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 Power! (Posted on 2006-03-09)
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?

 See The Solution Submitted by Joe Rating: 2.5000 (4 votes)

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 Puzzle Resolution Comment 2 of 2 |

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) = (1,3,5); (1,7,5); (5,1,8) ; (5,9,8) constitute all possible solutions to Part II.

Edited on October 14, 2006, 12:46 pm

Edited on October 14, 2006, 12:48 pm

Edited on October 14, 2006, 12:52 pm

Edited on October 14, 2006, 12:57 pm

Edited on October 14, 2006, 1:01 pm

Edited on October 14, 2006, 1:39 pm

Edited on December 9, 2006, 1:11 pm
 Posted by K Sengupta on 2006-10-14 12:44:06

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