probability questions

[echo cairn]

a random 6 digit number is made from only the digits 1 & 8, then the probability of it being divisible by 21 is p, what is 96p?

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[drowsy stream]

What have you tried?

[vernal rivet]

a number being divisible by 21 means a number is divisible by both 3 and 7.

• divisibility of 3 can check if all the digits add up to a number divisible by 3
  e.g. 188811 divisibly by 3 since 1 + 8 + 8 + 8 + 1 + 1 = 27 which is divisible by 3

• divisibility of 7 can be checked for 6 digits numbers by finding the difference between the first 3 digit number and the last 3 digit number, and checking if the result is also divisible by 7 (if your result is 0, then it is also divisible)
  e.g. 888181 is divisible by 7 since 888 - 181 = 707 which is divisible by 7.

maybe you can use these tools to help solve your problem. There's also 2^6 = 64 possible numbers that can be formed from only 1's and 8's

[orchid swallow]

8⁶

[timber creek]

there are 64 possible combinations

we can minimise the case by finding number of 1s and 8s possible. Hint:- using modular equations

[timber creek]

u talking about the answer? it's a 2 digit number

[vernal rivet]

idk why i thought 2^6 = 32 in my head lol ye 64

[timber creek]

i find the problem interesting.
we can like say that number of 1s is x and number of 8s is y.
x+y=6
and x+8y = 0 (mod 3)

and stuff

[orchid swallow]

No I was just correcting the former sentence.

[timber creek]

it is 2^6.

64

[orchid swallow]

Oh yeah I misread.

[vernal rivet]

maybe can show for divisibility of 3 to be true, need to have either all 1's, all 8's, or three 8's and three 1's

then narrow it down to divisibility of 7 from there

[timber creek]

we can find the possible values of number of 1s or 8s using modular equations

that will narrow down i assume atleast half of the cases

[vernal rivet]

yeah i honestly tho don't know what the OP knows and doesn't know so im doing this kinda in the 'dumb' way

all 1's work and all 8's work , but i dont know how to prove that any combination of three 8's and three 1's will always be divisible by 7 and thus 21

oh wait nvm im dumb i figured it out lol

[timber creek]

we arent trying out each and every combination

even though thats a way but i dont suggest doing that

[vernal rivet]

from what i originally wrote, you can show that literally any combination of 8's and 1's will always produce a 6 digit number that is divisible by 7
and all 1's is can divide 3, all 8's can divide 3, three 1's and three 8's can divide 3
so that's 1 + 1 + 6!/(3! * 3!)

so 22 ?

[timber creek]

yes

[vernal rivet]

ye tho i used divisibility rules of 3 and 7 for this. divisibility rule of 7 is weird and uncommon

curious what a modular way would go about doing that

[timber creek]

well as i thought, i would prolly find out number of 1s and 8s that work for 3 then check for those to work for 7

[drowsy stream]

10 == 3 (mod 7)
10^2 == 2 (mod 7)
10^3 == -1 (mod 7)```Therefore```
N = sum(n = 0, i) 10^n a_n
==> N == sum(n = 0, i/3) (-1)^n (a_(3n) + 3a_(3n + 1) + 2a_(3n + 2)) (mod 7)```

And in fact, 1 == 8 (mod 7) so if any of these strings is divisible by 7, they all are.

[timber creek]

1==8 (mod7). isnt it like 8==1(mod7)

[drowsy stream]

...congruence mod n is symmetric.

[timber creek]

i dont use them in daily basis i might be wrong

[timber creek]

oh

[drowsy stream]

It's an equivalence relation.

[timber creek]

okay.

[echo cairn]

+close