Consider all quadruples of real numbers ( x 1 , x 2 , x 3 , x 4 ) , satisfying x 1 ≤ x 2 ≤ x 3 ≤ x 4 , such that
⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ x 1 + x 2 x 3 x 4 x 2 + x 3 x 4 x 1 x 3 + x 4 x 1 x 2 x 4 + x 1 x 2 x 3 = 1 3 0 = 1 3 0 = 1 3 0 = 1 3 0
If S is the sum of all possible values of x 4 , what is S , rounded to the nearest integer ?
Details and assumptions
Each possible distinct value of x 4 should be counted only once, even if it appears in several different quadruples.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Nice solution! Did anyone notice how close this is to an integer? This is really not a coincidence.
Nice solution! I have another way to arrive at the cases: when we multiply each equation by their "alone" variable, we get: x 1 2 − 1 3 0 x 1 = x 2 2 − 1 3 0 x 2 = x 3 2 − 1 3 0 x 3 = x 4 2 − 1 3 0 x 4 = − x 1 x 2 x 3 x 4 . If the set of x 1 , x 2 , x 3 , x 4 form a three or four element set(three or four distinct numbers), then this would contradict the fact the a quadratic equation can have at most two distinct solutions.
Nice solution. I have one mathematical problem related to the differential equation. Can you help for that? my gmail address is jinjaladinesh@gmail.com.
If x i = 0 , then we must have x j + 0 = 1 3 0 for j = i . But then, we must also have 1 3 0 3 = x i + j = i ∏ x j = 1 3 0 which is impossible. Hence x i = 0 for all i . Let R = x 1 x 2 x 3 x 4 , the given condition in the problem means x i + x i R = 1 3 0 ⇒ x i 2 − 1 3 0 x i + R = 0 since x i = 0 . Now for i < j , we have x i 2 − 1 3 0 x i + R = x j 2 − 1 3 0 x j + R ( x i − x j ) ( x i + x j − 1 3 0 ) = 0 Thus, if we set i = 1 , we get to the conclusion that either x j = x 1 or x j = 1 3 0 − x 1 for j > 1 . Divide into cases :
i) x 1 = x 2 = x 3 = x 4 = a . Then we have the equation a + a 3 = 1 3 0 ( a − 5 ) ( a 2 + 5 a + 2 6 ) = 0 The quadratic equation doesn't have real roots since the discriminant is negative. x 4 = 5 is the only solution for this case.
For the next cases x 4 = 1 3 0 − x 1 > x 1 , and so x 1 < 6 5
ii) x 1 < x 2 = x 3 = x 4 = 1 3 0 − a , then we have the equation a + ( 1 3 0 − a ) 3 = 1 3 0 − a + a ( 1 3 0 − a ) 2 = 1 3 0 From the first equality, we have ( a − ( 1 3 0 − a ) ) + ( 1 3 0 − a ) 2 ( ( 1 3 0 − a ) − a ) = 0 ( 2 a − 1 3 0 ) ( 1 − ( 1 3 0 − a ) 2 ) = 0 2 ( a − 6 5 ) ( 1 + ( 1 3 0 − a ) ) ( 1 − ( 1 3 0 − a ) ) = 0 2 ( a − 6 5 ) ( a − 1 3 1 ) ( a − 1 2 9 ) = 0 We know a = 6 5 , so we either have a = 1 3 1 ⇒ 1 3 0 − a = − 1 < a or a = 1 2 9 ⇒ 1 3 0 − a = 1 < a . No solutions for this case
iii) a = x 1 = x 2 = x 3 < x 4 = 1 3 0 a . By letting b = 1 3 0 − a , we arrive to the same set of equations as case ii), and we get b = 1 2 9 ⇒ a = 1 which gives us ( x 1 , x 2 , x 3 , x 4 ) = ( 1 , 1 , 1 , 1 2 9 ) and b = 1 3 1 ⇒ a = − 1 which gives us ( x 1 , x 2 , x 3 , x 4 ) = ( − 1 , − 1 , − 1 , 1 3 1 ) . We can easily check these two solutions are valid and the sum of all possible x 4 in this case 1 2 9 + 1 3 1 = 2 6 0
