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04 Nov 2010, 18:15
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n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is
A. 8
B. 12
C. 16
D. 18
E. 24
A. 8
B. 12
C. 16
D. 18
E. 24
04 Nov 2010, 18:28
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shrive555
n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is
8
12
16
18
24
please share your method. Thanks
8
12
16
18
24
please share your method. Thanks
\(k\) is the product of all integers from 1 to \(n\) inclusive implies \(k=n!\);
\(k\) is a multiple of 1440 means \(n!=1440*p=2^5*3^2*5*p\), for some integer \(p\). Thus, \(n!\) must have at least 2 in power 5, 3 in power 2 and 5 as its factors.
Now, \(7!=2*3*(2^2)*5*(2*3)*7=2^4*3^2*5*7\) not enough power of 2. The next number, \(8!=2*3*(2^2)*5*(2*3)*7*(2^3)=2^7*3^2*5*7=1440*(2^2*7)\). Therefore, the smallest value of \(n\) is 8.
Answer: A (8).
04 Nov 2010, 18:35
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shrive555
n is a positive integer, and k is the product of all integers from 1 to n inclusive. If k is a multiple of 1440, then the smallest possible value of n is
8
12
16
18
24
please share your method. Thanks
8
12
16
18
24
please share your method. Thanks
When we say k is the product of all integers from 1 to n, we mean k = n!
Now 1440 = 1.2.3.4.5.6.2
Now if k is a multiple of 1440, it has to be at least 8!. This is so because it cannot be 6! due to the extra 2 factor at the end; it cannot be 7! either, again because the extra 2 in the end will be unaccounted for.
If k = 8! = 1.2.3.4.5.6.7.8, then k = 1440*7*4 i.e. k will be a multiple of 1440.
Answer A
