Beginner puzzle

This puzzle is intended to be suitable for people who are new to puzzle solving.

Clarification: Both experienced solvers and new solvers are welcome to post solutions to this puzzle.

Fill each square in this cross-number with a non-zero digit such that all of the conditions in the clues are fulfilled. The digits used are not necessarily distinct.

1 ACROSS. A composite factor of $$1001$$
3 ACROSS. Not a palindrome
5 ACROSS. $$pq^3$$ where $$p$$, $$q$$ prime and $$p \ne q$$

1 DOWN. One more than a prime, one less than a prime
2 DOWN. A multiple of $$9$$
4 DOWN. $$p^3q$$ using the same $$p$$, $$q$$ as 5 ACROSS

This puzzle comes from the Senior Kangaroo 2019 contest.

Solved grid:

Solution:

First, notice that for prime $$p\neq q$$, we require both $$pq^3$$ and $$p^3q$$ to be 2-digit numbers. This means that $$\{p,q\}=\{2,3\}$$, since the next largest prime, $$5^3=125$$ is already 3 digits long. So, 5 ACROSS and 4 DOWN are $$24$$ or $$54$$.

Now, note that the only 2-digit composite factors of $$1001$$ are $$77$$ and $$91$$. The grid now looks like:

1 DOWN tells us immediately that we are looking for a 2-digit even number in the middle of a pair of twin primes. We know that the even number must start with $$7$$ or $$9$$; the only twin primes in this vicinity are $$(71,73)$$, so 1 DOWN has answer $$\boxed{72}$$. This also allows us to figure out that 1 ACROSS is $$\boxed{77}$$. The grid now looks like:

3 ACROSS is not a palindrome, so 4 DOWN must be $$\boxed{54}$$, and 5 ACROSS must be $$\boxed{24}$$. This corresponds to $$p=3$$ and $$q=2$$. The grid now looks like:

The final observation is that 2 DOWN is a multiple of $$9$$. In particular, its digit sum must be divisible by $$9$$. This means the middle cell must contain a $$0$$ or a $$9$$. However, the puzzle states that all cells must contain nonzero numbers, so the middle cell contains a $$9$$. After filling this in, we have solved the puzzle!

Besides the answer of @DanDan面 here above, I found some more solutions, because

(1ACROSS) besides 77 and 91, other valid factors are 11 and 13 (their counterparts). Right? I suppose I'm wrong, but I'm new here.

This yields the following solutions:

1 1 █
2 6 5
█ 2 4

1 3 █
2 4 5
█ 2 4

1 1 █
8 6 5
█ 2 4

1 1 █
8 3 2
█ 5 4

1 3 █
8 4 5
█ 2 4

1 3 █
8 1 2
█ 5 4

EDIT: These solutions are not valid. My error here is that I mistaken "prime factors" for "composite factors"

• Welcome to Puzzling, Klaas :) The problem here is that 11 and 13 are prime, not composite, so although they are factors they are not composite factors. Sorry!
– Stiv
Commented Apr 29 at 8:23
• @Stiv, ah thanks! I learned something today! Commented Apr 29 at 8:57
• Welcome to PSE (Puzzling Stack Exchange)! Commented Apr 29 at 8:57
• Also, 122 is not divisible by 9, but I found that 162 is.
– stux
Commented Apr 29 at 12:26
• Oh, yes indeed. Commented Apr 29 at 12:44