A number n! is written in base 6 and base 8 notation. Its base 6…

Interpretation: trailing zeros in base 6 mean n! is divisible by 6^10 = 2^10·3^10 (exactly 10 zeros means divisible by 6^10 but not by 6^11). Trailing zeros in base 8 mean n! is divisible by 8^7 = 2^21 (exactly 7 zeros means divisible by 2^21 but not by 2^24).

• Find exponent of 3 in n!: v3(n!) = floor(n/3) + floor(n/9) + floor(n/27) + ...

• Smallest n with v3(n!) ≥ 10: test n = 24 gives floor(24/3)+floor(24/9)=8+2=10, so 24! has exactly 3^10. For n = 25 and 26 v3 remains 10; at 27 v3 jumps higher. Thus the n with v3 = 10 are 24, 25, 26.

• Find exponent of 2 in n!: v2(n!) = floor(n/2)+floor(n/4)+floor(n/8)+floor(n/16)+ ...

• We need 21 ≤ v2(n!) < 24 so n! is divisible by 2^21 but not by 2^24. Compute v2 for the candidate n values: v2(24)=12+6+3+1=22, v2(25)=22, v2(26)=13+6+3+1=23. All three lie between 21 and 23 inclusive, so each of 24, 25, 26 satisfies the base-8 condition exactly.

Conclusion: the smallest n is 24, and there are 3 values of n that satisfy both conditions (n = 24, 25, 26).