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If m is a positive integer, and n is product of all integers from 1 to

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If m is a positive integer, and n is product of all integers from 1 to [#permalink] If m is a positive integer, and n is product of all integers from 1 to   Updated on: 12 Oct 2024, 05:20
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Q57. 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


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Official Answer
C

Originally posted by QuantMadeEasy on 07 Aug 2020, 06:04.
Last edited by QuantMadeEasy on 12 Oct 2024, 05:20, edited 1 time in total.
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Re: If m is a positive integer, and n is product of all integers from 1 to [#permalink] If m is a positive integer, and n is product of all integers from 1 to   07 Aug 2020, 06:27
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Given

    • If m is a positive integer, and n is a product of all integers from 1 to m, inclusive.

To Find

    • The smallest possible value of m, such that n is a multiple of 2880.

Approach and Working Out

    • Product of all the integers from 1 to m is also denoted by m!
    • 2880 = \(2^6\) ×\( 3^2\) × 5
      o So, we must have six 2s, two 3s and one 5.
      o m must be 5 or more as we need a 5.
    • No of 2s in 5! is 2 + 1 = 3
      o The next multiple of 2 is 6 which has only one 2.
      o So we must go till 8.
    • 8! has 4 + 2 + 1 = 7 2s.
      o It has one 5.
      o It has two 3s.
    • It must be our answer.

Correct Answer: Option C
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Re: If m is a positive integer, and n is product of all integers from 1 to [#permalink] If m is a positive integer, and n is product of all integers from 1 to   07 Aug 2020, 08:15
IMO C

Basically the question stem says n = m! and we need to find the minimum value of m for which 2880 is a factor.

When prime factorized we get 2880 = 2^6 * 3^2 * 5

So we need a minimum of 6 2's , 2 3's and 1 5's.

Let's counts 2 , cause 3 and 5 will automatically get covered once we move past 5! (as egmat explained).

2! has one 2, 4! has two 2's, 6! has one 2, 8! has three 2's , so a total of 8 seven 2's sufficient.

So 8! is a multiple of 2880
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Re: If m is a positive integer, and n is product of all integers from 1 to [#permalink] If m is a positive integer, and n is product of all integers from 1 to   07 Aug 2020, 08:49
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'm' is an integer and n = m!

If n is multiple of 2,880 that means n = 2880 * something

First, find the prime factors of 2880: 2^6 * 3^2 * 5

From here, we know that m must be at least 5.

The first multiple of 2880 is itself. So, we require a minimum of six(2's), two(3's), and one(5).

Also, 6! = 720, 7! = 5040 which are not multiple of 2880.

The only option we are left is 8,9, or 10.

8! = 40,320 = 2^7*3^2*5*7 (2880 * 2 *7)

Hence, 8 will be the answer. Answer C
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Re: If m is a positive integer, and n is product of all integers from 1 to [#permalink] If m is a positive integer, and n is product of all integers from 1 to   08 Aug 2020, 00:27
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Q57. 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​

Solution


    • m and n are positive integers.
    • n is a multiple of 2880.
      o This means, \(n = 2880*k\), where k is a positive integer.
      o Now, prime factorization of \(2880 = 2^6*3^2*5\)
         Therefore, \(n = k*2^6*3^2*5\)
    • Also, \(n = 1*2*3*4*5*….*m\)
      o This means, \(n = 1*2*3*4*5*….*m = k*2^6*3^2*5 \)
      o \(⟹ n = k*2*3*(2^2)*5*(2*3)*(2^2)\)
      o \(⟹ n = k*2*3*4*5*6*(2^2)\)
    • From the above we can see that m must be an even integer greater than 6.
      o Now, the next even integer greater than 6 is 8 and \(8 = 2^3\), which can account for the remaining \(2^2\) in the expression of n.
      o Thus, the smallest value of \(m = 8\)
Thus, the correct answer is Option C.
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Re: If m is a positive integer, and n is product of all integers from 1 to [#permalink] If m is a positive integer, and n is product of all integers from 1 to   02 Sep 2022, 23:27
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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!
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Re: If m is a positive integer, and n is product of all integers from 1 to [#permalink] If m is a positive integer, and n is product of all integers from 1 to   14 Aug 2023, 07:32
Not understood after 2^6,3^2,5..pls help
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Re: If m is a positive integer, and n is product of all integers from 1 to [#permalink] If m is a positive integer, and n is product of all integers from 1 to   14 Aug 2023, 08:56
If m is a positive integer, and n is the 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? ​

n = m!

Let us go option one by one,

A. 6​ => 6! = 720 , hence incorrect

B. 7​=> 7! = 5040 , hence incorrect

C. 8​=> 8! = 8x7! = 8x5040 = (8x2x4)x(3x6)x(5)x7 = 64x18x5x7 = 5760x7 =2880x(2x7) , hence correct

D. 9​ , not the least, hence incorrect

E. 10​ not the least, hence incorrect
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Re: If m is a positive integer, and n is product of all integers from 1 to [#permalink] If m is a positive integer, and n is product of all integers from 1 to   25 Sep 2024, 08:30
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