BTGmoderatorLU wrote:Source: Manhattan Prep
If the number 200! is written in the form \(p \times 10^q\), where \(p\) and \(q\) are integers, what is the maximum possible value of \(q\)?
A. 40
B. 48
C. 49
D. 55
E. 64
The OA is C
We want to find the number of factors of 10 in 200! We do this by looking for 5-and-2 pairs (since each 5-and-2 pair = 10). Thus, we need to determine the number of 5-and-2 pairs in 200! Since there are fewer 5's than 2's, we can determine the number of 5's, and that will tell us the number of 5-and-2 pairs in 200!.
To determine the number of 5s n 200!, we can use the following shortcut in which we divide 200 by 5, then divide the quotient of 200/5 by 5 and continue this process until we no longer get a nonzero quotient.
200/5 = 40
40/5 = 8
8/5 = 1 (we can ignore the remainder)
Since 1/5 does not produce a nonzero quotient, we can stop.
The final step is to add up our quotients; that sum represents the number of factors of 5 within 200!.
Thus, there are 40 + 8 + 1 = 49 factors of 5 within 200!, and that means there are 49 5-and-2 pairs, which also means that the maximum value of q is 49.
Answer: C
