Let S denote the sum of three natural numbers. Let D denote the sum of the products of them taken two at a time. Let P denote their product. If 3S + D = 2P, how many combinations are there for the numbers?
a)3
b)5
c)6
d)7
ABCD is trapezoid,BC is parallel to AD.BC=18,AD=36,AB=7,CD=55.What would area of trapezoid?
@Torque024 said:Let S denote the sum of three natural numbers. Let D denote the sum of the products of them taken two at a time. Let P denote their product. If 3S + D = 2P, how many combinations are there for the numbers?a)3b)5 c)6 d)7
3
@Torque024 said:Let S denote the sum of three natural numbers. Let D denote the sum of the products of them taken two at a time. Let P denote their product. If 3S + D = 2P, how many combinations are there for the numbers?a)3b)5 c)6 d)7
3?
3(a+b+c) + ab+bc+ca = 2(abc)
Divide both sides by abc
a=b=c=3 comes out as 1 solution
Then we assume any one value on the LHS to be 1 and 2 since all the values on the LHS cannot be greater than 3 (value on RHS = 2 )
We get 3 solutions possible
Had read this solution somewhere..cant recollect where..but this is the way to go about it..
@gnehagarg said:ABCD is trapezoid,BC is parallel to AD.BC=18,AD=36,AB=7,CD=55.What would area of trapezoid?
Question correct hai?
@gnehagarg said:ABCD is trapezoid,BC is parallel to AD.BC=18,AD=36,AB=7,CD=55.What would area of trapezoid?
Trapezoid not possible.
@soumitrabengeri said:3?3(a+b+c) + ab+bc+ca = 2(abc)Divide both sides by abca=b=c=3 comes out as 1 solutionThen we assume any one value on the LHS to be 1 and 2 since all the values on the LHS cannot be greater than 3 (value on RHS = 2 )We get 3 solutions possibleHad read this solution somewhere..cant recollect where..but this is the way to go about it..
I proceeded this way.
Let numbers be n-1,n,n+1
after conditions we will get a cubic equation in n
2n^3-3n^2-11n+1=0; How to solve this fast?
Number of values of n=3; n may or may not be a natural number
So max possible combinations=3; If any option were
after conditions we will get a cubic equation in n
2n^3-3n^2-11n+1=0; How to solve this fast?
Number of values of n=3; n may or may not be a natural number
So max possible combinations=3; If any option were
@Torque024 said:I proceeded this way.Let numbers be n-1,n,n+1after conditions we will get a cubic equation in n2n^3-3n^2-11n+1=0; How to solve this fast?Number of values of n=3; n may or may not be a natural numberSo max possible combinations=3; If any option were
In this way you are assuming consecutive numbers..and as you said, if the options were
I do not know any fast way except the method i posted.
@Torque024 said:Let S denote the sum of three natural numbers. Let D denote the sum of the products of them taken two at a time. Let P denote their product. If 3S + D = 2P, how many combinations are there for the numbers?a)3b)5 c)6 d)7
3(x + y + z) + (xy + yz + zx) = 2xyz
3(1/xy + 1/yz + 1/zx) + (1/x + 1/y + 1/z) = 2
Without the loss of generality,
x ≥ y ≥ z
So, 3(3/z^2) + 3/z ≥ 2
2z^2 - 3z - 9 ≤ 0
(2z + 3)(z - 3) ≤ 0
So, z ≤ 3
When z = 1
4(x + y) + xy + 3 = 2xy
(x - 4)(y - 4) = 19
(x, y) = (5, 23)
When z = 2
5(x + y) + xy + 6 = 4xy
9xy - 15(x + y) = 18
(3x - 5)(3y - 5) = 43
(x, y) = (2, 12)
When z = 3
6(x + y) + xy + 9 = 6xy
25xy - 30(x + y) = 45
(5x - 6)(5y - 6) = 81
(x, y) = (3, 3)
So, three solutions
