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:

Find approx square root of the number. Then check if all the prime numbers below the square root are factors of the given number. if none are then the number is prime else not.

e.g. number 91. approx sq root is 10
Prime number below 10 are 2.3.5&7.

91 is not divisible by2,3 or 5. But it is divisible by 7.
Therefore 91 is not prime.
[/b]

x^2>x
You CAN'T divided both sides by x and say x>1. What you have to do is to move the right side to the left:
x^2-x>0
x(x-1)>0
Solution would be either both x and x-1 are greater than zero, or both x and x-1 are smaller than zero. So your solution is: x>1 or x<0
___________________________________________________
shoud X > 0, X > 1?

I recently read a question, what is the unit digit of the following number,

2 to the power of 11897. The question might sound difficult but it is not. Observe the following.

2 to the power of 1 is 2. (The unit digit is 2).
2 to the power of 2 is 4 (The unit digit is 4).
2 to the power of 3 is 8. (The unit digit is 8).
2 to the power of 4 is 16. (The unit digit is 6)
2 to the power of 5 is 32. (The unit digit is 2)
2 to the power of 6 is 64 (The unit digit is 4)
2 to the power of 7 is 128 (The unit digit is 8)
2 to the power of 8 is 256 (The unit digit is 6)
2 to the power of 9 is 512 (The unit digit is 2).

And so on. Hence from the above series, it should be obvious that the unit digit can be either 2 or 4 or 8 or 6. And now to find out what is the unit digit of 2 to the power of x, where x is any integer greater than zero.

1) Divide x by 4 and find out the remainder. (The remainder can be >= 0 and <= 3).
2) If the remainder of step 1 is 0 then the unit digit is 6
3) If the remainder of step 1 is 3 then the unit digit is 8
4) If the remainder of step 1 is 2 then the unit digit is 4
5) If the remainder of step 1 is 1 then the unit digit is 2

The only exception is when 2 to the power of zero, and which is 1.

Interesting properties: - Adding n consecutive integers will yield a sum that is divisible by n (i.e. n will be a factor of the sum)

E.g. Adding 3,4,5,6 will give a sum with 4 as a factor Adding 2,3,4 will give a sum with 3 as a factor

Basic principles dervied from adding, n, n+1, n+2 etc..

Ywilfred, i just noticed that adding 2,3,4 and 5 (consecutive nos) will not yield a sum that is a multiple of n=4. Kindly address.

Let's see ...
n+(n+1)+(n+2)+(n+3)=4n+6
You are right the sum of four integers will not be a mulitple of 4.

The sum of N consecutive integers would be:
S=(n+n+N-1)*N/2=n*N+N*(N-1)/2
S/N=n+(N-1)/2
If N is odd, then S would be a multiple of N. If N is even, then S would not be a multiple of N.

It may not be necessary to remember this as a rule as long as you know how to derive it.
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

Absolutely. (The sum of odd digits - The sum of even digits) should be divisible by 11.

OK. I got something you might like.

Here is an easy way to multiple by 11.

Step1: Write down the first and last digit of the Multiplicand (the number you are multiplying).

Step2: Add Successive digits and put them in between the above two digits.

eg1: 43 X 11

Step1: 4 ... 3
Step2: 4+3 = 7, answer is 473.

eg2: 13452

Step1: 1......2
Step2:

a. 1+3 = 4
b. 3+4 = 7
c. 4+5=9
d. 5+2=7.

putting it together, the answer is 147972.

eg3: (with a carry forward) 57 X 11

step1: 5....7
step2: 5+7 = 12 (2 + 1 carry).

putting it together 5 (12) 7
ie 627. the answer is 627.

eg4:

459 X 11

Answer 4(9)(14)9 ie 5049.

Where is this helpful?

Yes, GMAT might have a question or two where this can be applied.

eg1: A home bought for 123322, after 10% increase in the value. What is the new value?

Ans: Note that 10% increase is same as multiplying by 110 or 11.

So, the answer is 1356520.

-Srinivas

Caspace wrote:

and for 11 which i got on an OG problem recently

Start with the units digit, add every other digit and remember this number. Form a new number by adding the digits that remain. If the difference between these two numbers is divisible by 11, then the original number is divisible by 11.

Examples: Is the number 824472 divisible by 11? Starting with the units digit, add every other number:2 + 4 + 2 = 8. Then add the remaining numbers: 7 + 4 + 8 = 19. Since the difference between these two sums is 11, which is divisible by 11, 824472 is divisible by 11.

Is the number 49137 divisible by 11? Starting with the units digit, add every other number:7 + 1 + 4 = 12. Then add the remaining numbers: 3 + 9 = 12. Since the difference between these two sums is 0, which is divisible by 11, 49137 is divisible by 11.

Interesting properties: - Adding n consecutive integers will yield a sum that is divisible by n (i.e. n will be a factor of the sum)

E.g. Adding 3,4,5,6 will give a sum with 4 as a factor Adding 2,3,4 will give a sum with 3 as a factor

Basic principles dervied from adding, n, n+1, n+2 etc..

Ywilfred, i just noticed that adding 2,3,4 and 5 (consecutive nos) will not yield a sum that is a multiple of n=4. Kindly address.

oops... sorry this took so long, i've been slapped with an never-ending pile of work in the office. Anyway, this principle works only if you begin with a odd integer.

Mean is the average, median is the middle number, mode is the one that appears the most. Most likely they are not equal to each other. Standard deviation, don't worry about it. It means how much all the numbers vary from one another, basically. For two sets of numbers, if one of the four equals, it means nothing about the other three.
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

Set 1 contains {1,1,1,1} ; Set 2 contains {-1,1,-1,1}; Set 3 is the union of Set 1 and Set 2 along with the number 2

Hong/or anyone - Please answer the following: Mean, Median, Mode, Range of Set 1, Set 2, Set 3 and standard deviation of Set 1

Good example.

Set 1 = {1,1,1,1}; Mean=Median=Mode=1, Range=0
Set 2 = {-1,-1,1,1}; Mean=Median=0, Mode=-1,1 (A set of data can have more than one mode), Range = 2
Set 3 = {-1,-1,1,1,1,1,2}; Mean=4/7, Median=1, Mode=1, Range=3
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

A square root, also called a radical or surd, of x is a number r such that r^2=x. The function r=sqrt(n) is therefore the inverse function of f(x)=x^2 for x>=0.

Eg. if x<0, sqrt (x^2)=-x

Factors, Multiples, LCM, GCD The divisors (or factors) of a positive integer are the integers that evenly divide it. For example, the divisors of 28 are 1, 2, 4, 7, 14 and 28. Of course 28 is also divisible by the negative of each of these, but by "divisors" we usually mean the positive divisors.
The proper divisors of the integer n are the positive divisors of n other than n itself. Zero is NOT a divisor of any number.

A multiple of a number is the product of that number and any other whole number. Zero is a multiple of every number.

The least common multiple of two (or more) nonzero integers is the least positive integer divisible by all of them. This is usually denoted LCM.

The greatest common divisor (archaic: greatest common factor) of two integers a and b is the largest integer that divides them both. This is usually denoted by GCD(a,b).

Two integers are relatively prime if there is no integer greater than one that divides them both (that is, their greatest common divisor is one).

Some facts:
GCD(a,b)*LCM(a,b) = ab;
LCM(a,b) = ab if and only if a and b are relatively prime.
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

Last edited by HongHu on 28 Oct 2005, 20:58, edited 1 time in total.

I've compiled all info provided in this thread into a file and attached it in the OP for easy assess. Whoever is interested please take a look and let me know if you find any problems.
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...