n is a positive integer, and k is the product of all integers from 1

Question

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

Bunuel · Accepted Answer

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).

KarishmaB · Answer

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

MacFauz · Answer

Factorizing 1440, we get  2^5 * 3^2 * 5^1

So, k should have at least five 2s, two 3s and one 5.

a) 8! = 2*3*4*5*6*7*8. Above condition is satisfied. Hence answer is A

rockzom · Answer

Took me 0:37 to do 1440x2 and divide that by the smallest choice to see if it fit. 2880/8=360, so A.

(I know my methods are unorthodox, but what can you do.)

IanStewart · Answer

'Unorthodox' is one way of putting it.

I'm not sure I understand why you've done what you did, but we're not looking here for the smallest answer choice which divides some multiple of 1440; it's coincidence that you get the right answer if you try that. If '4' or '5' had instead been the first answer choice, you'd find that 2880/4 or 2880/5 is an integer, but those aren't correct answers to the question. There's a reason Bunuel and Karishma carried out the steps they did.

SOURH7WK · Answer

K= multiple of 1440 ie product of some integer P X 1440. &
K= n! = 1.2.3.4......n

Now lets factorize 1440 = 12 * 12 * 10
Now my objective is to find this multiple 12*12*10 with min possible numbers in n!. Lets go on now......
n= 1*  2  *(  3*4  )*  5  *  6  *7*8(4 X   2  )
Hence 8! is sufficient to get a multiple of 12*12*10

Hence Answer A.

JeffTargetTestPrep · Answer

Let’s break 1440 into prime factors:

1440 = 144 x 10 = 12 x 12 x 10 = 2^5 x 3^2 x 5^1

Thus, k/(2^5 x 3^2 x 5^1) = integer.

We also know that k is the product of all integers from 1 to n inclusive, or in other words, k = n!.

Let’s check our answer choices:

A. If n = 8, then k = 8! and 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 = 2^3 x 7 x 3 x 2 x 5 x 2^2 x 3 x 2 = 7 x 5 x 3^2 x 2^7 does contain five 2s, two 3s and one 5.

Answer: A

Tajmir · Answer

Hope, this illustration will work.

Posted from my mobile device

KalyanChamarthi · Answer

Factorise 1440 : = 10 X 144
2x5 x 12 x 12
2X5x 3 x 4 x 6 x 2
1 x 2x3x 5 x 6 x 8 ( to get consequtive number 1 to n = you need 4 and 7 
 Now K is the Multiple of 1440 : There fore K can be = 1440 x 28 
and also 1 x 2 x 3x4x5x6x7x8 = 1440 X 28 
So n should be latest : 8

MathRevolution · Answer

n is a positive integer, and k is the product of all integers from 1 to n inclusive which means k = n!

K is a multiple of 1440 => k = 1440 * p

1440 = 2*2*2*2*2*3*3*5

=> 1440 =  2^5  *  3^2  * 5

n! =  2^5  *  3^2  * 5 * p

We need at least five (2's), two(3's) and one(5)

=> 8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 = 2 * 2 * 2 * 7 * 2 * 3 * 5 * 2 * 2 * 3 * 2 =  2^5  *  3^2  * 5 *  2^2  * 7 = 1440 *  2^2  * 7

Answer A

MHIKER · Answer

The question tells k! is divided by  1440

So k have to have all the factors that  1440  will have.

The \ factors \ of \ 1440=2^5, 3^2\ & \ 5

From the answer choices  8!  contains  = 8 * 7 * 6 * 5 * 4 * 3 * 2 = 2 * 2 * 2 * 7 * 2 * 3 * 5 * 2 * 2 * 3 * 2=2^7, 3^2, 5, \ & \ 7

The answer is  A.

Rangerwarrior · Answer

Hey are these type of questions still relevant in 2024?

Bunuel · Answer

This particular question is very much in the syllabus of GMAT Focus.

DanTheGMATMan · Answer

Take the time to get it right. Have to see the factors of 1440 and then walk up the factorials until you get the lowest factorial that covers all the factors of 1440:

https://www.youtube.com/watch?v=LAwc12_D81M

georgia123 · Answer

I have a question here that i'm wondering if someone can help me answer.
In my head i knew that 6! = 720 which 720 * 2 = 1440 hence the smallest value of would be n! = (4 * 6!) 
Keeping this in mind, is there a way to get from (4 * 6!) to 8! without going through prime factorization?

Bunuel · Answer

6! = 720, so we need one more 2.

The next number, 7!, will add 7, which does not have a factor of 2. However, 8!, will have a factor of 2. That's how we get 8!.

Vibhatu · Answer

Prime Factorizing of 1440, we get 
2^5*3^2*5
There must have 5 2s. That is only possible in 8!
Answer:A

Posted from my mobile device

orangebicycle5 · Answer

First factor out the 1440 = 2^5 * 3^2 * 5

Now of the answer choices, which factorial not only shares the same primes but is the lowest option?

8! = 8*7*6*5*4*3*2*1 = 2^7 * 3^2 * 5 * 7

bumpbot · Answer

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n is a positive integer, and k is the product of all integers from 1

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