iv) a = x 1 = x 2 < x 3 = x 4 = 1 3 0 − a , then we have the equation a + a ( 1 3 0 − a ) 2 = 1 3 0 − a + a 2 ( 1 3 0 − a ) = 1 3 0 From the first equality, we have ( a − ( 1 3 0 − a ) ) + a ( 1 3 0 − a ) ( ( 1 3 0 − a ) − a ) = 0 ( 2 a − 1 3 0 ) ( 1 + a ( 1 3 0 − a ) ) = 0 2 ( a − 6 5 ) ( a 2 − 1 3 0 a + 1 ) = 0 Again we know that a = 6 5 , for the quadratic equation, let the roots be a 1 < a 2 . By Vieta's theorem, a 1 + a 2 = 1 3 0 and a 1 a 2 = 1 so both are them are positive (easy to check they are real numbers). Also we have a i ( 1 3 0 − a i ) = 1 Using the above's equality implies a i + a i ( 1 3 0 − a i ) 2 = a i + ( 1 3 0 − a i ) = 1 3 0 and also ( 1 3 0 − a i ) + a i 2 ( 1 3 0 − a i ) = ( 1 3 0 − a i ) + a i = 1 3 0 and so a 1 , a 2 satisfies the equation. But only a 1 < 6 5 since their sum is 1 3 0 . By quadratic roots formula, a 1 = 6 5 − 6 5 2 − 1 = 6 5 + 6 5 2 − 1 1 < 0 . 5 and so x 4 = 1 3 0 − a i ∈ ( 1 2 9 . 5 , 1 3 0 ) . Since this is the last case and there is no other non-integer solutions in other cases, the sum for this case is 1 3 0
Hence, the grand total is 5 + 2 6 0 + 1 3 0 = 3 9 5 .
This is not the most efficient solution, but it is very nicely written.
Common mistakes included being somewhat ambiguous about using the symmetry of the equations, disregarding the condition that the variables x i are increasing. While it is possible to do so, one has to word this carefully. One way is to say something like this: "We will find all real solutions of the system, regardless of the order of the numbers, and will then pick x 4 to be the largest of the four numbers").
Let x1=a : x2=b : x3=c : x4=d
a+bcd=130 ......(1)
b+cda=130 ......(2)
c+adb=130 ......(3)
d+abc=130 ......(4)
from 1 & 2 : a+bcd = b+cda i.e. (a-b)(1-cd) = 0
Case 1: a=b from 3 & 4 : (c-d)(1-ab)=0 i.e. (c-d)(1-a^2)=0
Case 1.1 : (1-a^2)=0 i.e. a= 1, -1
Case 1.1.1 : a=b=1 ; cd=129, c+d=130 ; c+129/c=130 ; c^2-130c+129=0 ; c=1,129 ; a=b=c=1, d=129 ;
Case 1.1.2 : a=b=-1 ; cd=-131, c+d=130 ; c-131/c=130 ; c^2-130c-131=0 ; c=-1, 131 ; a=b=c=-1, d=131 ;
Case 1.2 : c=d ; a(1+c^2)=130 ; c(1+a^2)=130 ; (a-c) - ac(a-c) = 0 ; (a-c)(1-ac) = 0 ;
Case 1.2.1 : a=c means a=b=c=d=5
Case 1.2.2 : ac=1 ; also, a(1+c^2)=130 ; so, c^2-130c+1=0 ; i.e. c=d=65+8 sqrt(66) , a=b=65-8 sqrt(66) ;
Case 2: cd=1 ; (3) c-(4) d ; (c-d)(c+d-130)=0
Case 2.1: c=d reduces to case 1.1
Case 2.2: c+d=130 ; ie c+1/c=130 ; ie c^2-130c+1=0 ; so, c=65-8 sqrt(66), d=65+8 sqrt(66) ; a+b=130 ; 65+8 sqrt(66) + ab (65-8 sqrt(66) ) = 130 ; ab=1 ; So, this reduces to Case 1.2.2 ;
So, in total, we have 4 solutions: a=b=c=-1, d=131 a=b=c=1, d=129 a=b=c=d=5 a=b=65-8 sqrt(66), c=d=65+8 sqrt(66)
So, S=131+129+5+65+8*sqrt(66) =394.9923 =395 (round off)
If x 4 = 0 , then x 1 = x 2 = x 3 = 1 3 0 . But 0 + 1 3 0 × 1 3 0 × 1 3 0 = 1 3 0 . So none of the x i can be 0 .
