For function M(x,y,z), where x, y, and z, are single-digit, positive integers, the following is true:
M(1,2,3)=32
M(3,2,1)=8
M(4,5,6)=245
M(6,5,4)=95
M(7,8,9)=800
M(9,8,7)=416
M(5,3,1)=15
M(1,3,5)=135
For what values of x,y,z, other than 1,3,5, does M(x,y,z)=100x+10y+z?
PART I:
Let us define a trivariate function P such that :
P(x,y,z) = M(x,y,z) � M(z,y,x).
Now, P(1,2,3) = 32-8 = 24, which is divisible by 3-1 = 2
P(4,5,6) = 150 which is divisible by 6-4 = 2
P(1,3,5) = 120, which is divisible by 5-1 = 4.
This leads us to conjecture that P(x,y,z) is divisible by (z-x).
Let, Q(x,y,z) = P(x,y,z)/(z-x).
Then, Q(1,2,3) = 12; Q(4,5,6) = 75 and Q(1,3,5) = 30.
Having regard to the title, we can surmise that the function M deals with exponents of x, y and z and accordingly , a little trial and error at this stage would reveal that:
Q(x,y,z) = z^2 + x^2 +xz -1; giving:
P(x,y,z)
= (z^3 � x^3) �(z �x)
= (x + L(y) + z^3) � (x^3 + L(y) + z)
= M(x,y,z) � M(z,y,x); where
M(x,y,z) = x + L(y) + z^3
Now,
M(1,2,3) = 32; M(4,5,6) = 245 ; M(1,3,5) = 135; and M(7,8,9) = 800 gives:
L(2) = 4 = 2^2; L(5) = 25 = 5^2 ; L(3) = 9 = 3^2 and :
L(8) = 64 = 8^2.
Accordingly, L(y) = y^2 and so:
M(x,y,z) = x + y^2 + z^3 is the required function.
PART II:
Since, M(x,y,z) = 100x+10y+z, it follows that each of x, y and z are between 1 and 9 ( both digits inclusive). Hence the quantity (10-y) is always positive.
From the conditions of the problem, we obtain:
100x + 10y + z = x + y^2 + z^3
Or, 99x = (z^3 � z) � y(10 � y)������(#)
From the above expression it is clear that if any given triplet (x,y,z) = (x', y', z') satisfies (#), then (x,y,z) = (x', 10-y', z') also corresponds a solution of equation (#).
Now, by trial and error, we observe that:
(y,z) = (1,8) gives 99x = 495, so that x= 5.
So, (x,y,z) = (5,1,8) ; (5,9,8).
(y,z) = (3,5) gives 99x =99, so that x=1.
So, (x,y,z) = (1,3,5); (1,7,5).
Concequently, (x,y,z) = (1,7,5); (5,1,8) ; (5,9,8) constitute all possible solutions to Part II.
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