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the product of 4 consecutive numbers will always be a multiple of 4...
the product of 5 consecutive numbers will always be a multiple of 5...and so on
the sum of 3 consecutive number is a multiple of 3...
the sum of 5 consecutvie number is a multiple of 5....
so sum of consecutive numbers is a multiple of the term N, if N is odd....
HongHu wrote:
Arsene_Wenger wrote:
ywilfred wrote:
adding more...
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.
Total = N(A) + N(B) + N(C) - N(A n B) - N(A n C) - N(C n B) + N(A n B n C)
If instead of numbers for (A n B) and (A n C) and (C n B), what is given is the total number of people who choose exactly two items, then the formula becomes:
Total = N(A) + N(B) + N(C) - (N(choose exactly two items)) - 2N(choose all three items)
Also,
Total = N(A) + N(B) + N(C) - (N(choose at least two items)) - N(choose all three items) _________________
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 24 Nov 2005, 19:56, edited 1 time in total.
Is there a useful formula or strategy of attacking remainder problems besides picking numbers? I only solve these mingaz by brute force..
e.g. question:
X is an integer, what is the remainder when it is divided by 10? 1) When divided by 5, the remainder is 2 2). When divided by 2, the remainder is 1
A useful strategy to help you thinking about reminder and multiple problems is to express your variable with an algebra expression.
In your example:
1) X=5m+2
Obiously if m is even then 5m is divided by 10 and the reminder of x/10 would be 2. If m is odd then the reminder would be 7. So insufficient.
2) X=2n+1
This says X is odd. Obviously all odd numbers divided by 10 yield many different reminders. So insufficient.
Combined, 2) would mean that m in 1) is odd and thus we would be able to determine the reminder. _________________
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.
Re: Cancelling out "common terms" on both sides [#permalink]
01 Nov 2005, 18:52
Guess inequality is one of my weak areas.
Could you further explain the following two examples?
Example:
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
Example:
x>1/x
Again you CAN'T multiply both sides by x because you don't know if x is positive or negative. What you have to do is to move the right side to the left:
x-1/x>0
(x^2-1)/x>0
If x>0 then x^2-1>0 =>x>1
If x<0 then x^2-1<0 =>x>-1
Therefore your solution is x>1 or 0>x>-1.
You could also break the original question to two branches from the beginning:
x>1/x
if x>0 then x^2>1 =>x>1
if x<0 then x^2<1 => x>-1
Therefore your solution is x>1 or 0>x>-1.[/quote]
The second method seems easier...not following the first one.
Care to explain further?
|y|>|y+1|
if y>=0, y+1>=0, y>y+1, no solution.
if y<0, y+1<0, -y>-(y+1), solution is y<-1
if y>=0, y+1<0, y>-(y+1), no solution.
if y<0, y+1>=0, -y>y+1, solution is -1<=y<-1/2
So your final solution is y<-1/2
You could also solve this question by going the square route.
y^2>(y+1)^2
y^2>y^2+2y+1
2y+1<0
y<-1/2[/quote]
The second method seems easier...not following the first one.
For your question about the first example, x^2>x, you could also adopt the second approach, that is to break the question into two parts. If x>0, then you can divide both side by x and you'll get x>1. If x<0, then obviously x^2 is always greater than x since x is negative and x^2 is positive. So combine the two you get the correct answer: x>1 or x<0. This approach is less mathematic, and more time consuming, but it may be more straight forward for some people. You need to evaluate the two methods yourself to see which approach you would like to take.
Quote:
Care to explain further?
|y|>|y+1| if y>=0, y+1>=0, y>y+1, no solution. if y<0, y+1<0, -y>-(y+1), solution is y<-1 if y>=0, y+1<0, y>-(y+1), no solution. if y<0, y+1>=0, -y>y+1, solution is -1<=y<-1/2 So your final solution is y<-1/2 You could also solve this question by going the square route. y^2>(y+1)^2 y^2>y^2+2y+1 2y+1<0 y<-1/2
Absolute value questions are definitely going to be on the test, and it is often easy to miss. Just remember that absolute values are always positive, but the things inside the absolute value sign may or may not be positive. This is why you need to break them into two or more parts, one being the inside is positive, another being the inside is negative.
If you are comfortable with the more algebra approach in the first example, then squaring absolute values may be a short cut for such questions sometimes. As you can see from the example, it took much less steps of reasoning. When I have time, I often try both ways, see if I get the same answer, then I'd know that I've got the correct answer. _________________
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.
Re: Cancelling out "common terms" on both sides [#permalink]
03 Nov 2005, 21:17
Thanks, HH...How about the following example. I was not clear on these 2 parts:
(x^2-1)/x>0 If x>0 then x^2-1>0 =>x>1 If x<0 then x^2-1<0 =>x>-1 Therefore your solution is x>1 or 0>x>-1.
AND
You could also break the original question to two branches from the beginning:
x>1/x if x>0 then x^2>1 =>x>1 if x<0 then x^2<1 => x>-1 Therefore your solution is x>1 or 0>x>-1.
