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
If the number 200! is written in the form \(p \times 10^q\),
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We have to get the value of \(q\) where \(q\) is the exponent of \(10^q\). Thus, we have to count the number of 10s in 200!. Note that 10 = 2*5; thus, we must count the number of 2s and the number of 5s in 200!. Since 5 > 2, the number of 5s will be fewer than the number of 2s in 200!. So, the number of 5s will decide the maximum possible value of \(q\).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
To count the number of 5s in 200!, we must keep dividing 200 by 5 and counting the quotients upon the divisions.
"¢ 200/5 = 40;
"¢40/5 = 8;
"¢8/5 = 1
Since 1 < 5, we must stop here.
Thus, the number of 5s in 200! = 40 + 8 + 1 = 49; thus, the maximum possible value of \(q\) = 49.
The correct answer: C
Hope this helps!
-Jay
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Jay@ManhattanReview wrote:We have to get the value of \(q\) where \(q\) is the exponent of \(10^q\). Thus, we have to count the number of 10s in 200!. Note that 10 = 2*5; thus, we must count the number of 2s and the number of 5s in 200!. Since 5 > 2, the number of 5s will be fewer than the number of 2s in 200!. So, the number of 5s will decide the maximum possible value of \(q\).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
To count the number of 5s in 200!, we must keep dividing 200 by 5 and counting the quotients upon the divisions.
"¢ 200/5 = 40;
"¢40/5 = 8;
"¢8/5 = 1
Since 1 < 5, we must stop here.
Thus, the number of 5s in 200! = 40 + 8 + 1 = 49; thus, the maximum possible value of \(q\) = 49.
The correct answer: C
Hope this helps!
-Jay
_________________
Manhattan Review GMAT Prep
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Aren't there at least 100 instances of 10^2 being multiplied between 100 and 200 alone in this factorial ?
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I guess I need to constrain p to an integer, that must be the differenceregor60 wrote:Jay@ManhattanReview wrote:We have to get the value of \(q\) where \(q\) is the exponent of \(10^q\). Thus, we have to count the number of 10s in 200!. Note that 10 = 2*5; thus, we must count the number of 2s and the number of 5s in 200!. Since 5 > 2, the number of 5s will be fewer than the number of 2s in 200!. So, the number of 5s will decide the maximum possible value of \(q\).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
To count the number of 5s in 200!, we must keep dividing 200 by 5 and counting the quotients upon the divisions.
"¢ 200/5 = 40;
"¢40/5 = 8;
"¢8/5 = 1
Since 1 < 5, we must stop here.
Thus, the number of 5s in 200! = 40 + 8 + 1 = 49; thus, the maximum possible value of \(q\) = 49.
The correct answer: C
Hope this helps!
-Jay
_________________
Manhattan Review GMAT Prep
Locations: Geneva GMAT Prep | Free TOEFL Practice Questions | LSAT Prep Courses Toronto | SAT Prep Courses Hong Kong | and many more...
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Aren't there at least 100 instances of 10^2 being multiplied between 100 and 200 alone in this factorial ?
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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!.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
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
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