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If the number 200! is written in the form \(p \times 10^q\),

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by Jay@ManhattanReview » Thu Aug 01, 2019 2:49 am
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 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 \(q\).

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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by regor60 » Thu Aug 01, 2019 6:19 am
Jay@ManhattanReview wrote:
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 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 \(q\).

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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Aren't there at least 100 instances of 10^2 being multiplied between 100 and 200 alone in this factorial ?
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by regor60 » Thu Aug 01, 2019 6:22 am
regor60 wrote:
Jay@ManhattanReview wrote:
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 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 \(q\).

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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Aren't there at least 100 instances of 10^2 being multiplied between 100 and 200 alone in this factorial ?
I guess I need to constrain p to an integer, that must be the difference
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by Scott@TargetTestPrep » Tue Aug 06, 2019 4:32 pm
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

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