Now we can safely multiply each equation with its 'lone' x to get: ( 1 ) x 1 2 + x 2 x 3 x 4 = 1 3 0 x 1 ( 2 ) x 2 2 + x 1 x 3 x 4 = 1 3 0 x 2 ( 3 ) x 3 2 + x 1 x 2 x 4 = 1 3 0 x 3 ( 4 ) x 4 2 + x 1 x 2 x 3 = 1 3 0 x 4
( 4 ) − ( 1 ) , ( 4 ) − ( 2 ) , and ( 4 ) − ( 3 ) give us: x 1 = x 4 ∨ x 1 = 1 3 0 − x 4 x 2 = x 4 ∨ x 2 = 1 3 0 − x 4 x 3 = x 4 ∨ x 3 = 1 3 0 − x 4
There are now three different cases: All x i = x , three of the x i = x and the fourth equals 1 3 0 − x , and two of the x i = x and the other two equal 1 3 0 − x .
The first case, x 1 = x 2 = x 3 = x 4 = x , gives us: x + x 3 − 1 3 0 = 0 ( x − 5 ) ( x 2 + 5 x + 2 6 ) = 0 x = 5 ∨ x = − 2 5 + ι ∨ x = − 2 5 − ι x = 5 gives us x 1 = x 2 = x 3 = x 4 = 5 .
The second case, x 1 = x 2 = x 3 = 1 3 0 − x 4 = x , gives us: 1 3 0 − x + x 3 − 1 3 0 = 0 x 3 = x x = − 1 ∨ x = 0 ∨ x = 1 x = − 1 gives us x 1 = x 2 = x 3 = − 1 and x 4 = 1 3 1 .
x = 1 gives us x 1 = x 2 = x 3 = 1 and x 4 = 1 2 9 .
The third case, 1 3 0 − x 1 = 1 3 0 − x 2 = x 3 = x 4 = x , gives us: 1 3 0 − x + ( 1 3 0 − x ) x 2 − 1 3 0 = 0 x 3 − 1 3 0 x 2 + x = 0 x = 0 ∨ x 2 − 1 3 0 x 2 + x = 0 x = 0 ∨ x = 6 5 − 6 5 2 − 1 ∨ x = 6 5 + 6 5 2 − 1 If x 3 = x 4 = 6 5 − 6 5 2 − 1 , x 1 = x 2 = 6 5 + 6 5 2 − 1 > x 4 , so x 4 = 6 5 + 6 5 2 − 1 is the only viable solution in this case.
We have x 4 = 5 , 1 2 9 , 1 3 1 , or 6 5 + 6 5 2 − 1 .
5 + 1 2 9 + 1 3 1 + 6 5 + 6 5 2 − 1 ≈ 3 9 4 . 9 9 . So the answer to the question is 3 9 5 .
Nice solution!
assume S = sum of all distinct values x4 can take
For ease of typing let me refer to x1, x2,x3,x4as a,b,c,and d.
Now it is given that:
a+bcd = 130
b+acd = 130
c+abd = 130
d+abc = 130
Multiplying 1 by a and 2 by b and subtracting leads to
a^2-b^2 = 130(a-b)
Therefore either a+b=130 or a=b
Taking other combination of equations (other than 1 and 2) and performing similar operations would lead to a symmetric result of
b+c=130 or b=c
c+d=130 or c=d
a+d=130 or a=d
Using a=b, (Using a=b also covers using b=c, c=d etc. by symmetry)
The equations reduce to a+acd=130, c+a^2d=130 and d+a^2c=130
The last two equations again lead to c=d or a^2=1
c=d => a+ac^2=130 and c+ca^2 = 130 which gives either a=c (giving a=b=c=d ) or ac=1 which along with the now simplified a+c=130 gives
a=b=129.992307, c=d=0.0076927 or a=b=0.0076927 and c=d=129.992307
Solution set = (129.992,129.992,0.00769,0.00769)
d being the highest value
a^2=1 => c+d=130 and (cd=129 from a=1 or cd= -131 from a= -1 )
leading to the solution sets (1,1,1,129) and (-1,-1,-1,131)
Not using a=b, we have a+b=130, b+c=130, c+d=130 and a+d=130 which gives a=b=c=d.
