Sorry rightmost digit should be zero in base 5 ..abhi dekha 
@krum sir
@tomdickharry sir
@Aizen sir
Do integer solutions exist for type ax+by=c and ax-by=c ?
1. When will they have a unique solution ?
2. When will they have no solution or how to determine if a solution exist?
3. How to determine 1 solution ?
4. How to determine a positive solution?
5. How to determine a general solution ?
I know the answer only for the 2nd point. Could you pls help with the others ?
HOW TO DETERMINE IF A SOLUTION EXIST ?
If HCF(a,b)=q and if q divides "c"
or
when ax+by=1 where HCF(a,b)=1
when ax-by=1 where HCF(a,b)=1
or
when ax+by=d where HCF(a,b)=d
when ax-by=d where HCF(a,b)=d
Then there most certainly exists a solution for x and y.
NO SOLUTION: eg 3x-12y=7
1. If ax+by=1 or ax-by=1 and HCF(a,b) is not 1
2. If ax+by=q or ax-by=q and HCF(a,b) is not q
3. If ax+by=c or ax-by=c and HCF(a,b) =d and "d" doesn't divide "c"
Case 1: ax+by=c eg: 56x+72y=40
Case 2: ax-by=c eg: 1990x-173y=11
I'm not sure of the rest. Can you please post some gyan on this !!!
@tomdickharry sir
@Aizen sir
Do integer solutions exist for type ax+by=c and ax-by=c ?
1. When will they have a unique solution ?
2. When will they have no solution or how to determine if a solution exist?
3. How to determine 1 solution ?
4. How to determine a positive solution?
5. How to determine a general solution ?
I know the answer only for the 2nd point. Could you pls help with the others ?
HOW TO DETERMINE IF A SOLUTION EXIST ?
If HCF(a,b)=q and if q divides "c"
or
when ax+by=1 where HCF(a,b)=1
when ax-by=1 where HCF(a,b)=1
or
when ax+by=d where HCF(a,b)=d
when ax-by=d where HCF(a,b)=d
Then there most certainly exists a solution for x and y.
NO SOLUTION: eg 3x-12y=7
1. If ax+by=1 or ax-by=1 and HCF(a,b) is not 1
2. If ax+by=q or ax-by=q and HCF(a,b) is not q
3. If ax+by=c or ax-by=c and HCF(a,b) =d and "d" doesn't divide "c"
Case 1: ax+by=c eg: 56x+72y=40
Case 2: ax-by=c eg: 1990x-173y=11
I'm not sure of the rest. Can you please post some gyan on this !!!
@ScareCrow28 said:bhai btao kaise kia.. mjse ni hila
Sides ko extend karo to equilateral triangle ban jaega :)
@deedeedudu said:Sides ko extend karo to equilateral triangle ban jaega
yup..POONAM(pradeep) PANDEY QUANT CHALLENGE se he :P
@anytomdickandhary said:in base of 5 right most digit is 0, this implies that the number is a multiple of 5.left most digit is 1 in base 8 => N left most digit is 1 in base 11 => N >= 1*11^2 + 0*11+0 = 121based on the 3 conditions above we can say N = 125=>remainder when 125 divided by 7 is 6 ATDH.
sir isme base 11 wali condtn yahi cnfrm karti hai na ki no.is greater than 121..
and base 8 wali condtn tells k max jo ho skta hai 127 ho skta hai??
@rkshtsurana said:ABCDE is an irregular polygon such that ∠A = ∠B = ∠C = ∠D = 120° and ∠E = 60° . If AB = 6 cm, BC = 5 cm and CD = 4 cm, find the perimeter of ABCDE.1) 35 cm 2) 37 cm 3) 29 cm 4) 31 cm 5) Can't be determined
1) 35 ?
Dropped perpendiculars from vertices and calculated the sum and differences of the components of the sides.. Remaining sides come out be 9 and 11
@gautham87 said:@krum sir@tomdickharry sir@Aizen sirDo integer solutions exist for type ax+by=c and ax-by=c ?1. When will they have a unique solution ?2. When will they have no solution or how to determine if a solution exist?3. How to determine 1 solution ?4. How to determine a positive solution?5. How to determine a general solution ?I know the answer only for the 2nd point. Could you pls help with the others ?HOW TO DETERMINE IF A SOLUTION EXIST ?If HCF(a,b)=q and if q divides "c" orwhen ax+by=1 where HCF(a,b)=1when ax-by=1 where HCF(a,b)=1orwhen ax+by=d where HCF(a,b)=dwhen ax-by=d where HCF(a,b)=dThen there most certainly exists a solution for x and y.NO SOLUTION: eg 3x-12y=71. If ax+by=1 or ax-by=1 and HCF(a,b) is not 12. If ax+by=q or ax-by=q and HCF(a,b) is not q3. If ax+by=c or ax-by=c and HCF(a,b) =d and "d" doesn't divide "c"Case 1: ax+by=c eg: 56x+72y=40Case 2: ax-by=c eg: 1990x-173y=11I'm not sure of the rest. Can you please post some gyan on this !!!
