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## Comment on

Factorial Notation## Which of the following is a

a)2!+3

b) 4!+12

c)5!+6

d)5!+30

e)8!+12

How do i tackle this problem Brent? Do i have to consider prime numbers?

## Nice question!

Nice question!

If A and a multiple of B, then we can write A = kB for some integer k.

For example, we know that 24 is a multiple of 6, because we can write 24 = (4)(6)

4! + 6 = (4)(3)(2)(1) + 6

All MULTIPLES of 4! + 6 are in the form k(4! + 6), where k is an integers

k(4! + 6) = k[(4)(3)(2)(1) + 6]

= (k)(4)(3)(2)(1) + 6k

So, all multiples of 4! + 6 can be written in the form (k)(4)(3)(2)(1) + 6k

At this point, I SCAN the answer choices to see if any of them can be derived from the expression (k)(4)(3)(2)(1) + 6k

Notice that answer choice D, 5! + 30, can be derived if we let k = 5

That is, 5[4! + 6] = 5[(4)(3)(2)(1) + 6]

= (5)(4)(3)(2)(1) + 30

= 5! + 30

Answer: D

## Did you work backwards and

## Great question.

Great question.

I have edited my response above to address that question.

Cheers.

Brent

## Even option E can be written

## Be careful. 2(4! + 6) is not

Be careful. 2(4! + 6) is not equivalent to 8! + 12

When we take 2(4! + 6) and EXPAND it, we get:

2(4! + 6) = (2)(4!) + (2)(6)

= (2)(4!) + 12

So the 12 part is correct.

However, (2)(4!) does not equal 8!

(2)(4!) = (2)(4)(3)(2)(1)

8! = (8)(7)(6)(5)(4)(3)(2)(1)

Another way to show that 2(4! + 6) is not equal to 8! + 12 is to EVALUATE each expression.

(2)(4! + 6) = (2)[(4)(3)(2)(1) + 6] = 2[24 + 6] = 2[30] = 60

8! + 12 = (8)(7)(6)(5)(4)(3)(2)(1) + 12 = 40,320 + 12 = 40,332

Cheers,

Brent

## Hey Brent,

Came across this question and found that you didn't cover the particular formula for circular arrangements in the module. Can you explain using the FCP?

"How many ways can six friends be arranged around a circular dinner table?"

Thanks

## I'm not a big fan of

I'm not a big fan of questions involving people seated at a circular table. Those questions require some assumptions that are un-GMAT-like.

I've seen test-prep companies ask those question types, but I've never seen an official question with the same setup.

## Can this be asked on the GMAT

How many zeros does 100! end with?

## That would be a reasonable

That would be a reasonable question.

Check out my solution to a VERY SIMILAR question: https://gmatclub.com/forum/if-60-is-written-out-as-an-integer-with-how-m...

The question above involves finding the number of zeros at the end of 60!

See if you can extend that solution to find the number of zeros at the end of 100!

If you need more help, let me know.

Cheers,

Brent

## Hi Brent,

Need your help https://gmatclub.com/forum/find-the-number-of-zeroes-in-232903.html

## Link: https://gmatclub.com

Link: https://gmatclub.com/forum/find-the-number-of-zeroes-in-232903.html

This is a bad question for 2 reasons:

1) The question should be asking for the number of TRAILING zeros (zeros at the END of the integer). Otherwise, we should also count any other zeros in the number. For example, the number 106,400 has two TRAILING zeros at the end, but there are actually three zeros altogether.

2) The question involves way too many calculations.

KEY CONCEPT: We get ONE trailing zero for every pair of one 2 and one 5 that's hiding in the prime factorization of 1142! × 348! × 17!

So, for example 17! has three 5's in its prime factorization, and 17! has fifteen 2's in its prime factorization

So, in the prime factorization of 17!, there are 3 PAIRS consisting of one 2 and one 5.

As such, 17! has THREE trailing zeros.

This concept can be expanded to 1142! × 348! × 17!