Example:
x>1/x
Again you CAN'T multiply both sides by x because you don't know if x is positive or negative. What you have to do is to move the right side to the left:
x-1/x>0
(x^2-1)/x>0
If x>0 then x^2-1>0 =>x>1
If x<0 then x^2-1<0 =>x>-1
Therefore your solution is x>1 or 0>x>-1.
You could also break the original question to two branches from the beginning:
x>1/x
if x>0 then x^2>1 =>x>1
if x<0 then x^2<1 => x>-1
Therefore your solution is x>1 or 0>x>-1.[/quote]
Re: Cancelling out "common terms" on both sides [#permalink]
03 Nov 2005, 21:25
AMBA wrote:
Thanks, HH...How about the following example. I was not clear on these 2 parts:
(x^2-1)/x>0
Here you want to multiple x to solve the equation. Because x is in the determinator we know that x can't be zero. However we don't know if x is positive or negative. And this is important because it will determine whether the sign of the inequality gets changed. Therefore you do the following breaking it into two parts: positive x or negative x.
Quote:
If x>0 then x^2-1>0 =>x>1 If x<0 then x^2-1<0 =>x>-1 Therefore your solution is x>1 or 0>x>-1.
The other way of solving it is to simplify the equation first. So (x^2-1)/x=x-1/x. Then you evaluate x-1/x>0, or x>1/x>0, by breaking it into two parts according to whether x is positive or negative.
Quote:
You could also break the original question to two branches from the beginning: x>1/x if x>0 then x^2>1 =>x>1 if x<0 then x^2<1 => x>-1 Therefore your solution is x>1 or 0>x>-1.
_________________
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.
Not quite sure how to derive the following 4 'if' scenario:
|y|>|y+1|
if y>=0, y+1>=0, y>y+1, no solution. if y<0, y+1<0, -y>-(y+1), solution is y<-1 if y>=0, y+1<0, y>-(y+1), no solution. if y<0, y+1>=0, -y>y+1, solution is -1<=y<-1/2 So your final solution is y<-1/2
I'll add on some more over a period of time, but here are for starters:
Divisibility rules: - Know them well
Integer is divisible by: 2 - Even integer 3 - Sum of digits are divisible by 3 4 - Integer is divisible by 2 twice or Last 2 digits are divisible by 4 5 - Last digit is 0 or 5 6 - Integer is divisbile by 2 AND 3 8 - Integer is divisible by 2 three times 9 - Sum of digits is divisible by 9 10 - Last digit is 0
Some other things to note: - If 2 numbers have the same factor, then the sum or difference of the two numbers will have the same facor.
(e.g. 4 is a factor of 20, 4 is also a factor of 80, then 4 will be a factor of 60 (difference) and also 120 (sum))
- Remember to include '1' if you're asked to count the number of factors a number has
While counting the number of factors, we also need to count the number it self as a factor.
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
Adding 3, 4, 5, 6 will give a sum of 18 and 4 is not a factor of 18. Can you explain please? _________________
Yes, this question has been asked later in the thread.
HongHu wrote:
Arsene_Wenger wrote:
ywilfred wrote:
adding more...
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.
Is there a useful formula or strategy of attacking remainder problems besides picking numbers? I only solve these mingaz by brute force..
e.g. question:
X is an integer, what is the remainder when it is divided by 10? 1) When divided by 5, the remainder is 2 2). When divided by 2, the remainder is 1
Sure, nero44 I'll put in my share and post some remainder rules
There are several rules concerning remainder theory that you might consider to be useful in dealing with remainder problems. First, there are some rules about dividing odd and even numbers. I' post them.
1. When an odd number is divided by an even number, the remainder will be odd regardless. Irrespective whether the quotient is even or odd, the product of the divisor and divident is even (since when you multiply any number by an even number, the product is even). Therefore, the remainder which is the difference between the product(even) and the dividend (odd) will be odd. Example, when a(odd)/b(even), the remainder will always be odd.
2. When odd number is divided by odd number, the remainder might exist. If the quotient is even, the remainder will inevitably exist. if the quotient is odd, the remainder will be even. And vice versa, if the quotient is even, the remainder will be odd. e. g. 9/5 gives the quotient of 1 (odd) and the remainder of 1 (odd) or 29/7 will give 4 as an even quotient and the remainder of 1 (odd). this is true for any numbers.
3. Even by odd. If the quotient is odd, the remainder will necessary exist, if the quotient is odd, the remainder will be even and vice versa (same as in 2.)
4. Even by even. Remainder will exist irrespective of the quotient.
This maight not be helpful with the problem above, but it can be brilliant help with remainder problems including even/odd. I'll post some more remainder rules later. _________________
Is there a useful formula or strategy of attacking remainder problems besides picking numbers? I only solve these mingaz by brute force..
e.g. question:
X is an integer, what is the remainder when it is divided by 10? 1) When divided by 5, the remainder is 2 2). When divided by 2, the remainder is 1
Sure, nero44 I'll put in my share and post some remainder rules There are several rules concerning remainder theory that you might consider to be useful in dealing with remainder problems. First, there are some rules about dividing odd and even numbers. I' post them.
