2024/037) How do I find some natural number a, such that 2a is a perfect square, 3a is a perfect cube, 5a is the fifth power of some natural number?

 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

$a=2^{15}3^{20}5^{24}m^{30}$