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

Question

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

EgmatQuantExpert · Answer

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

Harsh9676 · Answer

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

MathRevolution · Answer

'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

GMATWhizTeam · Answer

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.

BrushMyQuant · Answer

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!

manmithakotian · Answer

Not understood after 2^6,3^2,5..pls help

Paras96 · Answer

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

bumpbot · Answer

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

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