1
$\begingroup$

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 satisfied. The digits used are not necessarily distinct.

1 ACROSS. A prime which is the sum of two squares
3 ACROSS. Twice the answer to 2 DOWN

1 DOWN. $p \times q$, where $p$, $q$ are prime and $q=p+4$
4 DOWN. $60$% of 5 ACROSS

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


This puzzle comes from the Senior Kangaroo 2021 contest.


A similar puzzle is here: A simple cross-number puzzle

$\endgroup$

2 Answers 2

5
$\begingroup$

1 Down:

$p$ cannot exceed 10 since $pq$ would be too large otherwise (not 2 digits). The possibilities are $p = 3, q = 7, pq = 21$ and $p = 7, q = 11, pq = 77$.

From 1 Down, 3 Across starts with either 1 or 7. Since 3 Across is twice 2 Down and the two are equal in length, 3 Across cannot start with 1, and 1 Down is 77.

7 ?
7 ? ?
  ? ?

1 Across:

By Fermat's two-square theorem, a prime that is a sum of two squares is precisely a prime that is 1 modulo 4. The only 2-digit prime that is 1 modulo 4 and starts with 7 is 73 (= 64 + 9).

7 3
7 ? ?
  ? ?

Alternatively, you can

deduce the top middle cell by seeing that half of 3 Across (7xx) must start with 3. This way you can simply ignore the clue for 1 Across.

4 Down & 5 Across:

4 Down is 3/5 times 5 Across, and the two share the last digit. 5 Across must be a multiple of 5 (so that 3/5 of that is an integer), and then so is 4 Down (since the last digit is either 0 or 5). Therefore, 5 Across is a multiple of 25, i.e. one of 25, 50, or 75. 4 Down can be one of 15, 30, or 45.

Since 3 Across is twice 2 Down, it is an even number, and it ends with an even digit. So 4 Down must start with an even digit, making it 45.

7 3
7 ? 4
  7 5

3 Across & 2 Down:

We can write down an equation with a single unknown being the missing digit: $2(307 + 10x) = 704 + 10x$. Solving this gives $x = 9$.

Solution:

7 3
7 9 4
  7 5

$\endgroup$
1
  • 1
    $\begingroup$ Well explained! $\endgroup$
    – Lezzup
    Commented May 13 at 5:12
0
$\begingroup$

A bit of different approach(more straightforward/bruteforced, less math knowledge) than Bubbler's answer:

1 ACROSS. A prime which is the sum of two squares

1 across is 2 digits, so we can have values between 13($3^2 + 2^2$) and 98($7^2 + 7^2)$

3 ACROSS. Twice the answer to 2 DOWN

2 down must be at most 499(otherwise 3 across would become 4 digits). That means that 1 across can only end in 1, 2, 3 or 4
Using Excel and conditional formatting, I limited the possible values for 1 across to the following:
13, 32, 34, 41, 52, 53, 61, 64, 72, 73, 74, 81, 82

1 DOWN. p×q, where p, q are prime and q=p+4

First few prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23
With difference of 4 we get a few groups: (3, 7), (7, 11), (13, 17), (17, 23)
We also see that $13*17=221$, but 1 down is only 2 digits, so we only have 2 options for 1 down: 21 or 77
But given that 1 across can not start with 2, the only possible answer for 1 down is 77.
That also limits 1 across to 72, 73 or 74
Also, hint 2 suggests that 3 across(that starts with 7) is double of 2 down. That leaves only 1 possibility of 1 across = 73(because highest double of 2xx is $299*2 = 598$ and lowest double of 4xx is $400*2 = 800$)

4 DOWN. 60% of 5 ACROSS

With both numbers being 2 digits and sharing the last digit, we can limit the possible values to:
(25, 15), (75, 45)
So first digit of 4 down is either 1 or 4, but 3 across is double of 2 down, so the last digit of 3 across must be even. That leaves us with only 1 possibility for 4 down: 45
That means that 5 across is 75

So far we have:

7 3
7 x 4
7 5

Now to find the remaining digits:

3 x 7 +
3 x 7
---
7 x 4

14
2x
6

Last digit of 2x+1 = x and 2x+1 must be between 11 and 19(otherwise the sum would be either 6x4 or 8x4)
Therefore x = 9(18+1 = 19)

So the answer is:

7 3
7 9 4
7 5

And fit it back to the hints:

1 ACROSS. A prime which is the sum of two squares

$8^2 + 3^2 = 64 + 9 = 73$

3 ACROSS. Twice the answer to 2 DOWN

$397 * 2 = 794$

1 DOWN. p×q, where p, q are prime and q=p+4

$p = 7, q = 11, p*q = 7 * 11 = 77$

4 DOWN. 60% of 5 ACROSS

$75 * 60% = 45$

$\endgroup$

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.