8
$\begingroup$

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

Line 1: 1 2 black; Line 2: 3 empty 4; Line 3: black 5 empty


This puzzle comes from the Senior Kangaroo 2019 contest.

$\endgroup$

2 Answers 2

11
$\begingroup$

Solved grid:

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:step one

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:step two

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:step three

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!

$\endgroup$
0
$\begingroup$

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"

$\endgroup$
5
  • 6
    $\begingroup$ 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! $\endgroup$
    – Stiv
    Commented Apr 29 at 8:23
  • 4
    $\begingroup$ @Stiv, ah thanks! I learned something today! $\endgroup$ Commented Apr 29 at 8:57
  • $\begingroup$ Welcome to PSE (Puzzling Stack Exchange)! $\endgroup$ Commented Apr 29 at 8:57
  • $\begingroup$ Also, 122 is not divisible by 9, but I found that 162 is. $\endgroup$
    – stux
    Commented Apr 29 at 12:26
  • $\begingroup$ Oh, yes indeed. $\endgroup$ Commented Apr 29 at 12:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.