If 30!/10! is written as the product of consecutive integers, the larg

Quite straight forward actually. The answer is 11.
30! = 1*2*3......*29*30
10! = 1*2*3......*9*10

So, 1 to 10 gets cancelled out and the remaining series begins with 11. (1 cannot be considered since the series has to be consecutive integers)

\(\frac{30!}{10!} = \frac{30 * 29 * 28........... 11 * 10!}{10!}\)

Smallest integer = 11

Answer = D

\(\frac{30!}{10!} = \frac{(30)(29)(28)(27)(26). . . . . (13)(12)(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)}{(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)}\)

\( = (30)(29)(28)(27)(26). . . . . (13)(12)(11)\)

Smallest consecutive integer is 11

Answer: D

Once you see the answer, the logic becames clear.
But can anyone suggest how to deal with such kind of problems?

THX!

Yeahhh, you are 100% right!

I accidentally thought that the fraction transforms to 20*21*...*30, so I was looking for some very complicated method to extract 11 from 22 and make another very smart sequence ) Probably I was too tired

But thanks anyway!!!

Hi,

Just did the GMAT test prep and this question popped up.
Basically, my first answer was the good one, 11, but after double checking, the smallest of the integers, after simplifying the 30*29*28*...*11, is actually 1, because 1*30*29*28... It is possible to breakdown any of the numbers and multiplying them by one.
So I went for 1.
Anyone could tell me from the question, how could I have decided between 11 and 1?
My understanding is that the question specifies "the product of consecutive integer" only to explain what means 30!

Thanks a lot.

Hi tsunagaru,

You are right that multiplying the sequence by 1 would still give the same answer. However, including 1 in the sequence would mean that the sequence is not a sequence of consecutive integers any more.

As you have mentioned, the sequence would then become 1*11*12*13.....29*30

As you can see this would not be a valid sequence.

30!/10!

\(= \frac{30*29*28...........13*12*11*10!}{10!}\)

Thus we are left with : 30*29*28...........13*12*11

The largest number here is 30 and thus the smallest number will be 11, answer will be (D) 11
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We can simplify 30!/10! to 30 x 29 x 28 x 27 x … x 13 x 12 x 11. Thus, the smallest integer is 11.

Answer: D
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Hi All,

When a prompt provides you with a 'complex-looking' calculation, it can often help to come up with a much simpler example of the math involved. Here, we're asked to consider 30!/10! - and that might look 'scary' at first glance.

Instead of starting with that example, consider the following - what would 5!/3! look like...?

(5)(4)(3)(2)(1) / (3)(2)(1) = 120/6 = 20

You probably already know that when you simplify a fraction, you divide "top" and "bottom" by the same number, so we can 'cancel out' the 3, 2 and 1 from both the numerator and the denominator. This leaves us with...

(5)(4)/1 = 20

That wasn't too difficult, so now we can apply the same logic to 30!/10!... All of the numbers (10), (9)....(2) and (1) will cancel out, leaving us with...

(30)(29)....(12)(11)/1

Final Answer:

GMAT assassins aren't born, they're made,
Rich

30!= 1*2*3*4*5*6*7*8*9*10*11*....................*30

10!=1*2*3*4*5*6*7*8*9*10

Therefore 30!/10!= 11*12*............30 (as all common terms i.e. 10! cancel out).
Thus the smallest integer is 11.

Hi Bunuel

Can you explain to me why the answer is not 1 please?

I was going to choose 11 at first and I get the logic of it as well ; however, I then thought that the smallest integer would always remain to be 1, as anything multiplied by 1 is the number itself. Where is my understanding of the question going wrong?

Hi sssjav,

On Test Day, the GMAT questions that you will face will all be specifically written, so you have to pay careful attention to the details in each prompt. Here, we're told that 30!/10! is to be written as the product of CONSECUTIVE integers and we're asked for the SMALLEST number in that 'string' of integers.

If you're comfortable "reducing" the fraction, then you know the result will be:

(30)(29)....(12)(11)/1

Remember that we're looking for the smallest integer in the CONSECUTIVE string of those integers. We have to 'work down' from 30...29....28... etc. The number '1' is not included because the numbers 2, 3, 4,....8, 9 and 10 are NOT in that string.

GMAT assassins aren't born, they're made,
Rich

This question confused me cuz it's too easy.. I spent unnecessary extra time to think that there might be a trap...

30!= 30 * 29 * 28 ...... *13*12*11*10*9.........3*2*1

10! = 10*9.........3*2*1

\(\frac{30! }{ 10!}\) = \(\frac{30 * 29 * 28 ...... *13*12*11*10*9.........3*2*1}{ 10*9.........3*2*1}\)

=> 30*29*28......*13*12*11


=> Smallest integer : 11

Answer D

Bunuel, looks like this Q misses the Gmatprep tag, please add. Thanks!

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Added the tag. Thank you.
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