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## Comment on

Introduction to Divisibility## Whoa! Ok, I'm going to need a

N = 10^40 + 2^40. 2^k is a divisor of N, but 2^(k+1) is not a divisor of N. If k is a positive integer, what is the value of k−2?

Thanks

## Happy to help, bertyy!

Happy to help, bertyy!

You're referring to my question here: https://gmatclub.com/forum/n-10-40-2-40-2-k-is-a-divisor-of-n-237383.html

This is a VERY tricky question (750+)

I have a step-by-step solution here: https://gmatclub.com/forum/n-10-40-2-40-2-k-is-a-divisor-of-n-237383.htm...

Rather than repeat the entire solution, can you take a look and tell me which parts you'd like me to cllarify?

## Hi Brent,

Factoring the first part is not the problem or seeing that "5" to the power of anything ends in five. The rest after that just throws me for a loop as to what to do or what is even going on. Usually I can force it after a while and come back to it, but this one has got me blocked. I can't figure out what i'm missing or that some concept that has escaped me.

Thanks for your help

## Okay, I'll elaborate on my

Okay, I'll elaborate on my solution at https://gmatclub.com/forum/n-10-40-2-40-2-k-is-a-divisor-of-n-237383.htm...

WE have : N = 2^40(5^40 + 1)

We know that 5^40 must end in 25. So, we can say 5^40 = XXXX25 (the X's represent other digits on the number, but we don't really care about them.

So, we can say: N = = 2^40(XXXX25 + 1)

XXXX25 + 1 = XXXXX26 (if we add 1 to a number that ends in 25, the resulting value will end in 26

So, N = (2^40)(XXXX26)

At this part, we need to recognize that since XXX26 is EVEN. So, we can rewrite XXX26 as (2)(XXXX3).

So, N = (2^40)[(2)(XXXX3)]

We can now combine 2^40 and 2. We have: (2^40)(2) = (2^40)(2^1) = 2^41

So, N = (2^41)[XXXX3]

Since XXXX3 is an ODD number, we cannot factor any more 2's out of it. In other words, since XXXX3 is ODD, there are no 2's hiding in the prime factorization of XXXX3.

However, we DO know that there are 41 2's hiding in the prime factorization of 2^41. This means that 2^41 IS a factor of N, but 2^42 is NOT a factor of N.

The question tells us that 2^k is a factor of N, but 2^(k+1) is NOT a factor of N

In other words, k = 41

What is the value of k-2?

Since k = 41, we can conclude that k - 2 = 41 - 2 = 39

Does that help?

Cheers,

Brent

## Hi Brent,

what does this factorial expression mean 13!/7!....

Thanks

Fatima-Zahra

## Good question, Fatima-Zahra

Good question, Fatima-Zahra

This concept appears later, in the counting module (here's the video that covers this notation: https://www.gmatprepnow.com/module/gmat-counting/video/780 )

In general, n! = (1)(2)(3)(4).....(n-1)(n)

For example: 4! = (1)(2)(3)(4) = 24

And 7! = (1)(2)(3)(4)(5)(6)(7) = 5040

Cheers,

Brent

## Hi Brent,

I am referring to question https://gmatclub.com/forum/for-any-positive-integer-x-the-2-height-of-x-is-defined-to-be-the-207706.html

I cannot wrap my mind around terminology here "2-height of x", English is not my native language, so when I read "2-height of x", I am thinking of height as in "height of a person", is it a special math terminology "2-height of a number", or we are getting that 2-height is a number of 2s in a positive integer x, because we are given that it is defined such that x=2^n.

In other words, can we have something like 3-height of an integer, 7-height of an integer? I see such terminology for the first time... Thanks a bunch!

## Question link: https:/

Question link: https://gmatclub.com/forum/for-any-positive-integer-x-the-2-height-of-x-...

The term "2-height" is something the test-makers made up to create this question. We can think of this question as a Strange Operator question, since we're presented with a new term and a definition of that term (more here: https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...)

GIVEN: For any positive integer x, the 2-height of x is defined to be the greatest non-negative integer n such that 2^n is a factor of x.

Example: If x = 24, what is the 2-height of 24?

According to the definition, the 2-height of 24 is the biggest value of n, so that 2^n is a factor of 24.

24 = (2)(2)(2)(3)

We can see that 2¹ is a factor of 24.

Also, 2² is a factor of 24.

And 2³ is a factor of 24.

However, 2⁴ is NOT a factor of 24

So, the 2-height of 24 is 3 since 3 is the biggest possible power of 2 that is a factor of 24

Does that help?

Cheers,

Brent

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

Or it could be the case that z = (3)(7)(13)(2) = 546,....... i could not understand this part (2); how can you multiply 2?

