Day of the week 10^10 days from Sunday

To determine the day of the week, evaluate the number of shifts modulo 7: ⟨MATH⟩10^10 ± od 7⟨/MATH⟩ . First, simplify the base: 10 \equiv 3 ± od 7. Therefore, 10^10 \equiv 3^10 ± od 7.

Analyze the cycles of powers of 3 modulo 7: 3^1 = 3 \equiv 3 ± od 7 3² = 9 \equiv 2 ± od 7 3³ = 6 \equiv 6 ± od 7 3^4 = 18 \equiv 4 ± od 7 3^5 = 12 \equiv 5 ± od 7 3^6 = 15 \equiv 1 ± od 7.

The remainders cycle every 6 steps (Fermat's Little Theorem confirming 3^6 \equiv 1 ± od 7). Divide the exponent 10 by the cycle length 6: 10 = 6(1) + 4 (remainder of 4). This matches the 4th position in the cycle, which is 4. Shifting Sunday forward by 4 days yields: Monday, Tuesday, Wednesday, Thursday.

Answer: (b).