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Re: Math: Number Theory [#permalink]
11 Aug 2010, 17:07

2

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Expert's post

utfan2424 wrote:

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. _________________

Re: Math: Number Theory [#permalink]
01 Nov 2010, 15:16

2

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shrive555 wrote:

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 _________________

Re: Math: Number Theory [#permalink]
27 Oct 2012, 01:34

2

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About Exponents and divisibility:

\((a + b)^2 = a^2+ 2ab + b^2\) 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 _________________

Re: Math: Number Theory [#permalink]
24 Jan 2010, 14:06

1

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Quote:

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.

Re: Math: Number Theory [#permalink]
19 Nov 2010, 00:53

1

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Expert's post

Araj wrote:

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.

Re: Math: Number Theory [#permalink]
06 Dec 2010, 00:08

1

This post received KUDOS

Expert's post

shrive555 wrote:

\((a^m)^n=a^{mn}\) ----------1

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

\(a^m^n=a^{(m^n)}\) 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: \((2^3)^4=2^{(3*4)}=2^{12}\).

Re: Math: Number Theory [#permalink]
03 Jan 2011, 08:52

1

This post received KUDOS

Expert's post

resh924 wrote:

Bunuel,

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.

Re: Math: Number Theory [#permalink]
25 Sep 2012, 09:43

1

This post received KUDOS

conty911 wrote:

Bunuel wrote:

NUMBER THEORY

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 -----------------------------------------------------------------------------------------------------

Re: Math: Number Theory [#permalink]
17 Sep 2014, 21:09

1

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lbnyc13 wrote:

Hi -

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)

Re: Math: Number Theory [#permalink]
31 Jan 2010, 09:00

Bunuel wrote:

• 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!

Re: Math: Number Theory [#permalink]
31 Jan 2010, 09:19

Expert's post

ariel wrote:

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 _________________

Re: Math: Number Theory [#permalink]
31 Jan 2010, 10:06

Appreciate the very prompt response, walker. To your point re divisibility by 7: I'm having a hard time proving this algebraically, is it a fair statement to say that the only non-prime numbers of the form 6n-1 and 6n+1 are the ones that are divisible by 7?

If so, a quick way to check whether a big number is prime would be to: 1) check whether it's of the form 6n-1 or 6n+1 2) check whether it's divisible by 7

Re: Math: Number Theory [#permalink]
31 Jan 2010, 10:34

Expert's post

ariel wrote:

Appreciate the very prompt response, walker. To your point re divisibility by 7: I'm having a hard time proving this algebraically, is it a fair statement to say that the only non-prime numbers of the form 6n-1 and 6n+1 are the ones that are divisible by 7?

If so, a quick way to check whether a big number is prime would be to: 1) check whether it's of the form 6n-1 or 6n+1 2) check whether it's divisible by 7

Is this correct?

Not so. Divisibility by 7 does not check whether the number is prime or not.

Actually this issue is covered in the post. First you should know that all prime numbers except 2 and 5 end in 1, 3, 7 or 9. So if it ends in some other digit it's not prime.

Next, if the above didn't help (meaning that number ends in 1, 3, 7 or 9) there is a way to check whether the number is prime or not. Walker gave an example how to do this, but here it is again:

Verifying the primality 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 \(m<\sqrt{n}\).

Examples: Verifying the primality of \(161\): \(\sqrt{161}\) is little less than \(13\). We should check \(161\) on divisibility by numbers from 2 to 13. From integers from \(2\) to \(13\), \(161\) is divisible by \(7\), hence \(161\) is not prime.

Verifying the primality of \(149\): \(\sqrt{149}\) is little more than \(12\). We should check \(149\) on divisibility by numbers from 2 to 12, inclusive. \(149\) is not divisible by any of the integers from \(2\) to \(12\), hence \(149\) is prime.

Verifying the primality of \(73\): \(\sqrt{73}\) is little less than \(9\). We should check \(73\) on divisibility by numbers from 2 to 9. \(73\) is not divisible by any of the integers from \(2\) to \(9\), hence \(149\) is prime.

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