## Link to question and my

Link to question and my solution: https://gmatclub.com/forum/is-positive-integer-z-greater-than-205456.htm...

If we know that z is a multiple of 21, then we know that 21 is hiding in the prime factorization of z.

That is, we know that z = (3)(7)(?)(?)(?)(?)

IMPORTANT: The ?'s represent other possible values that might be in the prime factorization of z.

For example, it could be the case that z = (3)(7)(2)(11)(53)

Or, it could be the case that z = (3)(7)(5)(5)

Etc.

That said, all we can be CERTAIN of is that there's a 3 and a 7 hiding in the prime factorization of z

Likewise, if we know that z is a multiple of 39, then we know that 39 is hiding in the prime factorization of z.

That is, we know that z = (3)(13)(?)(?)(?)(?)

When we COMBINE the statements, we know that z MUST have (at the very least) a 3, a 7 and a 13 in its prime factorization.

So, it COULD be the case that z = (3)(7)(13)

Or it COULD be the case that z = (3)(7)(13)(2)

Or it COULD be the case that z = (3)(7)(13)(11)(2)(5)...etc.

Notice that, if z = (3)(7)(13)(2), then it's still divisible by 21, AND it's still divisible by 13

So, z = (3)(7)(13)(2) also satisfies both statements.

Likewise, if z = (3)(7)(13)(11)(2)(5), then it's still divisible by 21, AND it's still divisible by 13

So, z = z = (3)(7)(13)(11)(2)(5) also satisfies both statements.

etc.

Does that help?

Cheers,

Brent

## Thanks a ton sir

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

most time i rely on your explanations as maximum are easy to understand. but this one is little bit tough to digest.................

IMPORTANT: we need to recognize that 5^b will end in 25 for all integer values of b greater than 1.

For example, 5^2 = 25

5^3 = 125

5^4 = 625

5^5 = 3125

5^6 = XXX25etc....( is it a kind of concept that 5^(any power,last digits will be 5)?

secondly, = 2^40[2(XXXX3)] [Since XXX26 is EVEN, we can factor out a 2]

= 2^41[XXXX3] ; how you make 2^41 ? ..thanks in advance

## Question link: https:/

Question link: https://gmatclub.com/forum/n-10-40-2-40-2-k-is-a-divisor-of-n-237383.html

KEY CONCEPT: A x (B x C) = (A x B) x C

For example, 3 x (2 x 5) = (3 x 2) x 5

Now, let's keep going from the part when I say that: N = 2^40[2(some number ending in 3)]

Take: N = 2^40[2(some number ending in 3)]

Use key concept to rewrite this as: N = (2^40 x 2)(some number ending in 3)

Rewrite 2 as 2^1 to get: N = (2^40 x 2^1)(some number ending in 3)

Simplify to get: N = (2^41)(some number ending in 3)

Does that help?

Cheers,

Brent

## Hi Brent, I have one question

## Question link: https:/

Question link: https://gmatclub.com/forum/each-entry-in-the-multiplication-table-above-...

This official solution doesn't say that it MUST be the case a = b = 1. It just note that this COULD be the case.

That is, if c = f, then it COULD be the case that c = f = 2 and a = b = 1. In this case, the answer to the target question is "c = 2."

However, it could also be the case that c = f = 3 and a = b = 1. In this case, the answer to the target question is "c = 3."

Does that help?

Cheers,

Brent

## Hi Brent, I have one question

## There's a nice divisibility

There's a nice divisibility property that says:

If J is divisible by n, and K is divisible by n, then J-K must also be divisible by n.

Since 20! = (20)(19)(18)(17)(16)(15)(14)(13)....(2)(1), we can see that 20! is divisible by 15.

So, if n is divisible by 15 then, according to the above property, n - 20! must also be divisible by 15.

For more on this divisible property watch: https://www.gmatprepnow.com/module/gmat-integer-properties/video/831

## Hi Brent,

If n and k are positive integers, is n divisible by 6 ?

(1) n = k(k + 1)(k – 1)

(2) k – 1 is a multiple of 3.

In the statement (1) do I have to worry about k=1, in which case the whole expression k(k + 1)(k – 1)=0,

Do I have to understand that 0 when divided by any existing number except 0 with no remainder, which implies that 0 is divisible by any number?

Thank you in advance,

## You're correct to say that

You're correct to say that zero is divisible by any integer.

For example we can say that 0 is divisible by 3.

That said, when it comes to divisibility questions, the GMAT usually (probably ALWAYS) restricts values to POSITIVE integers.

Cheers,

Brent

## Thank you very much Brent,

I understand now

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