We can use cramer's rule for finding the solution.. Check this out..I hope this is what you were asking for..
@nick_baba said:An integer N, 100
base 8 => 144 - 200 =>100 - 128 base 10
base 11 => 100 - 172 =>121 - 200 base 10
between 121 and 128 only 125 is divisible by 5
125mod7=6
base 11 => 100 - 172 =>121 - 200 base 10
between 121 and 128 only 125 is divisible by 5
125mod7=6
@deedeedudu said:35
@deedeedudu said:Sides ko extend karo to equilateral triangle ban jaega
sahi logic lagaye ho bhai...:)
@gautham87 said:@krum sir@tomdickharry sir@Aizen sirDo integer solutions exist for type ax+by=c and ax-by=c ?1. When will they have a unique solution ?2. When will they have no solution or how to determine if a solution exist?3. How to determine 1 solution ?4. How to determine a positive solution?5. How to determine a general solution ?I know the answer only for the 2nd point. Could you pls help with the others ?HOW TO DETERMINE IF A SOLUTION EXIST ?If HCF(a,b)=q and if q divides "c" orwhen ax+by=1 where HCF(a,b)=1when ax-by=1 where HCF(a,b)=1orwhen ax+by=d where HCF(a,b)=dwhen ax-by=d where HCF(a,b)=dThen there most certainly exists a solution for x and y.NO SOLUTION: eg 3x-12y=71. If ax+by=1 or ax-by=1 and HCF(a,b) is not 12. If ax+by=q or ax-by=q and HCF(a,b) is not q3. If ax+by=c or ax-by=c and HCF(a,b) =d and "d" doesn't divide "c"Case 1: ax+by=c eg: 56x+72y=40Case 2: ax-by=c eg: 1990x-173y=11I'm not sure of the rest. Can you please post some gyan on this !!!
@krum said:base 8 => 144 - 200 =>100 - 128 base 10base 11 => 100 - 172 =>121 - 200 base 10between 121 and 128 only 125 is divisible by 5125mod7=6
ek no. dost..bas 128 ki jagah 127 hoga...base 8 me 128 nai ho skta na..
@nick_baba said:sir isme base 11 wali condtn yahi cnfrm karti hai na ki no.is greater than 121..and base 8 wali condtn tells k max jo ho skta hai 127 ho skta hai??
yes friend that is what it confirms.
ATDH.
@ScareCrow28 The first link is cramers rule which is solving a 2 equations with 2 unknowns and 3 equations with 3 unknowns... I am not sure if this can be used.. as we have 1 equation in 2 unknowns or sometimes 1 equations in 3 unknowns ! They are called as Linear diophantine equations.. there is heavy theory out there in the net.. but i went bonkers reading them 😞 
The number of rational points x = p/5 satisfying log(2x-3/4)/logx > 2, where p is an integer and GCD(p, 5) = 1, is/are??
@gautam22 said:triangle ABC ....AB=AC...BAC=20 degree.....D is a point on AC and BC=AD .Find /_DBC
50
@nick_baba said:The number of rational points x = p/5 satisfying log(2x-3/4)/logx > 2, where p is an integer and GCD(p, 5) = 1, is/are??
2x - 3/4 >0 => x>3/8 and (x not equal to 1)
Assuming the base of the logarithm to be 10,
Case 1: log(x) > 0 => x > 1
2x-3/4 > x^2 => 4x^2 - 8x + 3 (2x-3)(2x -1)
=> x > 1/2 and x less than 3/2
x = p/5 , p is an integer
=> p > 5/2 and less than 15/2 => p = 6,7
Case 2: log(x) x
(2x - 3/4)
x less than 1/2 => p p = 2
PS: Calculation mistakes could be there :|
Edit: Rookie mistake.. while multiplying by log(x) didn't consider its sign :banghead:
@ScareCrow28 said:Ab method batao.. OA matches with your answer
take a, d, g as 1/3 and the rest as 0. thats the minimum possible!!