Consider an arithmetic series and a geometric series having four initial terms from the set $$\{11, 8, 21, 16, 26, 32, 4\}$$. If the last terms of these series are the maximum possible four digit numbers, then the number of common terms in these two series is equal to ________.

The set is $$\{11, 8, 21, 16, 26, 32, 4\}$$. We need to find four terms forming an AP and four terms forming a GP from this set.

For the AP, we look for four numbers in arithmetic progression. Trying $$\{11, 16, 21, 26\}$$ — these are all in the set with common difference $$d = 5$$. This is the AP: $$11, 16, 21, 26$$ with $$d = 5$$.

For the GP, we need four terms in geometric progression. Checking: $$\{4, 8, 16, 32\}$$ — all in the set with common ratio $$r = 2$$. This is the GP.

The AP with first term $$a = 11$$ and $$d = 5$$ gives general term $$a_n = 11 + (n-1) \cdot 5 = 5n + 6$$. The largest 4-digit number in this AP: $$5n + 6 \leq 9999$$ gives $$n \leq 1998.6$$, so the last term is $$5(1998) + 6 = 9996$$.

The GP with first term $$a = 4$$ and $$r = 2$$ gives general term $$a_n = 4 \cdot 2^{n-1} = 2^{n+1}$$. The largest 4-digit number: $$2^{n+1} \leq 9999$$, so $$n + 1 \leq 13$$ (since $$2^{13} = 8192$$ and $$2^{14} = 16384$$). The last term is $$2^{13} = 8192$$.

We need common terms. AP terms satisfy $$a = 5k + 6$$ for positive integer $$k$$, and GP terms satisfy $$a = 2^m$$ for $$m \geq 2$$. We need $$2^m = 5k + 6$$, i.e., $$2^m \equiv 1 \pmod{5}$$.

The powers of 2 modulo 5 cycle as: $$2^1 = 2, 2^2 = 4, 2^3 = 3, 2^4 = 1, 2^5 = 2, \ldots$$ with period 4. So $$2^m \equiv 1 \pmod{5}$$ when $$m \equiv 0 \pmod{4}$$.

The valid values of $$m$$ (where $$2 \leq m \leq 13$$ and $$m \equiv 0 \pmod 4$$) are: $$m = 4, 8, 12$$, giving terms $$16, 256, 4096$$.

Verification: $$16 = 5(2) + 6$$ ✓, $$256 = 5(50) + 6$$ ✓, $$4096 = 5(818) + 6$$ ✓.

The number of common terms is $$3$$.