# Maths help

Hi!
R has this set for his homework and we are baffled, as far as we can see we can do like a few answers but the rest are like am I missing something??

He’s only yr9, I have A level and am so stuck…

This is moreorless the response I am going to send to his teacher.

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Haven’t got a fibbonacci

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9A, 12A, 20A are the only ones you can fill in straight away.

I’d then list out all the possibilities for 1A, 17A, 1D, 2D, 5D, 11D, 16D, 18D

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Just a grind, this, innit? A fun challenge for the more able kids* who are that way inclined, maybe. But intensely off-putting and discouraging for the other kids.

*Or, more to the point, parents. Cos that’s how this stuff goes isn’t it? (Which is absolutely no judgement on Slicky or any other parent who pitches in with this sort of thing. No doubt if my kid came gone with it I’d make a point of completing it.) Parental encouragement and input is one thing, but this is the kind of task that’s either getting farmed out or ignored.

What’s the point? I hope nothings resting on it?

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

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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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

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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.

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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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.

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

pretty sure Fibonacci was the guy with the candelabra?

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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.

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.