Finally the case a=b=c=d reduces the equations to a+a^3=130
The only real root of which is a=5.
Therefore S= 129.992307 +129 +131 + 5 = 394.9923
Solution 2 on Quora (http://www.quora.com/Mathematics/Consider-all-quadruples-of-real-numbers-x1-x2-x3-x4-satisfying-x1≤x2≤x3≤x4-such-that-for-each-of-the-four-numbers-adding-it-to-the-product-of-the-other-three-numbers-equals-130-If-S-is-the-sum-of-all-possible-values-of-x4-what-is-S)
"Not using a=b, we have a+b=130, b+c=130, c+d=130 and a+d=130 which gives a=b=c=d." Not true as stated: we may have a=c, b=d, and a+b=130. However there are more equations there than just those listed. However, overall, the solution is correct, just not perfectly written. Would be +2X if not for the cheating.
a+bcd=130 ......(1) b+cda=130 ......(2) c+adb=130 ......(3) d+abc=130 ......(4)
from 1 & 2 : a+bcd = b+cda i.e. (a-b)(1-cd) = 0
Case 1: a=b from 3 & 4 : (c-d)(1-ab)=0 i.e. (c-d)(1-a^2)=0
Case 1.1 : (1-a^2)=0 i.e. a= 1, -1
Case 1.1.1 : a=b=1 cd=129, c+d=130 c+129/c=130 c^2-130c+129=0 c=1,129 a=b=c=1, d=129
Case 1.1.2 : a=b=-1 cd=-131, c+d=130 c-131/c=130 c^2-130c-131=0 c=-1, 131 a=b=c=-1, d=131
Case 1.2 : c=d a(1+c^2)=130 c(1+a^2)=130 (a-c) - ac(a-c) = 0 (a-c)(1-ac) = 0
Case 1.2.1 : a=c means a=b=c=d=5
Case 1.2.2 : ac=1 also, a(1+c^2)=130 so, c^2-130c+1=0 i.e. c=d=65+8 sqrt(66) , a=b=65-8 sqrt(66)
Case 2: cd=1 (3) c-(4) d (c-d)(c+d-130)=0
Case 2.1: c=d reduces to case 1.1
Case 2.2: c+d=130 ie c+1/c=130 ie c^2-130c+1=0 so, c=65-8 sqrt(66), d=65+8 sqrt(66) a+b=130 65+8 sqrt(66) + ab (65-8 sqrt(66) ) = 130 ab=1 So, this reduces to Case 1.2.2
So, in total, we have 4 solutions: a=b=c=-1, d=131 a=b=c=1, d=129 a=b=c=d=5 a=b=65-8 sqrt(66), c=d=65+8 sqrt(66)
So, S=131+129+5+65+8*sqrt(66) =394.9923
Let's consider the products
Multiplying the first equation by and the second equation by , and comparing, you get
Quite clearly, either , or , for any two indices.
Next, we show that you can't have three of the terms being different from each other, i.e. we can't have and at the same time. This is because if , then , and implies that . Clearly, we must have
This reduces the problem to 4 cases:
This is a straightforward cubic equation, and we get
This gives two simultaneous equations:
But don't forget that we also require , and that is the biggest of the lot. Substituting this requirement, you'll find that y = 1 and , or y = -1, .
Again, substituting in our requirement, we get .
You'll find that this is identical to case 2, except that there won't be any solution with being the largest.
