The number is of the form $2^x3^y5^zm^{30}$ because the it has 2 ,3 and 5 as factors and $m^{30}$ is a $30^{th}$ power because it is a square, cibe and fith power so multiplying by this shall satisfy the criteria.
$2^{x+1}$ has to be perfect square and $2^{x}$ is a perfect cube and a fifth power.
so $x+1 \equiv 0 \pmod 2\cdots(1)$
$x \equiv 0 \pmod 3\cdots(2)$
$x \equiv 0 \pmod 5\cdots(3)$
we can solve the same by chinese remainder theorem but we use a short method
From (2) and (3)
$x \equiv 0 \pmod {15} \cdots(4)$
from (1) and (4) we have checking multiple of 15 that is 0 and 15
$x \equiv 15 \pmod {30} \cdots(5)$
similarly we evaluate y
$3^{y+1}$ has to be perfect cube and $3^{x}$ is a perfect square and a fifth power.
so $y+1 \equiv 0 \pmod 3\cdots(6)$
$y \equiv 0 \pmod 2\cdots(7)$
$y \equiv 0 \pmod 5\cdots(8)$
From (7) and (8)
$y \equiv 0 \pmod {10} \cdots(9)$
from (6) and (9) we have checking mutiple of 10 that is 0,10,20
$y \equiv 20 \pmod {30} \cdots(10)$
similarly we evaluate z
$5^{z+1}$ has to be perfect fifth cube and $5^{x}$ is a perfect square and a perfect cube.
so $z+1 \equiv 0 \pmod 5\cdots(11)$
$z \equiv 0 \pmod 2\cdots(12)$
$z \equiv 0 \pmod 3\cdots(13)$
From (12) and (13)
$z \equiv 0 \pmod {6} \cdots(14)$
from (11) and (14) we have checking mutiple of 6 that is 0,6,12,18,24
$z \equiv 24 \pmod {30} \cdots(10)$
we can take principal vaues and get
$a=2^{15}3^{20}5^{24}m^{30}$
No comments:
Post a Comment