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## Number properties help needed

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toddg Just gettin' started!
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Number properties help needed Tue Feb 28, 2012 3:27 pm
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I have my exam in one week but I am still struggling with number properties. Specifically, questions relating to divisibility, factors, and even/odd. I've been going through the forums looking at threads for assistance but haven't had much luck. I'm wondering if anyone has any tips to help me improve in these areas. Thanks!

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Tue Feb 28, 2012 8:41 pm
toddg wrote:
I have my exam in one week but I am still struggling with number properties. Specifically, questions relating to divisibility, factors, and even/odd. I've been going through the forums looking at threads for assistance but haven't had much luck. I'm wondering if anyone has any tips to help me improve in these areas. Thanks!
Hey Todd,

I'm new to this site but NOT to GMAT math problems.

I will do my best to offer you some assistance, but it would help to get some specific questions and then work from there. In the meantime, though, here are some key things to keep in mind...

>>>>>>>>>>
1. All positive integers can be uniquely represented as a product of prime numbers (raised to integer-value powers); you'd be surpised how many problems actually boil down to this simple fact. Some examples:

2 = 2^1
4 = 2^2
36 = 2^2 * 3^2

Once you break down a positive integer into this form, you can easily figure out the number of ALL factors (both prime and non-prime). To calculate this, you add 1 to each power and multiply them together. Working with the above three examples again:

2 = 2^1. Since the base is to the first power, we add one to it, which gives us 2 distinct factors (1 and 2)

4 = 2^2. This time we have the base to the power of two. We add one to it, which gives us 3 distinct factors (1,2, and 4)

36 = 2^2 * 3^2. For this example, we have the first base and the second base to the power of two, which gives us 3*3 = 9 distinct factors (1,2,3,4,6,9,12,18,36)
<<<<<<<<<<<<

2. I'm not sure exactly what you mean by odd/even, but there are some basic formulas to figure out the sums of consecutive integers, odd/even integers, etc.

Sum(where i equals 1 through n) i = n*(n+1)/2
Sum(where i equals 1 through n) i^2 = n*(n+1)*(2n+1)/6
Sum(where i equals all even integers from 1 through n) = n/2*(n/2+1) {assumes n is even}
Sum(where i equals all odd integers from 1 through n) = ((n+1)/2)^2 {assumes n is odd}

>>>>>>>>>

3. Some divisibility tips and ideas (mod's, remainders, etc)

Determining divisibility by 3, 6, and 9:

a. The remainder of an integer when divided by 3 is the same as the remainder for the sum of its digits when divided by 3.

(ie. 13344 -> sum of digits is 15, which means that it is divisible by 3)

b. To determine whether an integer is divisible by 6, we need to see whether it's divisible by 2 and also divisible by 3, since 6 = 2*3. To do that we check to see whether it's even and also apply the divisibility by 3 check.

(ie. 13344 -> since it's even, it is divisible by 6 because we already know it's divisible by 3

c. As with 3, an integer's remainder when divided by 9 is the same as the remainder of the sum of its digits when divided by 9.

(ie. 13344 -> the sum is 15, which is NOT divisible by 9)

<<<<<<<<<

4. Some more involved remainder ideas that will save you TONS of time for the right types of problems...

(note: 6 mod 5 = 1, mod just means the remainder of 6 divided by 5.)

a. (A + B) mod C = [(A mod C) + (B mod C)] mod C

I call this the additive property of remainders, and it works regardless of how many numbers are added up. Here's an example to make it easier:

9999 mod 5 = [9000 mod 5 + 900 mod 5 + 90 mod 5 + 9 mod 5] mod 5 = 9 mod 5 = 4

b. (A * B) mod C = [(A mod C) * (B mod C)] mod C

I call this the multiplicative property of remainders, which really works just like the additive property. Another example:

What is the remainder of 22*77*98*115 when divided by 5? Keep in mind that as I write this, I have absolutely NO intention of multiplying all of that crap out. Let's see how...