So the final answer that you need is
I am using a,b,c and d for the notations, we need to find sum of all possible values of d. we have a+ bcd=130 ...(1) b + acd=130...(2) c +abd =130...(3) d+ abc =130...(4)
Subtracting (2) from (1), a-b + cd(b-a)=0 (b-a)(cd-1) =0
Similarly, taking difference of any 2 equations from (1),(2), (3) and (4) We will get the following eqns :
(b-a)(cd-1)=0 (c-b)(ad-1)=0 (d-c)(ab-1)=0 (c-a)(bd-1)=0 (d-b)(ac-1)=0 (d-a)(bc-1)=0
One obvious solution is then a=b=c=d. (Case 1) Another will be a=b=c (case 2) Another will be a=b and c=d. (case 3)
Using these 3 cases and the first 4 eqns, possible values of d are 129,131,5 and 130(approx)
It is given that: 1. a+bcd = 130 2. b+acd = 130 3. c+abd = 130 and 4. d+abc = 130
Multiplying 1 by a and 2 by b and subtracting leads to a^2-b^2 = 130(a-b)
Therefore either a+b=130 or a=b
Taking other combination of equations (other than 1 and 2) and performing similar operations would lead to a symmetric result of
b+c=130 or b=c c+d=130 or c=d a+d=130 or a=d
Using a=b, (Using a=b also covers using b=c, c=d etc. by symmetry)
The equations reduce to a+acd=130, c+a^2d=130 and d+a^2c=130
The last two equations again lead to c=d or a^2=1
c=d => a+ac^2=130 and c+ca^2 = 130 which gives either a=c (giving a=b=c=d ) or ac=1 which along with the now simplified a+c=130 gives
a=b=129.992307, c=d=0.0076927 or a=b=0.0076927 and c=d=129.992307
Solution set = (129.992,129.992,0.00769,0.00769)
d being the highest value
a^2=1 => c+d=130 and (cd=129 from a=1 or cd= -131 from a= -1 )
leading to the solution sets (1,1,1,129) and (-1,-1,-1,131)
Not using a=b, we have a+b=130, b+c=130, c+d=130 and a+d=130 which gives a=b=c=d.
Finally the case a=b=c=d reduces the equations to a+a^3=130
The only real root of which is a=5.
Therefore S= 129.992307 +129 +131 + 5 = 394.9923
Problem Loading...
Note Loading...
Set Loading...
The equations tell us that ( x 1 − x 2 ) ( 1 − x 3 x 4 ) = ( x 3 − x 4 ) ( 1 − x 1 x 2 ) = 0 There are four cases to consider:
If x 1 = x 2 = a and x 3 = x 4 = b then a + a b 2 = b + a 2 b = 1 3 0 , and hence ( a − b ) ( 1 − a b ) = 0 . If a = b then x 1 = x 2 = x 3 = x 4 = 5 . If a b = 1 then a + b = 1 3 0 and hence x 1 = x 2 = 6 5 − 8 6 6 , x 3 = x 4 = 6 5 + 8 6 6 .
If x 1 = x 2 and x 1 x 2 = 1 then either x 1 = x 2 = 1 , in which case x 3 + x 4 = 1 3 0 and x 3 x 4 = 1 2 9 , so that x 3 = 1 and x 4 = 1 2 9 , or else x 1 = x 2 = − 1 , in which case x 3 + x 4 = 1 3 0 and x 3 x 4 = − 1 3 1 , so that x 3 = − 1 and x 4 = 1 3 1 .
If x 3 x 4 = 1 and x 3 = x 4 then either x 3 = x 4 = 1 , in which case x 1 + x 2 = 1 3 0 and x 1 x 2 = 1 2 9 , so that x 1 , x 2 = 1 , 1 2 9 , or else x 3 = x 4 = − 1 , in which case x 1 + x 2 = 1 3 0 and x 1 x 2 = − 1 3 1 , so that x 1 , x 2 = − 1 , 1 3 1 . Neither of these cases is possible.
If x 3 x 4 = x 1 x 2 = 1 then x 1 + x 2 = x 3 + x 4 = 1 3 0 , and hence x 1 = x 3 = 6 5 − 8 6 6 , x 2 = x 4 = 6 5 + 8 6 6 , which is not possible.
Thus the sum of the possible values of x 4 is 5 + 6 5 + 8 6 6 + 1 2 9 + 1 3 1 = 3 9 4 . 9 9 2 3 0 7 which is 3 9 5 to the nearest integer.