Question

Find the largest natural number n, such that 9999...999 (999 times) is divisible by 9^n

Answer 1

\(9999\ldots (999 \text{times}) = 10^{999}-1\)

Answer 2

write 10 as 1+9 and expand using binomial theorem

Answer 3

oh, ok...
(9 + 1)^999 - 1
= 9^999 + 9^998 +9^997 + ... + 9
then ?

Answer 4

\(9999\ldots (999 \text{times}) \\~\\

= \color{red}{10^{999}}-1\\~\\
=\color{red}{(1+9)^{999}}-1\\~\\
=\color{red}{1 + \dbinom{999}{1}*9+ \dbinom{999}{2}*9^2+ \cdots +\dbinom{999}{999}*9^{999}}-1

\)

Answer 5

the answer is \(2\)
il let you figure out why :)

Answer 6

but is this same like yours ?
= 9^999 + 9^998 +9^997 + ... + 9

Answer 7

Looks you forgot the binomial coefficients

Answer 8

i know the formula of binomial newtown formula :
(a + b)^n = a^n + a^(n-1)b + a^(n-2)b^2 + ... + b^n
i supposed a = 9 and b = 1, that's why i get like above

Answer 9

your formula is wrong,
check again

Answer 10

Here is the correct formula

Answer 11

my net conection is lagging, cant read the last your coment. but yeah, i forgot the coefficient there. thanks

Answer 12

ok, got it. all terms is divisible by 9^2. :)
Thank you very much @ganeshie8

Answer 13

right, but that doesn't prove "2" is the largest exponent

Answer 14

what, .... you mean the answer is not n = 2 ???

Answer 15

answer is 2
but ur reason is wrong

Answer 16

oh.... okay i want to show you how the to get that :
(1 + 9)^999 - 1
= (999, 0) * 9^0 + (999 , 1) * 9 + (999 , 2) * 9^2 + .... + (999 , 998) * 9^998 + (999, 999) * 9^999 - 1
cancels 1's, get
= (999 , 1) * 9 + (999 , 2) * 9^2 + .... + (999 , 998) * 9^998 + (999, 999) * 9^999
= (999 * 9) + (999 * 998) * 9^2 + .... + (999 * 9^998) + (9^999)
faktor out all terms by take the GCD is 9^2
so,
= 9^2 (111 + 999 * 998 + .... + 999 * 9^996 + 9^997)

obvious 9^2 (111 + 999 * 998 + .... + 999 * 9^996 + 9^997) is divisible by 9^2
am i correct @ganeshie8 ?