Contest math

[pure raptor]

I got the answer 11025, I want to confirm if that would be correct or not.
5.
This one I am confused about, I am not sure how the conditions other than the second and maybe the first matter. I have also seen some peers say that all the conditions matter am I just understanding the question wrong?
6.
I found that a = all numbers that end in a 3 or an 8 and b = all numbers ending in a 2 so would that be 200 pairs? I feel that I did something wrong here as the number seems too perfect. Where did I go wrong?

[pure raptor]

@wide holly Please help 🙏

[wide holly]

4. Yes.
5. Yes.
6. Why can't b=7?

[somber parrot]

how do u do 4. btw?

[pure raptor]

Oh for 6 ur right. I had an oversight in my equation thanks.

[pure raptor]

Sorry, what do you mean by yes for question 5?

[wide holly]

It's as easy as you think it is and points 3 and 4 don't matter'

[pure raptor]

oh really... alright thanks

[austere wind]

6.324 is 18^2 so the whole thing can be written as 2^(2a+b+2)*3^(a+4b+4)=a^5 for some a
That means 2a+b+2 and a+4b+4 must be divisible by 5
Then you can use modulo 5 to find the congruence of a and b to find all of the possible pairs

[pure raptor]

yeah i did that i just did forgot to include all nums that ends in 7 works for b

[austere wind]

Personally i think q5 2 last conditions are the same

[pure raptor]

sorry, wdym by that?

[austere wind]

Oh sr i think it was the rightmost digit 🥲

[pure raptor]

o

[austere wind]

I think it might be this

[wide holly]

That's a lot of overkill. The first 4 digits sum to a multiple of 3, as do the last 4, therefore all 8 sum to a multiple of 3.

[austere wind]

I didnt realize that until i was done 🥲

[pure raptor]

The main issue for me with this question why is it a two line question...

[austere wind]

I dont understand why there exist a largest n
I can simply choose x=57 y=37 z=3*5 and then multiply the set by any integer i want then the product xyz just blows up to infinity

If minimum than n=(357)^2

[wide holly]

read the question again

n must be a factor of xyz

so n can't be larger than any number that xyz is allowed to equal

[pure raptor]

Oh also would 6 then be 400?

as I got a = all numbers that end in a 3 or an 8 and b = all numbers ending in a 2 or 7 so 20 each

[wide holly]

Yeah, I think so.

It's a bit of a weird question because you can just look at it, see a bunch of equations mod 5, and then realize that you don't have to solve any of them because they will have one solution and it doesn't matter what it is

[austere wind]

But xyz is allowed to be any big enough number
It just have to satisfy the condition
If i choose the set x,y,z to be (35q,21q,15q) them i can choose any integer q

[pure raptor]

@wide holly Also how would you approach proving 4 if it doesnt matter how many digits n has?

ecause thats what i think they are really asking

[wide holly]

Read the question again. xyz can be as large as it likes, but it can also be small and n must still divide xyz. Larger numbers don't divide smaller ones.

[wide holly]

Reduce all of the digits mod 3 and just play around with it.

I had didn't know how to solve it until I spent a few minutes writing a bunch of numbers down trying to break it.

[pure raptor]

What do you mean by this?

[wide holly]

The digts 1 and 4 are the same when all you care about is divisibility by 3

[austere wind]

Then why n cannot equal xyz ?
If xyz is a mult of n n is at least xyz

[wide holly]

600 is a multiple of 2.
2 is not at least 600

[pure raptor]

Oh i meant what do you mean by play around with it

or do you mean reduce all the digits for
51182442 it becomes 2 1 1 2 2 1 1 2?

[wide holly]

Try some numbers

Just write some numbers and check if they're happy

Then figure out what it takes for them to be happy

[pure raptor]

alright thanks 👍

oh it repeats doesnt it

[austere wind]

But 2 is not the largest

[wide holly]

xyz is a multiple of n for every value of q.

Specifically, that means that xyz is a multiple of n when q=1

So n is less than or equal to (35)(21)(15)