If the number A = 81 × 117 × 63 × 46 is divisible by 3b, then find the maximum…

To find the highest power of a prime that divides a product of numbers, factorize every factor into primes and add up the exponents of that prime across all the factors — prime-factorization exponents are additive under multiplication, since 3m × 3n = 3(m+n).

1. Factorize 81: 81 = 3 × 3 × 3 × 3 = 34.

2. Factorize 117: 117 = 9 × 13 = 32 × 13.

3. Factorize 63: 63 = 9 × 7 = 32 × 7.

4. Factorize 46: 46 = 2 × 23 — no factor of 3 at all.

5. Add the exponents of 3 from every factor: 4 + 2 + 2 + 0 = 8.

Independent check: 81 alone already contributes 34, while 117 and 63 each add another 32 (since 117 = 9 × 13 and 63 = 9 × 7 are both multiples of 9 but not 27), and 46 adds none. Summing these confirms A contains exactly 38 as a factor.

So the maximum value of b for which 3b divides A is 8.