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dracarys007
Joined: 04 Jun 2020
Last visit: 24 Feb 2022
Posts: 68
Location: India
Concentration: Strategy, General Management
Schools: Harvard '23 INSEAD Aug 22 intake Sloan '23
GPA: 3.4
WE:Engineering (Consulting)
Schools: Harvard '23 INSEAD Aug 22 intake Sloan '23
Posts: 68
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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30%
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If the number 100! is written in the form p × 2 ^q, where p and q are integers, what is
the maximum possible value of q?
A) 97
B) 98
C) 99
D) 100
E) Can not be determined
the maximum possible value of q?
A) 97
B) 98
C) 99
D) 100
E) Can not be determined
Posts: 4,849
To find the number of 2's , 3's or any prime number in a factorial say N!, we keep dividing N (not the factorial value of N, but only N) by the prime number. If the quotient obtained is greater than or equal to the divisor, we keep dividing each quotient by the prime number until we get a value of the quotient less than the divisor. The sum of all the quotients gives us the required value
For eg. how many 3's are there in 100!
Then 100 ÷ 3 = 33 (we need the quotient only, not the remainder.)
Since 33 is > 3, then 33 ÷ 3 = 11
Since 11 is > 3, then 11 ÷ 3 = 3
Since 3 = 3, then 3 ÷ 3 = 1
Therefore the number of 3's in 100! = 33 + 11 + 3 + 1 = 48
This method becomes very important when we need to find the number of trailing 0's in a factorial. In that case the divisor is 5. For eg how many trailing 0's are there in 1000!
1000 ÷ 5 = 200
240 ÷ 5 = 48
48 ÷ 5 = 9
9 ÷ 5 = 1
Therefore there are 249 0's at the end of 1000!.
________________________________________
In the question, we need to find the number of 2's in 100!
100 ÷ 2 = 50
50 ÷ 2 = 25
25 ÷ 2 = 12
12 ÷ 2 = 6
6 ÷ 2 = 3
3 ÷ 2 = 1
Therefore the number of 2's = 50 + 25 + 12 + 6 + 3 + 1 = 97
Option A
Arun Kumar
For eg. how many 3's are there in 100!
Then 100 ÷ 3 = 33 (we need the quotient only, not the remainder.)
Since 33 is > 3, then 33 ÷ 3 = 11
Since 11 is > 3, then 11 ÷ 3 = 3
Since 3 = 3, then 3 ÷ 3 = 1
Therefore the number of 3's in 100! = 33 + 11 + 3 + 1 = 48
This method becomes very important when we need to find the number of trailing 0's in a factorial. In that case the divisor is 5. For eg how many trailing 0's are there in 1000!
1000 ÷ 5 = 200
240 ÷ 5 = 48
48 ÷ 5 = 9
9 ÷ 5 = 1
Therefore there are 249 0's at the end of 1000!.
________________________________________
In the question, we need to find the number of 2's in 100!
100 ÷ 2 = 50
50 ÷ 2 = 25
25 ÷ 2 = 12
12 ÷ 2 = 6
6 ÷ 2 = 3
3 ÷ 2 = 1
Therefore the number of 2's = 50 + 25 + 12 + 6 + 3 + 1 = 97
Option A
Arun Kumar
General Discussion
dracarys007
\(Power of 2 in 100! = [\frac{100}{2}] + [\frac{100}{2^2}] + [\frac{100}{2^3}] + [\frac{100}{2^4}] + [\frac{100}{2^5}] + [\frac{100}{2^6}] = 50+25+12+6+3+1 = 97\)
\(100! = 2^{97}*p\)
ANswer: Option A
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dracarys007
we need maximum value of q
this means we have to consider p as 1
100/2+100/4+100/8+100/16+100/32+100/64=50+25+12+6+3+1=97
ANS:A
how to use this formula: https://gmatclub.com/forum/everything-a ... 85592.html
kawal27
P is definitely not 1 because it is other prime numbers apart from 2 to some powers...
Posted from my mobile device
Posts: 29
GMATinsight could you please explain the role of p here, if any? What could 'p' be and do we need to be mindful of anything w.r.t it or can we ignore it?
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