Find all triplets (X,Y,Z) of positive integers that satisfy this system of equations:
lcm(X,Y,Z) + Y +Z = 426, and:
X + gcd(X,Y,Z) + Z = 66, and:
X+Y+Z =90
Two possibilities:
- X, Y, Z are all even numbers. It is easy to see from the system. And then also is even gdc.
- Y, Z are odd numbers and gdc is also odd.
I'll explore just the first possibility which is easier.
lcm (X,Y,Z) is rather low (think that for 26*34*30, with gdc=2 lmc=6630). So gdc is quite probably higher than 2).
Y=24+gdc
Begin trying gdc=4 Y=28 X+Z=62 which is no mod 4 so gdc#4
With gcd=6 Y=30 X+Z=60 which is mod 6 so it is a possible set of solutions.
- As in ths case both, X and Z, must be mod 6 the possibilities are only five:
- (X,Z)=(6,54) (12,48) (18,42) (24,36) (30,30) (or viciversa)
- For (X,Z)=(24,36) [y=30] lmc = 360
Then 360 +30 + 26= 426
So there is a solution for (X,Y,Z)=(24,30,36)
With gdc=8 Y=32 X+Z=58 which is no mod 8 so gdc#8
With gdc=10 Y=34 X+Z=56 which is no mod 10 so gdc#10
With gdc=12 Y=36 X+Z=54 which is no mod 12 so gdc#12
With gdc=14 Y=38 X+Z=52 which is no mod 14 so gdc#14
With gdc=16 Y=40 X+Z=50 which is no mod 16 so gdc#16
With gdc=18 Y=42 X+Z=48 which is no mod 18 so gdc#18
With gdc=20 Y=44 X+Z=46 which is no mod 20 so gdc#20
With gdc=22 Y=46 X+Z=44 which is mod 22 but do not lead to solution because lmc =506
There are no other possibilities.
Edited on May 4, 2016, 5:23 am
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Posted by armando
on 2016-05-04 04:54:42 |
think and try
