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x^2-1<0
(x+1)(x-1)<0
Since x<0, therefore x-1<0>0
therefore x>-1

Another way to go about it now that you know x<0
x^2<1
You know that must mean that the value of x without sign must be something less than 1, i.e. |x|<1
You also know that x is negative, so -x<1>-1

Does this help? _________________

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.

I'm new to the GMAT community. I took the GMAT test late last week, unfortunately I did not get the results I was hoping for. I did intensive studying leading to this test (2-5 months), so I'm not sure what went wrong. My mistake was that I did not do practice test prior to the exam which was a terrible idea. I gave myself a few days to feel sorry for myself, and I'm now ready for studying. I must improve the math section. I was wondering if anyone have any advice for me. I am planning to take the test end of May.

Really helpful basic maths principles...though I am an Engineer and am used to seeing numbers all the time....but I have been lacking on the very basic Mathjs..this will really help

For any number: double the units digit and subtract it from the remaining digits. If the resultant number is divisible by 7 then the number is divisible by 7.

To illustrate take : 343 Double the units digit 3 ---- > 6 and subtract it from 34 ---- > 34 - 6 = 28, which is divisible by 7 therefore 343 is divisible by 7.

Any number is divisible by 7 y the sum of the unit digit and 3 times the others are divisible y 7. For example: 406 is divisible by 7 since 6 + 3(40) = 6 + 120 = 126 which is evenly divisible by 7.

Can someone explain the division test for 14. If there is one.

I had a test that wanted to know the sum of numbers <500 that had a remainder of 1 when divided by 14.

I am not aware of test of divisibility for 14 but as far as this problem goes I don't think we need to know one either... First of all we need to find number < 500 which when divided by 14 gives 1. We can use Arithmetic progression to solve this problem. The first number in the list would be 15 (15 / 14 gives reminder 1) and the other number would all be incremented by 14..So the list looks something like 15,29,43,.... The last number in this list would be 491 (490 is the last number less than 500 divisible by 14) full list is 15,29,43,....491. The number of elements can be found with the formula First number + (n -1 ) diff = last number substituting values here

15 + (n -1) 14 = 491 --> n =35

Sum of numbers in AP is given by n/2(First Num + Last Num) = 35/2(15+491) = 8855.

So the answer is 8855.

Please let me know if i have gone something wrong.

You wouldn't need to know more than up to 20 in my opinion but I'm going to learn up to 20 - the less time it takes to work out this stuff on the exam the better.

E.g. when you get a large number like 221 - ... which is divisible by 17, the rule for divisibility by 17 is: "Subtract 5 times the last digit from the rest. 221: 22 - (1 × 5) = 17."

What is everyone else's opinion on this - know divisibility rules for numbers up to and including 20 or only up until 11 or 12?

The sum of N consecutive integers: 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.

The sum of N consecutive integers: 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.

i don't understand this.....

Suppose you are adding up N consecutive integers from n+1 to m. In other words, m-n=N, or, m=N+n. And let the sum be S. Now, let's say S1=1+2+...+n S2=1+2+..+n+(n+1)+...+m Then S=S2-S1

You know that S1=n(n+1)/2 S2=m(m+1)/2 Therefore S=(m(m+1)-n(n+1))/2 =(m^2-n^2+m-n)/2 =(m-n)(m+n+1)/2 =N(N+2n+1)/2 =(N^2+2nN+N)/2 =N(N+1)/2+nN

therefore, S/N=(N+1)/2+n

We want to see if S can be divisible by N. If N is odd, then N+1 is divisible by 2 and S/N is an integer. In other words, S is divisible by N, or S is a multiple of N.

If N is even, then S/N is not an integer. S is not a multiple of N. _________________

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.

I spent a good chunk of yesterday binge-watching installments of “ Handicapping Your MBA Odds ” that John Byrne from Poets and Quants and Sandy Kreisberg with HBS Guru...

I recently returned from attending the London Business School Admits Weekend held last week. Let me just say upfront - for those who are planning to apply for the...