Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Does this relationship breakdown at some point? I thought this was great and was just experimenting and looked at 21! (calculated in excel) and it ends with 5 zeros. Using the methodology you described above it should have 4 zeros. Am I missing something or did I make a mistake somewhere?

You made everything right 21! ends with 21/5=4 zeros. It's excel: it makes rounding with such a huge numbers thus giving incorrect result. _________________

If n is a positive integer greater than 1, then there is always a prime number P with n<P<2n

n<p<2n can someone please explain this with example .

Thanks

The result you are referring to is a weak form of what is known as Bertrand's Postulate. The proof of this result is beyond the scope of the GMAT, but it is easy to show some examples.

Choose any n>1, you will always find a prime number between n & 2n.

Eg. n=5, 2n=10 ... p=7 lies in between n=14, 2n=28 ... p=19 lies in between n=20, 2n=40 ... p=23 lies in between _________________

\((a + b)^2 = a^2+ 2ab + b^2\) Square of a Sum \((a - b)^2 = a^2 - 2ab + b^2\) Square of a Difference

\(a^n - b^n\) is always divisble by a-b i.e. irrespective of n being odd or even Proof: \(a^2 - b^2 = (a-b)(a+b)\) \(a^3 - b^3 = (a-b)(a^2+ab+b^2)\)

Thus divisible by a- b in both cases where n = 2 i.e. even and 3 i.e. odd

\(a^n + b^n\) is divisble by a+b i.e. only if n = odd Proof: \(a^3 - b^3 = (a+b)(a^2-ab+b^2)\) Thus divisible by a + b as n = 3 i.e. odd _________________

• Verifying the primality (checking whether the number is a prime) of a given number \(n\) can be done by trial division, that is to say dividing \(n\) by all integer numbers smaller than \(\sqrt{n}\), thereby checking whether \(n\) is a multiple of \(m<\sqrt{n}\). Example: Verifying the primality of \(161\): \(\sqrt{161}\) is little less than \(13\), from integers from \(2\) to \(13\), \(161\) is divisible by \(7\), hence \(161\) is not prime.

A minor point, but the inequalities here should not be strict. If you want to test if some large integer n is prime, then you need to try dividing by numbers up to and including \(\sqrt{n}\). We must include \(\sqrt{n}\), in case our number is equal to the square of a prime.

And it might be worth mentioning that it is only necessary to try dividing by prime numbers up to \(\sqrt{n}\), since if n has any divisors at all (besides 1 and n), then it must have a prime divisor.

It's very rare, though, that one needs to test if a number is prime on the GMAT. It is, computationally, extremely time-consuming to test if a large number is prime, so the GMAT cannot ask you to do that. If a GMAT question asks if a large number is prime, the answer really must be 'no', because while you can often quickly prove a large number is not prime (for example, 1,000,011 is not prime because it is divisible by 3, as we see by summing digits), you cannot quickly prove that a large number is prime. _________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

ding is simplifying a number to a certain place value. To round the decimal drop the extra decimal places, and if the first dropped digit is 5 or greater, round up the last digit that you keep. If the first dropped digit is 4 or smaller, round down (keep the same) the last digit that you keep.

Example: 5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5. 5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5. 5.3485 rounded to the nearest thousandth = 5.249, since the dropped 5 is equal to 5.

I'm assuming it was just a typo for the last part of the example: Its entered as 5.3485 rounded to the nearest thousandth = 5.249, since the dropped 5 is equal to 5.

I guess you meant 5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5.

NUMBER THEORY • For GMAT it's good to memorize following values: \(\sqrt{2}\approx{1.41}\) \(\sqrt{3}\approx{1.73}\) \(\sqrt{5}\approx{2.24}\) \(\sqrt{7}\approx{2.45}\) \(\sqrt{8}\approx{2.65}\) \(\sqrt{10}\approx{2.83}\)

Anyone else notice that these are wrong? They should be: • For GMAT it's good to memorize following values: \(\sqrt{2}\approx{1.41}\) \(\sqrt{3}\approx{1.73}\) \(\sqrt{5}\approx{2.24}\) \(\sqrt{6}\approx{2.45}\) \(\sqrt{7}\approx{2.65}\) \(\sqrt{8}\approx{2.83}\) \(\sqrt{10}\approx{3.16}\)

Thanks. Edited. +1 for spotting this. _________________

Hello Bunuel - thank you so much for this fantastic post!

with regards to checking for primality:

Quote:

Verifying the primality (checking whether the number is a prime) of a given number can be done by trial division, that is to say dividing by all integer numbers smaller than , thereby checking whether is a multiple of . Example: Verifying the primality of : is little less than , from integers from to , is divisible by , hence is not prime.