Cheers,

Brent

## https://gmatclub.com/forum/if

We are told in algebra when you have the difference of two squares, trick is to factor it out quickly into a^2 - b^2 = (a + b) (a - b)

But that's not the case here.

17! ^ 2 - 16! ^ 2

which should equal 16! (17 + 1) (17 - 1) = 16! (16)(18)

is not the case. Or have I missed something here?

## Question link: https:/

Question link: https://gmatclub.com/forum/if-n-n-2-then-162389.html

You've only factored out ONE 16!, when you can actually factor out 16! TWICE.

We have: (17!)² - (16!)² = (17! + 16!)(17! - 16!)

= (16!)(17 + 1)(16!)(17 - 1)

= (16!)²(17 + 1)(17 - 1)

= (16!)²(17² - 1)

= (16!)²(289 - 1)

= (16!)²(288)

= (16!)²(288)

Since we can write: 288 = (12²)(2), we get:

= (¡16!)(12²)(2)

Answer: E

Does that help?

Cheers,

Brent

## Aren't we double factoring in

If we have 16! on the outside, doesn't that multiply to all the brackets?

So if we have (16!) on the outside, wouldn't that multiply to BOTH the brackets (17 + 1) and (17 - 1)? In multiplication, isn't it the case that we have to multiply across all the values?

Or have I misunderstood this?

## The easiest way to test these

The easiest way to test these kinds of theories is to use actual numbers.

For example, take 8 + 20 and factor it to get: 8 + 20 = 4(2 + 5)

When we evaluate both sides, we get: 28 = 28 (perfect!)

NOTE: the above example involves factoring a SUM. Now let's see what happens when we factor a PRODUCT.

Here's some factoring that's analogous to your factoring:

Take: (8)(20)

Factor 4 from EACH bracket to get: (8)(20) = 4[(2)(5)]

When we evaluate both sides, we get: 160 = 40 (oops!)

So, when it comes to PRODUCTS, we can't factor this way.

Instead, we must factor EACH COMPONENT of the product.

8 = (4)(2)

20 = (4)(5)

So, (8)(20) = (4)(2)(4)(5)

= (4)(4)(2)(5)

= (16)(2)(5)

We get: (8)(20) = (16)(2)(5)

When we evaluate both sides, we get: 160 = 160 (perfect!)

Does that help?

Cheers,

Brent

## Are there any rules for

Cheers,

Philipp

## When dividing factorials, we

When dividing factorials, we can often cancel out terms.

For example, 7!/5! = (7)(6)(5)(4)(3)(2)(1)/(5)(4)(3)(2)(1)

= (7)(6)

= 42

No nice rule for multiplying factorials.

When adding or subtracting factorials, you might be able to simplify things by factoring.

For example: 6! - 4! = (6)(5)(4)(3)(2)(1) - (4)(3)(2)(1)

= (4)(3)(2)(1)[(6)(5) - 1]

= (4)(3)(2)(1)[30 - 1]

= (4!)[30 - 1]

No nice rule for powers of factorials.

Cheers,

Brent

## Hi Brent,

Regarding this question, https://gmatclub.com/forum/if-n-1-n-2-n-440-what-is-the-sum-of-the-digits-of-n-295886.html

How did you go about factoring this quickly?

## Question link: https:/

Question link: https://gmatclub.com/forum/if-n-1-n-2-n-440-what-is-the-sum-of-the-digit...

Good question!

I wouldn't exactly say that I factored the quadratic quickly :-)

In fact, the difficult values in the given quadratic make this question a little less GMAT-like.

Given: n² + 4n - 437 = 0

It doesn't take long to realize that there aren't many pairs of values that have a product of 437.

I initially tested whether 437 is divisible by the first five primes (2, 3, 5, 7, and 11), but none of these values work.

So I just kept going (not really any other alternatives).

13...NO GOOD

19...Finally!!!

So, I was able to get: (n + 23)(n - 19) = 0

Cheers,

Brent

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