Discussion
What is the highest power of 45 that divides 100! ?
Answer: A
The prime factorization of 45 is \(3^2 \times 5\). We need to find the number of pairs of (two 3s) and (one 5) in 100!. Let's find the powers of 3 and 5. Power of 5 = \(\lfloor 100/5 \rfloor + \lfloor 100/25 \rfloor = 20+4=24\). Power of 3 = \(\lfloor 100/3 \rfloor + \lfloor 100/9 \rfloor + \lfloor 100/27 \rfloor + \lfloor 100/81 \rfloor = 33+11+3+1 = 48\). We need two 3s for each 45. We have 48 threes, so we can make \(\lfloor 48/2 \rfloor = 24\) pairs of 3s. We have 24 fives. The number of 45s is limited by the smaller count, which is 24 (from both the 3s and the 5s). So the highest power is 24.
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