(22*77*98*115) mod 5 = (22 mod 5 * 77 mod 5 * 98 mod 5 * 115 mod 5) mod 5 = (2*2*3*0) mod 5 = 0

c. (A^B) mod C = [(A mod C)^B] mod C

I call this the power property of remainders. This one really saves time when used properly and is actually pretty powerful. An example to show you what I mean:

7^112234 mod 2 = [(7 mod 2)^112234] mod 2 = 1^112234 mod 2 = 1 mod 2 = 1

Hopefully that gets you started.

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toddg Just gettin' started!
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Tue Feb 28, 2012 9:01 pm
Wow! Thanks for taking the time and effort to explain some great little tricks. I'll see if I can post some sample problems I am having trouble with later.

toddg Just gettin' started!
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Tue Feb 28, 2012 11:51 pm
Any tricks/tips for the following:

1)If y is a prime number greater than 5 and x = 10y, how many different positive even divisors does x have, including x ?

2)If x = 5y + 4 and y is an integer, then which of the following cannot be a factor of x ?

3)Consider a sequence of numbers given by the expression 5 + (n - 1) * 3, where n runs from 1 to 85. How many of these numbers are divisible by 7?

Thanks!

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Wed Feb 29, 2012 2:43 pm
toddg wrote:
Any tricks/tips for the following:

1)If y is a prime number greater than 5 and x = 10y, how many different positive even divisors does x have, including x ?

2)If x = 5y + 4 and y is an integer, then which of the following cannot be a factor of x ?

3)Consider a sequence of numbers given by the expression 5 + (n - 1) * 3, where n runs from 1 to 85. How many of these numbers are divisible by 7?

Thanks!
1)If y is a prime number greater than 5 and x = 10y, how many different positive even divisors does x have, including x ?

OK, great question involving prime numbers. Let's break it down one step at a time...

a. y is a prime number (greater than 5)

Remember that all prime numbers are divisible by themselves and 1. That's IT! Also, all prime numbers are odd (other than 2), which means that y must be ODD, since it is greater than 5.

So, following the formula I gave you dealing with numbers and their factors, we can represent y as such:

y = y^1

And, since y is odd, we know that y has ZERO even divisors. But what about x?

b. x = 10y

Well, we know that y is prime and greater than 5, so can break down x as follows:

x = 10y = 2^1 * 5^1 * y^1

This brings us to another formula I gave you: how to find the total number of unique factors. We simply add 1 to the power of each base and multiply them together: (2*2*2) = 8 factors

But how do we generate a list of all of the factors? I didn't tell you how to do that in my notes, but it is easy once you do it a few times. We simply enumerate through all combinations of powers for each of x's prime factors (from 0 through 1 in this case).

Here's a chart to explain what I'm describing:

By going through all possible powers of the three factors of x -- and then multiplying the results of the bases to those powers together -- we are able to compute each distinct factor.

Also notice that the last four rows are the only factors that are even, because they have 2 as a factor (multiplying an integer by an even number creates an even product).

So, the answer is that their are 4 even divisors of x.

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Wed Feb 29, 2012 2:44 pm
toddg wrote:
Any tricks/tips for the following:

1)If y is a prime number greater than 5 and x = 10y, how many different positive even divisors does x have, including x ?

2)If x = 5y + 4 and y is an integer, then which of the following cannot be a factor of x ?

3)Consider a sequence of numbers given by the expression 5 + (n - 1) * 3, where n runs from 1 to 85. How many of these numbers are divisible by 7?

Thanks!
2)If x = 5y + 4 and y is an integer, then which of the following cannot be a factor of x ?

I will need the choices for this one.

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Wed Feb 29, 2012 3:00 pm
3)Consider a sequence of numbers given by the expression 5 + (n - 1) * 3, where n runs from 1 to 85. How many of these numbers are divisible by 7?

This is an interesting problem; the first thing I'd like to do is simplify the formula above:

5+(n-1)*3 = 3n+2

We're dealing with a relatively small divisor here (7), so at most we will have seven remainders until it resets (the divisor, by the way, will ALWAYS be the max number of remainders before repeat occurs with a linear function like 3n+2):

So, we can see by the chart that n = 4 is divisible by 7. We also see that the pattern repeats itself at n = 8, which is exactly 7 more than its previous value (n = 1). So we have the following values that are divisible by 7 (starting at 4 and increasing by 7's):

4,11,18,25,32,39,46,53,60,67,74,81

Counting them, we get 12.

