
=> (9-a)*v >= (9-c)*u
Will Do!!
for 10,
=> (9-a)*v - (9-c)*u >= 111
Not Feasible!!
v-u > 0 => v>u ----- (5)
[EDIT] -> This Condition is Wrong!!
Taking values for (3), we have -
[au-cv=0]
a*u - c*v, v-u, a*v-c*u, b
9*2 - 2*9, 7, 77, 11
9*1 - 1*9. 8, 80, 10
8*2 - 2*8
7*1 - 1*7, 6, 48, 8
6*3 - 3*6, 3, 27, 9
6*2 - 2*6, 4, 32, 8
6*1 - 1*6, 5, 35, 7
5*4 - 4*5, 1, 9, 9
5*3 - 3*5, 2, 16, 8
5*2 - 2*5, 3, 21, 7
5*1 - 1*5, 4, 24, 6
4*5 - 5*4, -1, -9, 9
4*3 - 3*4, 1, 7, 7
4*2 - 2*4, 2, 12, 6
4*1 - 1*4, 3, 15, 5
3*6 - 6*3, -3, -27, 9
3*5 - 5*3, -2, -16, 8
3*4 - 4*3, -1, -7, 7
3*2 - 2*3, 1, 5, 5
3*1 - 1*3, 2, 8, 4
2*7 - 7*2, -5, -45, 9
2*6 - 6*2, -4, -32, 8
2*5 - 5*2, -3, -21, 7
2*4 - 4*2, -2, -12, 6
2*3 - 3*2, -1, -5, 5
2*1 - 1*2, 1, 3, 3
1*8 - 8*1, -7, -63, 9
1*7 - 7*1, -6, -48, 8
1*6 - 6*1, -5, -35, 7
1*5 - 5*1, -4, -24, 6
..and so on till.. 5,4,
1*2 - 2*1, -1, -3, 3
@Logrhythm Ye dekho.. edit kar diya.. 32 aaye hain abhi.. I guess kuchh miss ho gye honge cases!!
@koyal1990 Please verify the answer, I may be wrong as it is 4:30 in the morning!! :P