Power of 3 with a string of 2008 consecutive zeros?

Yes, it does:

Lemma 1: 9^10 mod 100 = 1

Proof: By binomial expansion,

9^10 = (10-1)^10 = sum[n=0 to 10] (10Cn)*10^n*(-1)^(10-n)

The terms for n > 1 are all multiplies of 100 due to the 10^n. So, let's drop them to get

9^10 == (10C1) * 10^1 * (-1)^9 + (10C0) * 10^0 * (-1)^10 = -100 + 1 == 1 (mod 100)

QED.

Note: It is equal to 3486784401.

Lemma 2: If a == 1 (mod 10^k), then a^10 == 1 (mod 10^(k+1)).

Proof: Let a = 10^k*l+1. Use binomial expansion once more to get

a^10 = sum[n=0 to 10] (10Cn) * 10^(nk)

Again, the terms for n > 1 are multiplies of 10^(2k) and thus also of 10^(k+1). Dropping them, we obtain

a^10 == (10C1) * 10^k + (10C0) * 10^0 = 10*10^k + 1 = 10^(k+1) + 1 == 1 (mod 10^(k+1))

QED

Theorem: The number 9^(10^k) = 3^(2*10^k) is congruent to 1 modulo 10^(k+1).

Proof: By mathematical induction.

Step 1: Let k = 1. Then 9^10 == 1 (mod 10^2) is exactly what Lemma 1 says.

Step 2: Let's assume that 9^(10^k) == 1 (mod 10^(k+1)) for some k in N. We'll need to show that 9^(10^(k+1)) == 1 (mod 10^(k+2)).

Plug a = 9^(10^k) and k+1 instead of k into Lemma 2. This lemma then says that a^10 == 1 (mod 10^(k+1+1)). Rewriting, a^10 is (9^(10^k))^10 = 9^(10*10^k) = 9^(10^(k+1)), so we got

9^(10^(k+1)) == 1 (mod 10^(k+2))

exactly as we needed.

Note: Aside from mathematical language, the above meant:

9^10 = (...)01

9^100 = (...)001

9^1000 = (...)0001

and so on. Let's continue until there are 2008 zeros before the last 1, obtaining the:

Corollary: 9^(10^2008), or 3^(2*10^2008), contains a string of 2008 consecutive zeros.

Proof: out Theorem above tells us that 9^(10^2008) == 1 (mod 10^2009), which means its end is really 2008 zeros followed by a single 1.

So the answer is YES, an example of such number is 3^(2*10^2008). Smaller answers may of course exist, but we are asked for a yes/no answer.

Really interesting!

EDIT: as you did not explicitly say "3^n where n is a positive integer", you could obtain various funny answers like 3^(2008*ln(10)/ln(3)).

This was a joke, of course.

Yes but it's pretty big. The key is that 3^20 = 3486784401. Now any multiple of the power 20 (40,60,80,100...) will also retain the 01 at the end. Now 3^100 will end in ...52001 so two zeros. So any multiple of 100 will end 001.

3^500 ends with ...0001

3^5000 ends in 00001.

so 3^5with 2007 zeros will end with at least 2008 zeros and a one.

I see that above answers are about making 2008 zeros near the tail, in the least significant digits.

Alternatively one can easily show that 3^n will for some n begin with

10000(2008 zeros)000xyz...

in most significant digits, simply because log(3) is irrational number.

Allah in his wisdom inserts zeros from either direction.

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