Find the maximum value of $n$ such that $61 \times 42 \times 26 \times 81 \times 64 \times 49 \times 47 \times 51 \times 63 \times 16 \times 67$ is perfectly divisible by $56^n$

Break 56 into primes: $56=2^3\times7$ So we only need to know how many 2s and how many 7s the big product has. | Number | 2s it gives | 7s it gives | |---|---|---| | $42=2\times3\times7$ | 1 | 1 | | $26=2\times13$ | 1 | 0 | | $64=2^6$ | 6 | 0 | | $16=2^4$ | 4 | 0 | | $49=7\times7$ | 0 | 2 | | $63=7\times9$ | 0 | 1 | The remaining numbers $61$, $81$, $47$, $51$ and $67$ contain neither 2 nor 7. $\text{Total 2s}=1+1+6+4=12$ $\text{Total 7s}=1+2+1=4$ Every single 56 needs three 2s and one 7. The 2s allow $\dfrac{12}{3}=4$ such groups. The 7s allow $\dfrac{4}{1}=4$ such groups. We must take the smaller of the two, so $n=4$