help

[orchid mortar]

Hello!

[orchid mortar]

Number Theory is very easy at learning, let me interpret the answer to this question

We can begin by considering the prime factorization of 9999, which is 3^2 * 11 * 101.

That is, we let n = 2^a * 5^b, where a and b are non-negative integers.

Then, the prime factorization of 9999n is given by:

9999n = 3^2 * 11 * 101 * 2^a * 5^b

So, we need to find the smallest non-negative integers a and b such that:

a < 2
b < 2
2^a * 5^b > 1

Which gives:

n = 2^1 * 5^0 = 2

Therefor the smallest positive integer is 2

@ivory mist

[ivory mist]

but it says 9999n does not contain any digit 9 but 9999x2 is 19,998

[modern crescent]

@orchid mortar

I checked up until 30, it doesnt work

[orchid mortar]

Oh wait, sorry sorry, let me rephrase my solutions

[ivory mist]

9999 x 1112 does not contain 9

[orchid mortar]

Oh wait your right,

we can see that 9999 divided by 13 gives a quotient of 769, and the product 9999 x 769 = 7699231 does not contain the digit 9.

[modern crescent]

Bro did you brute force it?

[ivory mist]

no lol

[modern crescent]

Ok

[ivory mist]

9999 x 769 = 7689231

[modern crescent]

There is a 9

[bleak holly]

bro is using chatgpt

[modern crescent]

How did you find it out🤨

[bleak holly]

"and the product 9999 x 769 = 7699231 does not contain the digit 9" who else but chatgpt would say something like that

[ivory mist]

chat gpt gave me 3 answers the 3 times i ran it. all were wrong

[modern crescent]

Lol

@orchid mortar did you watch aot s4p3e1?

[orchid mortar]

that's sad

[modern crescent]

Yeah

@ivory mist 1112

Is your answer

You were right

I made a program to calculate that

[ivory mist]

how did u confirm it?

[modern crescent]

I basically calculated 1 to 2000 * 1112

And checked if 9 was in the digit

[orchid mortar]

Wait wait

[modern crescent]

The first number which didn't have 9 was 1112 * 9999

[orchid mortar]

nvm nvm

continue

[modern crescent]

It has 3 9s

[ivory mist]

calculators are not allowed in the past paper so their must be some way to do this on paper

[modern crescent]

Oh

[ivory mist]

im guessing this can be done with number theory divisibility

[bleak holly]

maybe you can construct an argument where 9999n = 10000n - n

then consider a few possibilities

if n is 2 digits large, then there always be a 9 there

because then some 10000 - n will have 99xx

if it's 3 digits large, then by the same logic, it'll have a 9xxx

so it has to be 4 digits large or bigger

now for 4 digits, if we try any of the 10xx, we will get a 9 from the carry of 10000-10xx

so it has to be at least 11xx

and likewise, trying to avoid carries, we get 111x

if the x is a 1, then we will get a 9, but a 2 works

so 1112

@ivory mist

[craggy swan]

Never mind

My method gives but not the very smallest

[ivory mist]

@bleak holly thanks ur method is definitely the simplest

[rigid light]

.solved