1. When an odd number is divided by an even number, the remainder will be odd regardless. Irrespective whether the quotient is even or odd, the product of the divisor and divident is even (since when you multiply any number by an even number, the product is even). Therefore, the remainder which is the difference between the product(even) and the dividend (odd) will be odd. Example, when a(odd)/b(even), the remainder will always be odd.
2. When odd number is divided by odd number, the remainder might exist. If the quotient is even, the remainder will inevitably exist. if the quotient is odd, the remainder will be even. And vice versa, if the quotient is even, the remainder will be odd. e. g. 9/5 gives the quotient of 1 (odd) and the remainder of 1 (odd) or 29/7 will give 4 as an even quotient and the remainder of 1 (odd). this is true for any numbers.
3. Even by odd. If the quotient is odd, the remainder will necessary exist, if the quotient is odd, the remainder will be even and vice versa (same as in 2.)
4. Even by even. Remainder will exist irrespective of the quotient.
This maight not be helpful with the problem above, but it can be brilliant help with remainder problems including even/odd. I'll post some more remainder rules later.
For Even by Odd:
Quotient ODD, Remainder ODD (always)
Quoteint EVEN, remainder EVEN (if exist)
Even by Even:
Remainder might not exist if numerator is perfectly divisible by denominator. 6/3: quotient 3, remainder 0 _________________
Whether you think you can or think you can't. You're right! - Henry Ford (1863 - 1947)
I'll add on some more over a period of time, but here are for starters:
Divisibility rules: - Know them well
Integer is divisible by: 2 - Even integer 3 - Sum of digits are divisible by 3 4 - Integer is divisible by 2 twice or Last 2 digits are divisible by 4 5 - Last digit is 0 or 5 6 - Integer is divisbile by 2 AND 3 8 - Integer is divisible by 2 three times 9 - Sum of digits is divisible by 9 10 - Last digit is 0
Some other things to note: - If 2 numbers have the same factor, then the sum or difference of the two numbers will have the same facor.
(e.g. 4 is a factor of 20, 4 is also a factor of 80, then 4 will be a factor of 60 (difference) and also 120 (sum))
- Remember to include '1' if you're asked to count the number of factors a number has
While counting the number of factors, we also need to count the number it self as a factor.
Yeah, there is very time-saving method to use when you try to solve a problem with finding the number of factors a specific number can have.
1. You have to write the number as the product of primes as a^p*b^q*c^r, where a, b, and c are prime factors and p,q, and r are their powers.
then, the number of factors the product contains will be expressed by the formula (p+1)(q+1)(r+1). e.g. Find the number of all factors of 1435.
1. 1435 can be expressed as 5^1*17^1*19^1
2. total number of factors of 1435 including 1 and 1435 itself is (1+1)*(1+1)*(1+1)=2*2*2=8 factors
However, the number of ways the number can be expressed as the product of of two other numbers can be found using 1/2*(p+1)(q+1)(r+1)
e.g. In how many ways 1620 can be expressed as the product of two numbers.
1. Express the number as the product of primes. 2^2*3^4*5^1
2. 1/2*(2+1)*(4+1)*(1+1)=15 ways _________________
Is there a useful formula or strategy of attacking remainder problems besides picking numbers? I only solve these mingaz by brute force..
e.g. question:
X is an integer, what is the remainder when it is divided by 10? 1) When divided by 5, the remainder is 2 2). When divided by 2, the remainder is 1
Sure, nero44 I'll put in my share and post some remainder rules There are several rules concerning remainder theory that you might consider to be useful in dealing with remainder problems. First, there are some rules about dividing odd and even numbers. I' post them.
1. When an odd number is divided by an even number, the remainder will be odd regardless. Irrespective whether the quotient is even or odd, the product of the divisor and divident is even (since when you multiply any number by an even number, the product is even). Therefore, the remainder which is the difference between the product(even) and the dividend (odd) will be odd. Example, when a(odd)/b(even), the remainder will always be odd.
2. When odd number is divided by odd number, the remainder might exist. If the quotient is even, the remainder will inevitably exist. if the quotient is odd, the remainder will be even. And vice versa, if the quotient is even, the remainder will be odd. e. g. 9/5 gives the quotient of 1 (odd) and the remainder of 1 (odd) or 29/7 will give 4 as an even quotient and the remainder of 1 (odd). this is true for any numbers.
3. Even by odd. If the quotient is odd, the remainder will necessary exist, if the quotient is odd, the remainder will be even and vice versa (same as in 2.)
4. Even by even. Remainder will exist irrespective of the quotient.
This maight not be helpful with the problem above, but it can be brilliant help with remainder problems including even/odd. I'll post some more remainder rules later.
For Even by Odd: Quotient ODD, Remainder ODD (always) Quoteint EVEN, remainder EVEN (if exist)
Even by Even: Remainder might not exist if numerator is perfectly divisible by denominator. 6/3: quotient 3, remainder 0
Thanks for correcting me in 3, duttsit, you are right
Apologize for that. _________________
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