2016/061)Find the smallest positive integer m such that 5m is an exact 5th power, 6m is an exact 6th power, and 7m is an exact 7th power. (26th Irish)

The number has to be of the form $5^a6^b7^c$
now $5^{a+1}6^b7^c$ is a 5th power so  $a+1 \equiv 0 \pmod 5$ ,$b \equiv 0 \pmod 5$,$c \equiv 0 \pmod 5$
$5^a6^{b+1}7^c$ is a 6th power so  $a \equiv 0 \pmod 6$ ,$b+1  \equiv 0 \pmod 6$,$c \equiv 0 \pmod 6$
$5^a6^b7^{c+1}$ is a 7th power so  $a \equiv 0 \pmod 7$ ,$b \equiv 7 \pmod 5$,$c+1 \equiv 0 \pmod 7$
so we need to solve for
$a+1 \equiv 0 \pmod 5$ ,$a \equiv 0 \pmod 42$ giving a = 84 (taking multiples of 42 adding 1 to be divsible by 5)
$b+1 \equiv 0 \pmod 6$ ,$b \equiv 0 \pmod 35$ giving b = 35 (taking multiples of 35 adding 1 to be divsible by 6)
$c+1 \equiv 0 \pmod 7$ ,$c \equiv 0 \pmod 30$ giving c = 90 (taking multiples of 30 adding 1 to be divsible by 7)
so the number is $5^{84}* 6^{35}*c^{90}$