Xiao Zhang has three watches. The first watch is 2 minutes fast per hour, the second watch is 6 minutes fast, and the third watch is 16 minutes fast. If the minute hands of all three watches are currently pointing in the same direction, how many hours later will the three minute hands point in the same direction again?

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Clock problem.
The first watch is 2 minutes fast per hour compared to standard time, and the second watch is 6 minutes fast per hour; thus, the second watch is $6-2=4$ minutes faster per hour than the first watch;
Every $60 \div 4=15$ hours, the minute hands of the first and second watches point in the same direction;
The first watch is 2 minutes fast per hour, and the third watch is 16 minutes fast per hour; thus, the third watch is $16-2=14$ minutes faster per hour than the first watch;
Every $60 \div 14=\frac{30}{7}$ hours, the minute hands of the first and third watches point in the same direction;
$\left[15, \frac{30}{7}\right]=\frac{[15,30]}{(1,7)}=30$ hours later, the three minute hands point in the same direction again.