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Re: Math: Number Theory [#permalink]
14 Sep 2010, 04:33

Bunuel wrote:

NUMBER THEORY Consecutive Integers

Consecutive integers are integers that follow one another, without skipping any integers. 7, 8, 9, and -2, -1, 0, 1, are consecutive integers.

• Sum of \(n\) consecutive integers equals the mean multiplied by the number of terms, \(n\). Given consecutive integers \(\{-3, -2, -1, 0, 1,2\}\), \(mean=\frac{-3+2}{2}=-\frac{1}{2}\), (mean equals to the average of the first and last terms), so the sum equals to \(-\frac{1}{2}*6=-3\).

• If n is odd, the sum of consecutive integers is always divisible by n. Given \(\{9,10,11\}\), we have \(n=3\) consecutive integers. The sum of 9+10+11=30, therefore, is divisible by 3.

• If n is even, the sum of consecutive integers is never divisible by n. Given \(\{9,10,11,12\}\), we have \(n=4\) consecutive integers. The sum of 9+10+11+12=42, therefore, is not divisible by 4.

• The product of \(n\) consecutive integers is always divisible by \(n!\). Given \(n=4\) consecutive integers: \(\{3,4,5,6\}\). The product of 3*4*5*6 is 360, which is divisible by 4!=24.

Evenly Spaced Set

Evenly spaced set or an arithmetic progression is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. The set of integers \(\{9,13,17,21\}\) is an example of evenly spaced set. Set of consecutive integers is also an example of evenly spaced set.

• If the first term is \(a_1\) and the common difference of successive members is \(d\), then the \(n_{th}\) term of the sequence is given by: \(a_ n=a_1+d(n-1)\)

• In any evenly spaced set the arithmetic mean (average) is equal to the median and can be calculated by the formula \(mean=median=\frac{a_1+a_n}{2}\), where \(a_1\) is the first term and \(a_n\) is the last term. Given the set \(\{7,11,15,19\}\), \(mean=median=\frac{7+19}{2}=13\).

• The sum of the elements in any evenly spaced set is given by: \(Sum=\frac{a_1+a_n}{2}*n\), the mean multiplied by the number of terms. OR, \(Sum=\frac{2a_1+d(n-1)}{2}*n\)

• Special cases: Sum of n first integers: \(1+2+...+n=\frac{1+n}{2}*n\)

Sum of n first odd numbers: \(a_1+a_2+...+a_n=1+3+...+a_n=n^2\), where \(a_n\) is the last, \(n_{th}\) term and given by: \(a_n=2n-1\). Given \(n=5\) first odd integers, then their sum equals to \(1+3+5+7+9=5^2=25\).

Sum of n first positive even numbers: \(a_1+a_2+...+a_n=2+4+...+a_n=n(n+1)\), where \(a_n\) is the last, \(n_{th}\) term and given by: \(a_n=2n\). Given \(n=4\) first positive even integers, then their sum equals to \(2+4+6+8=4(4+1)=20\).

• If the evenly spaced set contains odd number of elements, the mean is the middle term, so the sum is middle term multiplied by number of terms. There are five terms in the set {1, 7, 13, 19, 25}, middle term is 13, so the sum is 13*5 =65.

There seems to be a discrepancy in what some study guides consider consecutive integers. In Kaplan Premier 2011 consecutive integers are defined as numbers that occur at a fixed interval or exhibit a fixed pattern. However, on the Kaplan Free Practice Test I got a DS question wrong because it didn't consider evenly spaced numbers to necessarily be consecutive. Your definition also separates the two. Could anyone clarify which is correct so I know for the actual GMAT. Thanks! _________________

The Brain Dump - From Low GPA to Top MBA(Updated September 1, 2013) - A Few of My Favorite Things--> http://cheetarah1980.blogspot.com

Re: Math: Number Theory [#permalink]
14 Sep 2010, 05:38

Expert's post

cheetarah1980 wrote:

There seems to be a discrepancy in what some study guides consider consecutive integers. In Kaplan Premier 2011 consecutive integers are defined as numbers that occur at a fixed interval or exhibit a fixed pattern. However, on the Kaplan Free Practice Test I got a DS question wrong because it didn't consider evenly spaced numbers to necessarily be consecutive. Your definition also separates the two. Could anyone clarify which is correct so I know for the actual GMAT. Thanks!

When we see "consecutive integers" it ALWAYS means integers that follow each other in order with common difference of 1: ... x-3, x-2, x-1, x, x+1, x+2, ...

-7, -6, -5 are consecutive integers.

2, 4, 6 ARE NOT consecutive integers, they are consecutive even integers.

3, 5, 7 ARE NOT consecutive integers, they are consecutive odd integers.

2, 5, 8, 11 ARE NOT consecutive integers, they are terms of arithmetic progression with common difference of 3.

All sets of consecutive integers represent arithmetic progression but not vise-versa.

Re: Math: Number Theory [#permalink]
19 Sep 2010, 23:27

7 - Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7 (including 0), then the number is divisible by 7.

Is this the only way to check divisibility by 7? For huge numbers there is no big difference to divide the number directly by 7 or to use the algorithm above.

Re: Math: Number Theory [#permalink]
19 Sep 2010, 23:35

Expert's post

Kronax wrote:

7 - Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7 (including 0), then the number is divisible by 7.

Is this the only way to check divisibility by 7? For huge numbers there is no big difference to divide the number directly by 7 or to use the algorithm above.

Note that you can perform this operation number of times. Also you won't need to check divisibility by 7 for huge numbers on GMAT. _________________

Re: Math: Number Theory [#permalink]
19 Sep 2010, 23:45

Bunuel wrote:

Kronax wrote:

7 - Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7 (including 0), then the number is divisible by 7.

Is this the only way to check divisibility by 7? For huge numbers there is no big difference to divide the number directly by 7 or to use the algorithm above.

Note that you can perform this operation number of times. Also you won't need to check divisibility by 7 for huge numbers on GMAT.

Thank you for the notice! I haven't thought about doing the operation multiple times.

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

2

This post received KUDOS

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]
18 Nov 2010, 22:47

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 ...?

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

1

This post received KUDOS

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]
19 Nov 2010, 01:18

Quote:

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.

Hope it's clear.

Understood Sir! .. i'll just use it one way; i.e, if i'm told that n is a prime number>3, then i can express it as 6n+1 or 6n+5

I think I just got a bit too excited about it that I forgot to thoroughly test it thru...

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

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