Find the remainder when 777^777 is divided by 1000
@shattereddream said:Find the remainder when 777^777 is divided by 1000
1000 = 8*125
N = 777^777
Using Euler's theorem,
E(8) = 4
E(125) = 100
777 mod 8 = 1 => 777^777 mod 8 = 1 => N = 8k + 1
777^777 mod 125 = 27^777 mod 125 = 27^77 mod 125 = 3^231 mod 125 = 3^31 mod 125
= 3* (243)^6 mod 125 = 3*(-7)^6 mod 125 = 3* 343 * 343 mod 125 = 3*(-1) mod 125
=> N mod 125 = -3 => N = 125m - 3
So, 8k + 1 = 125m - 3
8k + 4 = 125m
Dividing both the sides by 8 => 4 = (5)*(m mod 8) => m = 8a + 4 type
So, N = 125*(8a + 4) - 3 = 1000a + 500 - 3 = 1000a + 497
So, N mod 1000 = 497 ?
There might be some calculation mistakes 😐
@shattereddream said:Find the remainder when 777^777 is divided by 1000
@YouMadFellow said:1000 = 8*125N = 777^777Using Euler's theorem,E(8) = 4E(125) = 100777 mod 8 = 1 => 777^777 mod 8 = 1 => N = 8k + 1777^777 mod 125 = 27^777 mod 125 = 27^77 mod 125 = 3^231 mod 125 = 3^31 mod 125= 3* (243)^6 mod 125 = 3*(-7)^6 mod 125 = 3* 343 * 343 mod 125 =3*(-1)mod 125
There might be some calculation mistakes
Striked part should be 3*24 (mod 125) or 72 (mod 125)
N = 8k + 1 = 125n + 72
8k = 120n + 72 + (5n - 1)
so, n = 8a + 5
N = 1000a + 697
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A person lent out some money for 1year at 6% per annum simple interest and after 18months he again lent out the same money at a simple interest of 24% per annum. In both cases he got Rs.4704. What was the amount that he lent out if interest is paid half yearly?
a rs 4000 b rs 4200 c rs 4400 d rs 3600
There are 2 different quantities of oil,with the masses differing by 2 kg.The same amount of heat,equal to 96kcal was imparted to each mass,and the larger oil of mass was found to be 4 degrees cooler than the smaller mass.Find the mass of oil in each of the two quantities.
solution is 96/x-96/(x+2)=4. i.e x=6 and 8. i.e they have mass of 6 and 8 kg.
It means we are comparing the heat(96kcal) applied to 1/6 th of a 6kg substance with heat applied to (1/8th) of 8kg i.e 96(1/6)-96(1/8). But i don't get the equation because wherever i touch it should have same temperature .I mean if i compare 1% of 6kg with 1% of 8 kg ,or say 32% of 6kg with 32% of 8 kg,the heat difference should remain same right? But then it violates the same condition given in the question that temperature varies with mass.So i am confused.
Also in the solution they are comparing 1/6th of something with 1/8th of something.Shouldn't they compare the same percentage amount?
solution is 96/x-96/(x+2)=4. i.e x=6 and 8. i.e they have mass of 6 and 8 kg.
It means we are comparing the heat(96kcal) applied to 1/6 th of a 6kg substance with heat applied to (1/8th) of 8kg i.e 96(1/6)-96(1/8). But i don't get the equation because wherever i touch it should have same temperature .I mean if i compare 1% of 6kg with 1% of 8 kg ,or say 32% of 6kg with 32% of 8 kg,the heat difference should remain same right? But then it violates the same condition given in the question that temperature varies with mass.So i am confused.
Also in the solution they are comparing 1/6th of something with 1/8th of something.Shouldn't they compare the same percentage amount?
@nole said:There are 2 different quantities of oil,with the masses differing by 2 kg.The same amount of heat,equal to 96kcal was imparted to each mass,and the larger oil of mass was found to be 4 degrees cooler than the smaller mass.Find the mass of oil in each of the two quantities.
6kg and 8kg
Let m and (m+2) be the masses of the two quantities of oil.
(m+2)(t-4) = 96
=> t - 4 = 96/(m+2)
(m)(t) = 96
=> t = 96/m
96/m - 4 = 96/(m+2)
96/m - 96/(m+2) = 4
192 = 4 * (m^2 + 2m)
m^2 + 2m - 48 = 0
(m+8)(m-6) = 0
m = -8 or m = 6
So the masses are 6kg and 8kg.
Here you can see that the temperature increase is lesser for the larger mass for the same amount of heat.
So the mass and temperature increase are inversely proportional which makes sense logically.
So mass * temperature increase = constant
The constant here is the heat imparted to both the quantities.
PS: No physics involved.
Let m and (m+2) be the masses of the two quantities of oil.
(m+2)(t-4) = 96
=> t - 4 = 96/(m+2)
(m)(t) = 96
=> t = 96/m
96/m - 4 = 96/(m+2)
96/m - 96/(m+2) = 4
192 = 4 * (m^2 + 2m)
m^2 + 2m - 48 = 0
(m+8)(m-6) = 0
m = -8 or m = 6
So the masses are 6kg and 8kg.
Here you can see that the temperature increase is lesser for the larger mass for the same amount of heat.
