what is the smallest number which whwn increased by 6 is divisible by 36, 63 & 108?
@kumoh02 said:what is the smallest number which whwn increased by 6 is divisible by 36, 63 & 108?
LCM(36, 63, 108) - 6 => 756 - 6 = 750 ?
@kumoh02 said:what is the smallest number which whwn increased by 6 is divisible by 36, 63 & 108?
750?
x + 6 = LCM(36,63,108) = 756
x = 750
x + 6 = LCM(36,63,108) = 756
x = 750
750 (756-6)
@kumoh02 said:what is the smallest number which whwn increased by 6 is divisible by 36, 63 & 108?
750?
@kumoh02 LCM of 36, 63 and 108 is
36 = 9 × 4
63 = 9 × 7
108 = 36 × 3 = 9 × 4 × 3
LCM = 9 × 7 × 4 × 3
= 756
so required number = 756 – 6 = 750
36 = 9 × 4
63 = 9 × 7
108 = 36 × 3 = 9 × 4 × 3
LCM = 9 × 7 × 4 × 3
= 756
so required number = 756 – 6 = 750
@kumoh02 said:what is the smallest number which whwn increased by 6 is divisible by 36, 63 & 108?
LCM of (38,63,108) -6 =750
@grkkrg said:again from figure this isn't the right approach though..
Usually the approximation and assumption approaches are better than the correct ones... 
I will try to figure out the 90 degree.... 
I will try to figure out the 90 degree.... Find out the remainder when 123 ^ 300 is divided by 37
Please do this without the pattern / cyclicity approach
also teach the approach in which you do
thanks ,,..
@deepaknrn said:Find out the remainder when 123 ^ 300 is divided by 37 Please do this without the pattern / cyclicity approach also teach the approach in which you do thanks ,,..
26?
123^300 = 12^300 mod 37
36*8 = 288
12^12 mod 37 = 144^6 mod 37
= 33^6 mod 37
= (-4)^6 mod 37
= (256 * 16) mod 37
= (-3 * -21) mod 37
= 63 mod 37
= 26
123^300 = 12^300 mod 37
36*8 = 288
12^12 mod 37 = 144^6 mod 37
= 33^6 mod 37
= (-4)^6 mod 37
= (256 * 16) mod 37
= (-3 * -21) mod 37
= 63 mod 37
= 26
@deepaknrn said:Find out the remainder when 123 ^ 300 is divided by 37 Please do this without the pattern / cyclicity approach also teach the approach in which you do thanks ,,..
I am using Euler's method, if you are not aware, then google it, it is a very famous method
123^300 mod 37 = ( 111 + 12 )^300 mod 37 = 12^300 mod 37
Now E(37) = 36 => 12^36 mod 37 = 1
So, 12^300 mod 37 = 12^12 mod 37 = (12^2)^6 mod 37 = 144^6 mod 37 = 33^6 mod 37
= 4^6 mod 37 = 2^12 mod 37 = 4096 mod 37 = 26 ?
@deepaknrn said:Find out the remainder when 123 ^ 300 is divided by 37 Please do this without the pattern / cyclicity approach also teach the approach in which you do thanks ,,..
26??
The mean propotion between two numbers is 12. The 3rd propotion of the same numbers is 96. Find the greater of the two?
@gupanki2 said:The mean propotion between two numbers is 12. The 3rd propotion of the same numbers is 96. Find the greater of the two?
Let the numbers be a, b
Mean proportional = 12 => a*b = 12^2 = 144
Third proportional = 96 => a/b = b/96 => b^2 = 96*a
Combing both -> b^3 = 96*144 = 24^3 => b = 24 , a = 6
So, greater of the two is 24 ?
The value of a diamond varies direclty with the square of its weight. A diamond broke into 3 pieces whose weights are in ratio 32:24:9. The loss caused due to the breakage was Rs 25.44 lakh. Find inital value of diamond ( in lakhs)?
@gupanki2 said:The value of a diamond varies direclty with the square of its weight. A diamond broke into 3 pieces whose weights are in ratio 32:24:9. The loss caused due to the breakage was Rs 25.44 lakh. Find inital value of diamond ( in lakhs)?
P=k(65x)^2 ; p1= k(32x)^2 ; p2= k(24x)^2 ; p3=k(9x)^2
k= P/(65x)^2
put value of k in p1,p2,p3
p1= P (32/65)^2
p1= P (24/65)^2
p1= P (9/65)^2
acc to ques ::
P - {P (32/65)^2 + P (24/65)^2+ P (9/65)^2} =2544000
on solving from here we get
P ~ 42.25 lakkh
@gupanki2 said:The value of a diamond varies direclty with the square of its weight. A diamond broke into 3 pieces whose weights are in ratio 32:24:9. The loss caused due to the breakage was Rs 25.44 lakh. Find inital value of diamond ( in lakhs)?
antodya bhai__/\__
42.25
42.25