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Re: Math: Number Theory [#permalink]
21 Jul 2010, 17:38

Thank you so much for compiling this! I have a few questions on the reasoning behind the rules... Sorry for the very long post.

Why is:

Perfect Square • The sum of distinct factors of a perfect square is ALWAYS ODD. • A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. • Perfect square always has even number of powers of prime factors.

Consecutive Integers • 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 • 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)

Terminating Decimal Why must the denominator be 2^n5^m?

Exponents Why are: a^n-b^n is ALWAYS divisible by a-b. a^n + b^n is divisible by a+b if n is odd, and not divisible by a+b if n is even.

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]
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 ...?

-Raj

gmatclubot

Re: Math: Number Theory
[#permalink]
18 Nov 2010, 22:47

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...