So the mass and temperature increase are inversely proportional which makes sense logically.
So mass * temperature increase = constant
The constant here is the heat imparted to both the quantities.
PS: No physics involved.
Subscribed!!!
@nole said:There are 2 different quantities of oil,with the masses differing by 2 kg.The same amount of heat,equal to 96kcal was imparted to each mass,and the larger oil of mass was found to be 4 degrees cooler than the smaller mass.Find the mass of oil in each of the two quantities.solution is 96/x-96/(x+2)=4. i.e x=6 and 8. i.e they have mass of 6 and 8 kg.It means we are comparing the heat(96kcal) applied to 1/6 th of a 6kg substance with heat applied to (1/8th) of 8kg i.e 96(1/6)-96(1/8). But i don't get the equation because wherever i touch it should have same temperature .I mean if i compare 1% of 6kg with 1% of 8 kg ,or say 32% of 6kg with 32% of 8 kg,the heat difference should remain same right? But then it violates the same condition given in the question that temperature varies with mass.So i am confused. Also in the solution they are comparing 1/6th of something with 1/8th of something.Shouldn't they compare the same percentage amount?
Q(heat) = m(mass)*c(specific heat of oil)*Dt(change in temp)
Let both oil be at same temperature
let mass of one be x and let its temperature raised by y
96=x*c*y-------1,
96=(x-2)*c*(y+4)----------2
{We should convert kcal to joules for calculating c but, kcal will not affect values of x and y}
solve
put y from 1 to 2
96=(xc-2c)((96/xc) +4))
x=(c+-sqrt(c(c+48)))/c
integral solutions, which we shouldn't assume :/ so ambiguous question
c=48, x=1, y=-2, not possible as temperature will raise
c-=1, x=-6 not possible as mass can't be negative
c=1, x=8, y=12, possible, so masses will be 8 and 6
c=2, x=-4, y=-12, not possible as temperature will raise
Let both oil be at same temperature
let mass of one be x and let its temperature raised by y
96=x*c*y-------1,
96=(x-2)*c*(y+4)----------2
{We should convert kcal to joules for calculating c but, kcal will not affect values of x and y}
solve
put y from 1 to 2
96=(xc-2c)((96/xc) +4))
x=(c+-sqrt(c(c+48)))/c
integral solutions, which we shouldn't assume :/ so ambiguous question
c=48, x=1, y=-2, not possible as temperature will raise
c-=1, x=-6 not possible as mass can't be negative
c=1, x=8, y=12, possible, so masses will be 8 and 6
c=2, x=-4, y=-12, not possible as temperature will raise
Doodhnath, a milkman, has 100 litres of milk and water solution that contains 70% milk. What quantity of water (in litres) should he add to the solution to bring down the concentration of milk to 40%?
@Cat.Aspirant123 said:Doodhnath, a milkman, has 100 litres of milk and water solution that contains 70% milk. What quantity of water (in litres) should he add to the solution to bring down the concentration of milk to 40%?
75 liters?
Currently solution contains---> 70 liters milk and 30 liters water
40% of 175 = 70
So quantity of water to be added = 75
Use Alligation Principle :
100 X
0.7 0
.....0.4
0.4 0.3
100/X =4/3
X=75
@Cat.Aspirant123 said:Doodhnath, a milkman, has 100 litres of milk and water solution that contains 70% milk. What quantity of water (in litres) should he add to the solution to bring down the concentration of milk to 40%?
75 liters?
70/(100+x) = 0.4
70 = 40 + 0.4x
0.4x = 30
x = 75
70/(100+x) = 0.4
70 = 40 + 0.4x
0.4x = 30
x = 75
@Cat.Aspirant123 said:Doodhnath, a milkman, has 100 litres of milk and water solution that contains 70% milk. What quantity of water (in litres) should he add to the solution to bring down the concentration of milk to 40%?
70/(30+x) = 2/3
x = 75
@Cat.Aspirant123 said:Doodhnath, a milkman, has 100 litres of milk and water solution that contains 70% milk. What quantity of water (in litres) should he add to the solution to bring down the concentration of milk to 40%?
70/(100+x) =.4, x=75
X is a three-digit positive integer in which each digit is either 1 or 2. Y has the same digits as X, but in reverse order. What is the remainder when X is divided by 3?
(1) The hundreds digit of XY is 6.
(2) The tens digit of XY is 4.
(1) The hundreds digit of XY is 6.
(2) The tens digit of XY is 4.
a) using 1 alone
b) using 2 alone
c) using both
d) either
e) CBD
@bullseyes said:X is a three-digit positive integer in which each digit is either 1 or 2. Y has the same digits as X, but in reverse order. What is the remainder when X is divided by 3?(1) The hundreds digit of XY is 6.(2) The tens digit of XY is 4.a) using 1 aloneb) using 2 alonec) using bothd) eithere) CBD
CBD?
@bullseyes said:X is a three-digit positive integer in which each digit is either 1 or 2. Y has the same digits as X, but in reverse order. What is the remainder when X is divided by 3?(1) The hundreds digit of XY is 6.(2) The tens digit of XY is 4.a) using 1 aloneb) using 2 alonec) using bothd) eithere) CBD
using either?