@Torque024 said:Let S denote the sum of three natural numbers. Let D denote the sum of the products of them taken two at a time. Let P denote their product. If 3S + D = 2P, how many combinations are there for the numbers?a)3b)5 c)6 d)7
3(x+y+z)+(xy+yz+xz)=2xyz
Divide by xyz
3(1/yz+1/xz+1/xy)+(1/z+1/x+1/y)=2
put x=1
3(1/yz+1/z+1/y)+1+1/z+1/y=2
3/yz+4/z+4/y =1
Multiply by yz
3+4y+4z=yz
yz-4y-4z=3
(y-4)*(z-4)=19=1*19 or 19*1
y=5,z=23
put x=2 in above equation
3(1/yz+1/2*z+1/2*y)+1/z+1/y+1/2=2
3/yz+5/2*z+5/2*y=3/2
multiply by 2yz
6+5y+5z=3yz
multiply by 3
9yz-15y-15z=18
(3y-5)*(3z-5)=43(1*43)
y=2,z=16
(1,5,23)
(2,2,16)
(3,3,3)
x=3
5yz-6y-6z=9
25yz-30y-30z=45
(5y-6)*(5z-6)=81
y=3,z=3
Sorry,calculation mistake previously.
The number of integers n for which the value of log(base10)[n(10-n)/16] becomes negative is
Easy question, but conceptual.
Easy question, but conceptual.
@Torque024 said:The number of integers n for which the value of log(base10)[n(10-n)/16] becomes negative isEasy question, but conceptual.
n(10-n) / 16
n(10 - n)
- n^2 + 10n - 16
n^2 - 10n + 16 > 0
(n-8)(n-2) >0
(-inf , 2) U ( 8 , inf)
also
n( 10 - n) > 0
n( n - 10 )
(0,10)
(0,2) U (8,10)
@Torque024 said:The number of integers n for which the value of log(base10)[n(10-n)/16] becomes negative isEasy question, but conceptual.
0
0
=> 0
and n(10 - n)/16
n^2 - 10n + 16 > 0
n 8
So, (0, 2) U (8, 10)
@Torque024 said:The number of integers n for which the value of log(base10)[n(10-n)/16] becomes negative isEasy question, but conceptual.
1 and 9
?
361 pages 
I will be regular now on this thread
X and Y are playing a game. There are 11 coins on the table and each player must pick up at least 1 coin but not more than 5 coins. The person picking up the last coin loses. X starts. How many coins should he pick up to ensure a win no matter what strategy Y employs.
A. 4
B. 3
C. 2
D. 5
What is correct approach to solve such problems???
@mohnish_khiani said:X and Y are playing a game. There are 11 coins on the table and each player must pick up at least 1 coin but not more than 5 coins. The person picking up the last coin loses. X starts. How many coins should he pick up to ensure a win no matter what strategy Y employs.A. 4B. 3C. 2D. 5What is correct approach to solve such problems???
4
@mohnish_khiani said:X and Y are playing a game. There are 11 coins on the table and each player must pick up at least 1 coin but not more than 5 coins. The person picking up the last coin loses. X starts. How many coins should he pick up to ensure a win no matter what strategy Y employs.A. 4B. 3C. 2D. 5What is correct approach to solve such problems???
Always try to find a pattern in these kind of question.
If there is just 1 coin on the table, then the person whose turn it is will loose.
In case of 2, 3, 4, 5, 6 coins, person whose turn it is can pick 1, 2, 3, 4, 5 coins and then win the game.
In case of 7 coins no matter what a person does, he will always loose (as he will leave 2 or 3 or .. or 6 coins for the other person)
Then for 8 or 9 or 10 or ... or 12, person whose turn it is will always win (as he can leave 7 for the other person)
Now, here we have 11 coins on the table. So, if X can manage to leave 7 coins for Y, then he is certain of winning the game
Hence, X will pick 4 coins to win the game
How many numbers n less than 1000 are there such that number of even factors of n is 3 times the number of odd factors of n.