Would it be accurate to say that a number is prime ONLY if it gives a remainder of 1 or 5 when divided by 6? i.e, for eg. 10973/6 gives a remainder of 5, so it has to be prime...

i found the reasoning behind this in one of the OG solutions: prime numbers always take the form: 6n+1 or 6n+5 ....

the only possible remainders when any number is divided by 6 are [0,1,2,3,4,5] ... A prime number always gives a remainder of 1 or 5, because: a) if the remainder is 2 or 4, then the number must be even b) if the remainder is 3, then it is divisible by 3 ...

hence, if a number divided by 6 yields 1 or 5 as its remainder, then it must be prime ...?

-Raj

First of all there is no known formula of prime numbers.

Next: Any prime number \(p>3\) when divided by 6 can only give remainder of 1 or 5 (remainder can not be 2 or 4 as in this case \(p\) would be even and remainder can not be 3 as in this case \(p\) would be divisible by 3).

So any prime number \(p>3\) could be expressed as \(p=6n+1\) or\(p=6n+5\) or \(p=6n-1\), where n is an integer >1.

But: Not all number which yield a remainder of 1 or 5 upon division by 6 are prime, so vise-versa of above property is not correct. For example 25 yields a remainder of 1 upon division be 6 and it's not a prime number.

\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\) ------------------2

\(2^2^2 = 2^(2^2) = 2^4\)

If above example is correct then whats the difference 1 & 2. Please clarify thanks

If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus: \(a^m^n=a^{(m^n)}\) and not \((a^m)^n\), which on the other hand equals to \(a^{mn}\).

So: \((a^m)^n=a^{mn}\);

\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\).

Now, there are some specific values of \(a\), \(m\) and \(n\) for which \(a^m^n\) equals to \(a^{mn}\). For example: \(a=1\): \(1^{m^n}=1=1^{mn}\);

\(m=0\): \(a^0^n=a^0=1\) and \(a^{0*n}=a^0=1\);

\(m=2\) and \(n=2\) --> \(a^{2^2}=a^4\) and \(a^{2*2}=a^4\);

\(m=4\) and \(n=\frac{1}{2}\) --> \(a^{4^{\frac{1}{2}}}=a^2\) and \(a^{4*{\frac{1}{2}}}=a^2\); ...

So, generally \(a^m^n\) does not equal to \((a^m)^n\), but for specific values of given variables it does.

shrive555 wrote:

In question would that be given explicitly ... i mean the Brackets ( )

\(a^m^n\) ALWAYS means \(a^{(m^n)}\), so no brackets are needed. For example \(2^{3^4}=2^{(3^4)}=2^{81}\);

If GMAT wants the order of operation to be different then the necessary brackets will be put. For example: \((2^3)^4=2^{(3*4)}=2^{12}\).

For determining last digit of a power for numbers 0, 1, 5, and 6, I am not clear on how to determine the last digit.

Your post says: • Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base.

What is the last digit of 345^27 ---is the last digit 5? What is the last digit of 216^32----is the last digit 6? What is the last digit of 111^56---is the last digit 1?

Any clarification would be helpful.

Thanks for all your help.

First of all: last digit of 345^27 is the same as that of 5^27 (the same for 216^32 and 111^56);

Next: 1 in any integer power is 1; 5^1=5, 5^2=25, 5^3=125, ... 6^1=6, 6^2=36, 5^3=216, ...

So yes, integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base: thus 0, 1, 5, and 6 respectively.

Trailing zeros: Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

125000 has 3 trailing zeros;

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer \(n\), can be determined with this formula:

\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that \(5^k<n\).

It's easier if you look at an example:

How many zeros are in the end (after which no other digits follow) of \(32!\)? \(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)

Hence, there are 7 zeros in the end of 32!

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.

