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08 Aug 2010, 08:49
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
64% (01:34) correct
36%
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How many powers of 900 are in 50!
A) 2
B) 4
C) 6
D) 8
E) 10
A) 2
B) 4
C) 6
D) 8
E) 10
Can one explain this answer clearly? This is actually a post in the GMATClub Math Tutorial (I don't know how to paste the link, sorry, am new). It says at the end that...
"We need all the prime {2,3,5} to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50!"
I did not understand this. What does "5 can provide us with only 6 pairs" mean? Is the answer only driven by that? What about 2 and 3? And if the powers had been all different for the original number say X = 2^4 3^7 5^9, then what?
"We need all the prime {2,3,5} to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50!"
I did not understand this. What does "5 can provide us with only 6 pairs" mean? Is the answer only driven by that? What about 2 and 3? And if the powers had been all different for the original number say X = 2^4 3^7 5^9, then what?
08 Aug 2010, 08:59
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mainhoon
This is from my topic: math-number-theory-88376.html or everything-about-factorials-on-the-gmat-85592.html
If you have a problem understanding it don't worry, you won't need it for GMAT.
There is a following solution:
How many powers of 900 are in 50!
\(900=2^2*3^2*5^2\)
Find the power of 2:
\(\frac{50}{2}+\frac{50}{4}+\frac{50}{8}+\frac{50}{16}+\frac{50}{32}=25+12+6+3+1=47\)
= \(2^{47}\)
Find the power of 3:
\(\frac{50}{3}+\frac{50}{9}+\frac{50}{27}=16+5+1=22\)
=\(3^{22}\)
Find the power of 5:
\(\frac{50}{5}+\frac{50}{25}=10+2=12\)
=\(5^{12}\)
We need all of them (2,3,5) to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50!
900^6
To elaborate:
\(50!=900^xa=(2^2*3^2*5^2)^x*a\), where \(x\) is the highest possible value of 900 and \(a\) is the product of other multiples of \(50!\).
\(50!=2^{47}*3^{22}*5^{12}*b=(2^2*3^2*5^2)^6*(2^{35}*3^{10})*b=900^{6}*(2^{35}*3^{10})*b\), where \(b\) is the product of other multiples of \(50!\). So \(x=6\).
Below is another example:
Suppose we have the number \(18!\) and we are asked to to determine the power of \(12\) in this number. Which means to determine the highest value of \(x\) in \(18!=12^x*a\), where \(a\) is the product of other multiples of \(18!\).
\(12=2^2*3\), so we should calculate how many 2-s and 3-s are in \(18!\).
Calculating 2-s: \(\frac{18}{2}+\frac{18}{2^2}+\frac{18}{2^3}+\frac{18}{2^4}=9+4+2+1=16\). So the power of \(2\) (the highest power) in prime factorization of \(18!\) is \(16\).
Calculating 3-s: \(\frac{18}{3}+\frac{18}{3^2}=6+2=8\). So the power of \(3\) (the highest power) in prime factorization of \(18!\) is \(8\).
Now as \(12=2^2*3\) we need twice as many 2-s as 3-s. \(18!=2^{16}*3^8*a=(2^2)^8*3^8*a=(2^2*3)^8*a=12^8*a\). So \(18!=12^8*a\) --> \(x=8\).
Again don't worry about this examples too much.
21 Jan 2017, 23:44
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To quickly solve such a problem, you need to factorise 900 (in this case). 900=2^2*3^2*5^2. You just need to find out the highest power of the largest prime number in 50!. As all other smaller primes will obviously be there in 50! if the largest prime is present. So 50! has 12 5's. So we have 6 pairs of 5's in 50!. Job done! 50! has 6 900s. :D
