Question:

<p><strong>Step 1:</strong> General term in expansion of $(5^{1/6} + 2^{1/8})^{100}$ is:</p><p>$$T_{r+1} = \binom{100}{r}(5^{1/6})^{100-r}(2^{1/8})^r = \binom{100}{r} \cdot 5^{(100-r)/6} \cdot 2^{r/8}$$</p><p><strong>Step 2:</strong> For a term to be rational, both exponents must be integers:</p><p>• $(100-r)/6$ is an integer ⟹ $100-r \equiv 0 \pmod{6}$ ⟹ $r \equiv 4 \pmod{6}$</p><p>• $r/8$ is an integer ⟹ $r \equiv 0 \pmod{8}$</p><p><strong>Step 3:</strong> Using Chinese Remainder Theorem, find r satisfying both conditions:</p><p>From $r \equiv 4 \pmod{6}$: $r = 6k + 4$</p><p>Substitute into $r \equiv 0 \pmod{8}$: $6k + 4 \equiv 0 \pmod{8}$</p><p>$6k \equiv -4 \equiv 4 \pmod{8}$</p><p>$6k \equiv 4 \pmod{8}$ ⟹ $k \equiv 6 \pmod{4}$ ⟹ $k \equiv 2 \pmod{4}$</p><p>So $k = 4m + 2$, giving $r = 6(4m+2) + 4 = 24m + 16$</p><p><strong>Step 4:</strong> Find valid values: $0 \leq 24m + 16 \leq 100$</p><p>$-16 \leq 24m \leq 84$ ⟹ $m = 0, 1, 2, 3$</p><p>Rational terms: $r = 16, 40, 64, 88$ (4 terms)</p><p><strong>Step 5:</strong> Irrational terms = Total terms − Rational terms = $101 - 4 = 97$</p><p>∴ Answer: <strong>97</strong></p>