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Joined: 03 Jan 2005
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x^21<0
(x+1)(x1)<0
Since x<0, therefore x1<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?
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Re: Basic Mathematical Principles: [#permalink]
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13 Apr 2009, 06:02
Hi,
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 (25 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.
Than you.



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Re: Basic Mathematical Principles: [#permalink]
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05 May 2009, 06:20
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



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Re: Basic Mathematical Principles: [#permalink]
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06 May 2009, 08:00
Good one really helpful to brush up the basics



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Re: Basic Mathematical Principles: [#permalink]
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05 Aug 2009, 11:14
3
This post received KUDOS
Here's one :
Divisibility test for 7
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.



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Re: Basic Mathematical Principles: [#permalink]
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28 Oct 2009, 05:19
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.



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21 Dec 2009, 09:13
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.



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Re: Basic Mathematical Principles: [#permalink]
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31 Jan 2010, 06:44
a>b,d<c we can write (a/b)>1, (c/d)>1 now, we can multiply, (ac/bd)>1 => (a/b)>(d/c) =>(a/d)>(b/c) =>ac>bd =>(c/b)>(d/a) and so on... can we?
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ywilfred wrote: adding more...
Adding on to the odd/even rule honghu wrote...
Adding/subtracting two odds or two evens > even Add/ Subtract an odd and an even > od
Multiplication with at no even number > odd ** Even number in a multiplication will always ensure an even product
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[strike]4[/strike]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..
 Adding a consecutive set of odd integers will result in sum that is a multiple of the number of integers
E.g. Adding 2,4,6,8,10 > sum will be multiple of 5 (sum=30) Adding 1,3,5 > sum will be multiple of 3 (sum=9)  An even integer in a multiplication > product divisible by 2  2 even integers in a multiplication > product divisible by 4 etc s=3+4+5+6=18 18 is not divided by 4 where am i wrong?
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sanjayism wrote: ywilfred wrote: adding more...
Adding on to the odd/even rule honghu wrote...
Adding/subtracting two odds or two evens > even Add/ Subtract an odd and an even > od
Multiplication with at no even number > odd ** Even number in a multiplication will always ensure an even product
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[strike]4[/strike]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..
 Adding a consecutive set of odd integers will result in sum that is a multiple of the number of integers
E.g. Adding 2,4,6,8,10 > sum will be multiple of 5 (sum=30) Adding 1,3,5 > sum will be multiple of 3 (sum=9)  An even integer in a multiplication > product divisible by 2  2 even integers in a multiplication > product divisible by 4 etc s=3+4+5+6=18 18 is not divided by 4 where am i wrong? The sum of ODD number of consecutive integers is a multiple of number of integers. 2+3+4=9, 3 terms (odd), sum=9 is divisible by 3. The sum of EVEN number of consecutive integers is not a multiple of number of integers. 4+5=9, 2 terms (even), sum=9 is not divisible by 2. Hope it's clear.
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Re: Basic Mathematical Principles: [#permalink]
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01 Feb 2010, 11:10
Thank you! Very useful for me as i´m a beginner with all the GMATThing



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14 Jun 2010, 12:24
s3017789 wrote: 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. Thanks Mani Natarajan



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Re: Basic Mathematical Principles: [#permalink]
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09 Jul 2010, 04:39
http://en.wikipedia.org/wiki/Divisibility_ruleHas all the divisibility rules up to 989 lol 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?



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Re: Basic Mathematical Principles: [#permalink]
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09 Jul 2010, 04:42
btw wikipedia has some interesting stuff on number properties, fractions, factorials etc.
obviously it gets very complex but you should be able to pick out the stuff that is relevant to your GMAT preparation



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ywilfred wrote: Quote: 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.. This does not apply universally look at following example of four consecutive integers 27,28,29,30 > sums to 114 but the sum is not a multiple of 4



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18 Aug 2010, 23:41
excellent ! I´m starting my preparation in quant and I think it's going to be extremely helpfull. Thanks !



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Re: Basic Mathematical Principles: [#permalink]
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01 Oct 2010, 11:23
The sum of N consecutive integers: S=(n+n+N1)*N/2=n*N+N*(N1)/2 S/N=n+(N1)/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.....
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12 Nov 2010, 20:08
shrive555 wrote: The sum of N consecutive integers: S=(n+n+N1)*N/2=n*N+N*(N1)/2 S/N=n+(N1)/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, mn=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=S2S1 You know that S1=n(n+1)/2 S2=m(m+1)/2 Therefore S=(m(m+1)n(n+1))/2 =(m^2n^2+mn)/2 =(mn)(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.
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04 Jan 2011, 20:40



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11 Jan 2011, 13:20
thanks man. great post




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