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If m is a positive integer, and n is product of all integers
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07 Aug 2020, 05:07
07 Aug 2020, 05:07
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Q5. If m is a positive integer, and n is product of all integers from 1 to m, inclusive. What is the smallest possible value of m, such that n is a multiple of 2880?
A. 6
B. 7
C. 8
D. 9
E. 10
A. 6
B. 7
C. 8
D. 9
E. 10
Re: If m is a positive integer, and n is product of all integers
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07 Aug 2020, 05:52
07 Aug 2020, 05:52
1
n=m!
Prime factorization of 2880 is (2^6)(3^2)(5)
We need the smallest m! such that all these factores in the prime factorization are included.
We'll start composing a factorial until we use all the factors we have and including factors we don't have
We have six 2's, two 3's, one 5
1*(2)*(3)*(2*2)*(5)*(3*2)*7*(2*2*2)=
1*2*3*4*5*6*7*8
=8!
So the smallest factorial containing all the factors in the prime factorization of 2880 is 8!
Final Answer: C
Prime factorization of 2880 is (2^6)(3^2)(5)
We need the smallest m! such that all these factores in the prime factorization are included.
We'll start composing a factorial until we use all the factors we have and including factors we don't have
We have six 2's, two 3's, one 5
1*(2)*(3)*(2*2)*(5)*(3*2)*7*(2*2*2)=
1*2*3*4*5*6*7*8
=8!
So the smallest factorial containing all the factors in the prime factorization of 2880 is 8!
Final Answer: C
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Re: If m is a positive integer, and n is product of all integers
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02 Sep 2022, 22:27
02 Sep 2022, 22:27
1
Given that n is product of all integers from 1 to m, inclusive and we need to find what is the smallest possible value of m, such that n is a multiple of 2880
Lets start by simplifying 2880
2880 = 2*1440 = 2*10*144 = 2*2*5*12*12 = \(2^2 * 5 * 3^2 * 4^2\) = \(2^6 * 3^2 * 5\)
Now, n = product of all integers from 1 to m = 1*2*3...*m = m!
Now for m! to be a multiple of 2880, m! must have \(2^6 * 3^2 * 5\)
We need two 3's, one 5 and six 2's
If we take m = 6 then we will have
two 3's one in 3 and one in 6
one 5 in 5
but only four 2's -> one in 2, two in 4 and one in 6
But if we take m = 8 then we will have
two 3's one in 3 and one in 6
one 5 in 5
more than six 2's -> one in 2, two in 4, one in 6 and 3 in 8
So, Answer will be C
Hope it helps!
Lets start by simplifying 2880
2880 = 2*1440 = 2*10*144 = 2*2*5*12*12 = \(2^2 * 5 * 3^2 * 4^2\) = \(2^6 * 3^2 * 5\)
Now, n = product of all integers from 1 to m = 1*2*3...*m = m!
Now for m! to be a multiple of 2880, m! must have \(2^6 * 3^2 * 5\)
We need two 3's, one 5 and six 2's
If we take m = 6 then we will have
two 3's one in 3 and one in 6
one 5 in 5
but only four 2's -> one in 2, two in 4 and one in 6
But if we take m = 8 then we will have
two 3's one in 3 and one in 6
one 5 in 5
more than six 2's -> one in 2, two in 4, one in 6 and 3 in 8
So, Answer will be C
Hope it helps!
