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Jan 10 at 10:48 comment added dEmigOd @pts, now I understand the answer.
Jan 10 at 9:45 comment added pts An even simpler proof: Let's suppose it's possible. The two products are equal, so they are equal modulo 4 as well. 7 is 3 modulo 4, all other numbers are 1 modulo 4 (because they end in 77). Thus one of the products containing 7 is $3\cdot1\cdot1\cdot\ldots=3$ modulo 4, and the other product is $1\cdot1\cdot1\cdot\ldots=1$ modulo 4. This is a contradiction, so it's impossible to have such products.
Jan 8 at 21:18 vote accept Iulian Serbanoiu
Jan 8 at 4:43 history edited Timothy Chow CC BY-SA 4.0
improved formatting
Jan 7 at 21:50 history answered Aleksei Kulikov CC BY-SA 4.0