Expert: Scott A Wilson - 2/28/2010

Question
QUESTION: "find a,b,c and d where they satisfy all the following

a+b+c+d=2    ////  abcd=0 ////  b^c=a ////d^b=a /// 3a -c = b -d  //

c/a =b(a+d)  ////   tan(c)= b-2a ////  d>a^a>b  ///  5c>d

a^b^c^d=3b-5a   ///  10c-a = 2d/b  ///

make sure that a,b,c and satisfy all equations and inequalities "

I think that the case where c =0 is right then a=1 left is  d and b said that since d^b=1 (which is a) d must be 1 right? well what if its -1^(2,4,6,8...) then it will also be d can be -1 and b an even number so lest try b=2
1+2+0-1 =3-1 = 2
2^0=1
3*1-0=2+1 I guess I was able to do it by guessing do you know if there is a more solid way of doing it ?

ANSWER: If c = 0, then b^c = a says that a is 1.

If a is 1, this says that d^b = 1, so b must also be 0.

Since a = 1, b = 0, and c = 0, a+b+c+d=2 means that d = 1 as well.

Now that we have a, b, c, and d, as, respectively, 1, 0, 0, and 1, lets check them all.

1) It says a+b+c+d=2, and we already did that.
2) It says abcd=0, and since b=0 or c=0, this is true.
3) It says b^c=a, and we used that.
4) It says d^b=a, and 1^0 = 1, so that checks.
6) It says that c/a = b(a+d), and 0/1 = 0(1+1), and that checks.

5) It says that 3a-c=b-d, bt 3-0 is not the same as 0-1 since 3-0=3 and 0-1=-1.
7) It says that tan(c) = b - 2a, and tan(0) is not 0-2=-2, so this is not true.
8) It says that d>a^a>b, and 1 is not greater that 1^1=1, though 1>0.
9) It says that 5c>d, and 5(0) is not greater that 1.
11) It says that 10c-a=2d/b, and 10(0)-1=2d/0 is definitely not true since /0 is unefined.

10) The value of a^b^c^d is 1 since both b and c are 0.
The value of 3b - 5a is 3(0) - 5, and that is -5, not 0.


Now (2) says that abcd = 0, and that means at least one of a, b, c, or d is 0.

It can be shown that none of the eqations will work for a=0, b=0, c=0, or d=0.
Set one of them equal to 0 and start going through the rest.
Eventually, one will be found not to work.


---------- FOLLOW-UP ----------

QUESTION: But I am saying that b =2 not 0 and d =-1 it will work in that case but I only said so because it will work like that I just don't have another method of proving that they are equal to that
1+2+0-1=2
1*2*0*-1=0
2^0=1
-1^2=1
0/1= (2(1-1)))
3-0= 2+1
tan(0) =2 - 2*1
-1>1^1 >2
5*0 > -1
-1 =2 *-1/2
1^2^0^-1 =3*2 -5

Answer
a+b+c+d=2
abcd=0
b^c=a
d^b=a
3a -c = b -d
c/a =b(a+d)
tan(c)= b-2a
d>a^a>b
5c>d
a^b^c^d=3b-5a
10c-a = 2d/b

If we take b as 2 and d as -1, what happens?
You say that a=1, b=2, c=0, and d=-1.

The only problem here is in -1>1^1 >2.  It is not true.  If that were d < a^a < b, it would work.
Note > is "greater than" and < is "less than".

If this were the case, c=0 would have produced correct results.