#7

I think that the idea is that you treat it like a sudoku, working your way around the board.

Your answers for 9A, 12A , 20A and 11D are correct.

Your two options for 16D are correct, but it’s tied to 15D. This is where you probably have to take a leap and see where it takes you. 15D is unlikely to be EXACTLY the same as 16D. Therefore we could assume that 16D is 91, and 15D is 13 (it has to be a two-digit number). Which then tells you 14A (2197)…

17A can be either 89 or 83, which affects 4A.

I’ll come back to this later, but you might be able to solve it from there

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#8

We got this far then gave up, and I think part of this is wrong because 6 across is 17 or 19 and that doesn’t work with… something else I don’t care I’m going back to my game of Crusader Kings 3 where I accidentally impregnated my sister fuck off

bums
bums1518×2024 430 KB

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#9

6A has to be either 17 or 19, which means
13A has to be either 51 or 57
5D is therefore either 2401 or 6561 (there are no fourth powers of four digits ending in 7), and 13A is therefore 51, and 6A is therefore 17.

#10

4A = 17A-13A,

so is either 83-51 (32), or 89-51 (38)

There’s only one option here, as we’ve already established that 5D is either 2401 or 6561. It’s the former (first digit has to be a 2 or an 8), with 4A being 32 and 17A being 83

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#11

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#12

listen if marckee can google the answer anyone can

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#13

We can work out 4D because we know 6A and 4A, which means that it is 34

7D is a multiple of 6A (17), in the form _19. A quick check shows that it has to be 119.

We can now say that 7A, which is a multiple of 9A (162), in the form 1_ _4, has to be 1944.

#14

I very much hope you are enjoying doing this. Thank you so much, I knew it was a logic thing of working out

Poor R who didn’t seem to even know what a Fibonacci number was etc without Google? I will be able to help him do it a step at a time

#15

pretty sure Fibonacci was the guy with the candelabra?

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#16

19A a prime factor of 10D (657), so either 3 or 3 or 73. It has to be the latter

8D is a multiple of 7D (119) in the form of _ _27. Going through them, it has to be 3927

This leaves five to get:
1A, which is a four-digit multiple of 135
1D, which is a two-digit prime number, in the form _1. So either 31, 41, 61, 71.
2D is a four-digit triangular number in the form _ 7 _ 1.
3D is a prime factor of 2D
8A is a prime factor of 1A in the form 3 _

I think from here you have to work through all the options of 1D and see if any of them click. There may even be multiple examples that fit.

#17

I’m trying to remember when we did Fibonacci. I think it was the same year we did prime numbers, factors etc, so about 10 years old. A bit older than R, then, which seems a little unfair on him.

#18

No, Y9 is about 13-14.

#19

Thing is he has been off school for months, back for a day then this which is just sent as with with no input or video lesson etc.

He is 13 so knows primes, factors etc but not necessarily how to use them to complete this kind of thing.

#20

Oh, oops. I thought it was 9 years old, not Y9.

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#21

STRETCH AND CHALLENGE!

#22

Something to get the cogs moving again…

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#23

I’m starting to think about applying for maths phd programmes and looking at this homework I want to give up and go back to school.

#24

Extension, extension, extension.

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#25

You’re going to take that back to school all filled in correctly for the teacher to say oops this was for our mensa recruits.

Your child will be elevated to genius.

#26

On what fucking planet is a 9 year old doing this??
:grinning:

(Fibonacci was my gcse coursework so I did it when I was 15-16)

(I didn’t actually do it. Just copied it off my mate)

(Got an a)

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