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27 Jul 2019, 06:33
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I am reading the club math book.. and i dont understand the below. especially the last line. can some kind soul explain why we got 6 as the answer.. i thought we are trying to find powers of non prime. why did we only choose 5 and not 3 and 5
------- begin text-------
Finding the power of non-prime in n!:
How many powers of 900 are in 50!
Make the prime factorization of the number: 900=22∗32∗52900=22∗32∗52, then find the powers of these prime numbers in the n!.
Find the power of 2:
50/2+50/4+50/8+50/16+50/32 =25+12+6+3+1=47
= 2^47
Find the power of 3:
50/3+50/9+50/27=16+5+1=16+5+1=22
=3^22
Find the power of 5:
50/5+50/25=10+2=12
=5^12
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 do not understnad the last line conclusion.
------- begin text-------
Finding the power of non-prime in n!:
How many powers of 900 are in 50!
Make the prime factorization of the number: 900=22∗32∗52900=22∗32∗52, then find the powers of these prime numbers in the n!.
Find the power of 2:
50/2+50/4+50/8+50/16+50/32 =25+12+6+3+1=47
= 2^47
Find the power of 3:
50/3+50/9+50/27=16+5+1=16+5+1=22
=3^22
Find the power of 5:
50/5+50/25=10+2=12
=5^12
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 do not understnad the last line conclusion.
Updated on: 27 Jul 2019, 07:09
Originally posted by TestPrepUnlimited on 27 Jul 2019, 06:55.
Last edited by TestPrepUnlimited on 27 Jul 2019, 07:09, edited 1 time in total.
Last edited by TestPrepUnlimited on 27 Jul 2019, 07:09, edited 1 time in total.
Kudos
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Basically for each factor of 900, you need two 2s, two 3s and two 5s.
Since 50! has 47 2s, 22 3s, and 12 5's,
the number of 5's is the limiting factor.
50! has 12 5s, and for each 900 you need 2 5s.
12/2=6
Thus there are 6 powers of 900 in 50!.
Let us know if you have further questions!
Since 50! has 47 2s, 22 3s, and 12 5's,
the number of 5's is the limiting factor.
50! has 12 5s, and for each 900 you need 2 5s.
12/2=6
Thus there are 6 powers of 900 in 50!.
Let us know if you have further questions!
27 Jul 2019, 07:03
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TestPrepUnlimited
Basically for each factor of 900, you need two 2s, two 3s and two 5s.
Since 50! has 47 2s, 22 3s, and 12 5's,
the number of 5's is the limiting factor.
50! has 12 5s, and for each 900 you need 12 5s.
12/2=6
Thus there are 6 powers of 900 in 50!.
Let us know if you have further questions!
Thanks for your kind explanation Since 50! has 47 2s, 22 3s, and 12 5's,
the number of 5's is the limiting factor.
50! has 12 5s, and for each 900 you need 12 5s.
12/2=6
Thus there are 6 powers of 900 in 50!.
Let us know if you have further questions!
50! has 12 5s, and for each 900 you need 12 5s.
12/2=6
is there a typo here.. did you mean each 900 you need 2 5s? and not 12 5s.