I noticed in case the number (n) is multiple of \(5^k\) and we have to find number of trailing zero zeroes, then it will be \(5^k<=n\) rather \(5^k<n\)

no of trailing zeros in 25! =6

\(\frac{25}{5}+\frac{25}{5^2}= 5+1\); Please correct me, clarify if i'm wrong. Thanks

The highest power of a prime number "k" that divides any number "n!" is given by the formula n/K + n/k^2+n/k^3.. (until numerator becomes lesser than the denominator). Remember to truncate the remainders of each expression

E.g : The highest number of 2's in 10! is 10/2 + 10/4 + 10/8 = 5 + 2 + 1 = 8 (Truncate the reminder of each expression)

As a consequence of this, the number of zeros in n! is controlled by the presence of 5s. Why ? 2 reasons

a) 10 = 5 x 2, b) Also in any n!, the number of 5's are far lesser than the number of 2's.

Think about this example. The number of cars that you make depends on the number of engines. You can have 100 engines and 1000 cars, but you can only make 100 cars (each car needs an engine !)

10 ! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 Lets factorize each term ... 10! = (5 x 2) x(3x3)x(2x2x2)x7x(2x3)x(5)X(2x2)x1 the number of 5s = 2 The number of 2s = 7 The number of zeros in 10! = the total number of 5s = 2 (You may use a calc to check this10! = 3628800)

hence in any n! , the number of 5's control the number of zeros.

As a consequence of this, the number of 5's in any n! is n/5 + n/25 + n/125 ..until numerator becomes lesser than denominator.

Again, i want to emphasize that this formuala only works for prime numbers !! So to find the number of 10's in any n!, DO NOT DIVIDE by 10 ! (10 is not prime !) i.e DONT do n/10 + n/100 + n/1000 - THIS IS WRONG !!! _________________

----------------------------------------------------------------------------------------------------- IT TAKES QUITE A BIT OF TIME AND TO POST DETAILED RESPONSES. YOUR KUDOS IS VERY MUCH APPRECIATED -----------------------------------------------------------------------------------------------------

Can somebody explain where the '00' part of the 9900 in the denominator is coming from? I understand where the '99' is coming from. I copied this from the number theory section.

Thanks

" Example #2: Convert 0.2512(12) to a fraction. 1. The number consisting with non-repeating digits and repeating digits is 2512; 2. Subtract 25 (non-repeating number) from above: 2512-25=2487; 3. Divide 2487 by 9900 (two 9's as there are two digits in 12 and 2 zeros as there are two digits in 25): 2487/9900=829/3300. "

let x =0.2512(12) 100x = 25.(12) 10000x = 2512.(12)

• All prime numbers except 2 and 5 end in 1, 3, 7 or 9, since numbers ending in 0, 2, 4, 6 or 8 are multiples of 2 and numbers ending in 0 or 5 are multiples of 5. Similarly, all prime numbers above 3 are of the form \(6n-1\) or \(6n+1\), because all other numbers are divisible by 2 or 3.

Awesome post, thank you so much! +1

What is the quickest way to figure out whether a number is prime? I usually check if it's odd or even, then sum its digits to figure out if it's divisible by 3, then look if it ends in 5 and if all else fails divide it by 7. Is this the recommended approach?

What might be a bit confusing is that while all prime numbers are of the form 6n-1 or 6n+1, not all numbers of that form are in fact prime. I think this is crucial. For instance, the number 49 is 6n+1, but is not prime.

Any insight on a quicker check (if one exists) would be much appreciated and thank you again for your efforts. They make a real difference!

What is the quickest way to figure out whether a number is prime?

Unfortunately, there is no such quick way to say that this number is prime. You can remember all numbers till 50 and then use rule:

Rule: To check whether a number is prime or not, we try to divide it by 2, 3, 5 and so on. You can stop at \(\sqrt{number}\) - it is enough. Why? Because if there is prime divisor greater than \(\sqrt{number}\), there must be another prime divisor lesser than \(\sqrt{number}\).

Example,

n = 21 -- > \(\sqrt{21}\)~ 4-5 So, we need to check out only 2,3 because for 7, for instance, we have already checked out 3.

n = 101 --> 2,3,5 is out (the last digit is not even or 5 and sum of digits is not divisible by 3). we need to check out only 7 _________________

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

As you leave central, bustling Tokyo and head Southwest the scenery gradually changes from urban to farmland. You go through a tunnel and on the other side all semblance...

Ghibli studio’s Princess Mononoke was my first exposure to Japan. I saw it at a sleepover with a neighborhood friend after playing some video games and I was...