3²=9=10-1, so 3²ⁿ=(3²)ⁿ=(10-1)ⁿ=10ⁿ-10ⁿ⁻¹n+10ⁿ⁻²n(n-1)/2!-... for any n.
The terms at the end of this expansion are 100n(n-1)/2!-10n+1.
1, 3, 9, 27, ... is a series of ascending powers of 3 starting with 3⁰=1. Therefore, if the last two digits of a power of 3 are 01, we have a cyclic series of pairs of digits. That is, 3ˣ=...01 for some x=2n, where n is a non-negative integer. From the above binomial expansion we can see that if n=10 we have the last terms:
4500-100+1=4401. n=10 is the minimum value for which the last two digits can be 01. No other terms can alter these 4 digits because the next term would be -1000n(n-1)(n-2)/3!=-120000, which cannot affect the last 4 digits. x=2n=20, so 3²⁰ ends in 01. We can write:
3²⁰=100X+1, where X is some integer (X=34867844 in fact).
3³=27 (t₁), t₂=3²⁷=3²⁰⁺⁷=3²⁰3⁷=2187(100X+1)=218700X+2187. Clearly, the last two digits of this must be 87. (1+100X)ⁿ=1+100nX+n(n-1)/2!(100X)²+... The second and subsequent terms are integers involving powers of 10 of at least 2, so the last two digits cannot be affected by these terms and must also be 01, and since 1+100X=3²⁰, 3²⁰ⁿ=(1+100X)ⁿ also ends in 01.
So, t₃’s exponent ends in 87, but 87=4×20+7, so this exponent is 7 more than a multiple of 20, making t₃ also end in 87.
Consider (1+100X)¹⁰ⁿ=1+1000nX+10(100X)²n(10n-1)/2!+...
The third and subsequent terms of this expansion cannot affect the last three digits of (3²⁰)¹⁰ⁿ=3²⁰⁰ⁿ, which must be 001 for n>0. But let’s also look at (1+100X)⁵ⁿ=1+500nX+... remembering that we already know the last 4 digits of 3²⁰ are 4401, so X ends in 44, so X is divisible by 4. So 500nX=2000n(X/4) and X/4 is an integer. Therefore, when n=1, (3²⁰)⁵=3¹⁰⁰ ends 001 and so does 3¹⁰⁰ⁿ by implication.
(1+100X)⁵ⁿ=1+2000n(X/4)+5n(5n-1)(100X)²/2!+...
5n(5n-1)/2! is always an integer whether n is even or odd.
Let’s expand the third term:
(5n(5n-1)/2)(10000X²).
When n=1, we have 10⁵X² which cannot affect the last 5 digits, so, if we expand the first two terms we get 1+2000(X/4)=1+1000(X/2)=...22001 or ...72001 (in fact, it’s ...22001). We can be sure of the last 4 digits: 2001.
First, let’s look at the last 3 digits of t₃. We know the last two are 87 and we know the last 4 digits of 3²⁰ are 4401. We need to know the last three digits of 3⁸⁰. The last 2 are 01. Consider 3²⁰⁽ⁿ⁺¹⁾-3²⁰ⁿ=3²⁰ⁿ(3²⁰-1) after factorising. We can write this 4400(100X+1)=440000X+4400. So we can work out the last three digits of 3⁸⁰ by progressively adding 4400. 3²⁰ ends 401, 3⁴⁰=401+400=801, 3⁶⁰=801+400=201, 3⁸⁰=201+400=601. We need to multiply this by 3⁷=2187=367 in the last 3 digits, which are the last 3 digits of t₃.
Let’s recap on what we’ve got so far, using a slight variation in notation.
3²⁰ⁿ=9¹⁰ⁿ=(10-1)¹⁰ⁿ=(1-10)¹⁰ⁿ=1+100X₀,
t₂=3²⁷=3⁷(3²⁰)=2187(1+100X₀)=2187+218700X₀=...87=...80+7,
t₃=3^t₂=3^(...80+7)=2187(3²⁰ⁿ)=2187(1+100X₁)=2187+218700X₁=...87,
X₀=...44, an even number,
(1+100X₀)⁵ⁿ=1+1000(X₀/2)+...=1+1000X₁,
3¹⁰⁰ⁿ=(1+100X₁)¹⁰ⁿ=1+1000X₂.
The process can be continued: (1+1000X₂)¹⁰ⁿ=1+10000X₃ for which we can find a t term close to 3¹⁰⁰⁰ⁿ.
More to follow...