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toddg Just gettin' started!
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Wed Feb 29, 2012 4:05 pm
I can not thank you enough, these examples and tips are really helping me grasp some of the number properties. Sorry I forgot to include the answer choices on #2, it was quite late when I posted it lol. I'll see if I can find that problem again so I can add in the answer choices. If it is ok with you, I'd like to post a few more examples as I come across them.

krusta80 Really wants to Beat The GMAT!
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Wed Feb 29, 2012 4:52 pm
toddg wrote:
I can not thank you enough, these examples and tips are really helping me grasp some of the number properties. Sorry I forgot to include the answer choices on #2, it was quite late when I posted it lol. I'll see if I can find that problem again so I can add in the answer choices. If it is ok with you, I'd like to post a few more examples as I come across them.
No problem!

toddg Just gettin' started!
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Thu Mar 01, 2012 9:21 pm
1) If x and y are positive integers, which of the following CANNOT be the greatest common divisor of 35x and 20y?

-5
-5(x - y)
-20x
-20y
-35x

2) If integer k is equal to the sum of all even multiples of 15 between 295 and 615, what is the greatest prime factor of k?

-5
-7
-11
-13
-17

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Thu Mar 01, 2012 10:17 pm
toddg wrote:
If x and y are positive integers, which of the following CANNOT be the greatest common divisor of 35x and 20y?

5
5(x - y)
20x
20y
35x
Rule of thumb: substitution is great when they ask which of the following CAN be something but not so much for CANNOT, so let's start by breaking down 35x and 20y as the product of factors.

35x = 5^1 * 7^1 * x
20y = 2^2 * 5^1 * y

From the breakdown, we can see that 35 and 20 have a GCD of 5. Therefore, the GCD of 35x and 20y will be 5*GCD(7x,4y)

Let's look at the choices now...

A) 5

This holds true if GCD(7x,4y) = 1, which works for x not equal to any factor or multiple of 4 or y and for y not any multiple of 7 or x. x = y = 1 works nicely.

B) 5*(x-y)

This holds true if GCD(7x,4y) = x-y, which works if we find a value for x-y that is not a multiple or factor of 4 or 7, while also making sure that x is not a multiple of factor of 4 and that y is not a multiple of 7 AND that x and y are both multiples of x-y. Let's try x=9 and y=6, which gives x-y = 3.

C) 20*x = 5*(4x)

This holds true if GCD(7x,4y) = 4x, which will NEVER work because x can never have 4x as a factor!

D) 20*y = 5*(4y)

This holds true if GCD(7x,4y) = 4y, which will work if x = 4y and y not equal a multiple of 7.

E) 35*x = 5*(7x)

This holds true if GCD(7x,4y) = 7x, which will work if y = 7x and x not a factor or multiple of 4.

OK, so maybe I'm missing a good shortcut with this problem, but hopefully my explanations help build a deeper understanding of these sorts of problems. I welcome anyone else to offer a simpler method of solution.

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Thu Mar 01, 2012 10:27 pm
toddg wrote:
2) If integer k is equal to the sum of all even multiples of 15 between 295 and 615, what is the greatest prime factor of k?

-5
-7
-11
-13
-17
An even multiple of 15 is the same thing as a multiple of 30. Since the jumps are so big, let's add up the numbers manually:

+300
+330
+360
+390
+420
+450
+480
+510
+540
+570
+600
----
4950

We can either perform the division the long way, or we can factorize the total first. Since we know that it's a multiple of 30, let's divide it by that and keep going:

4950 = 30 * 165 = 30 * 5 * 33 = 30 * 5 * 3 * 11 = 2*3*5*5*3*11

11 is the biggest factor

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toddg Just gettin' started!
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Fri Mar 02, 2012 10:51 am
[quote="krusta80"]
toddg wrote:
OK, so maybe I'm missing a good shortcut with this problem, but hopefully my explanations help build a deeper understanding of these sorts of problems. I welcome anyone else to offer a simpler method of solution.
Trust me, they are very helpful. Your explanations have been helping me grasp some of the more abstract number property questions.

Also, any shortcuts to see if a number is divisible by 7? I have all of the other divisibility rules memorized.

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Fri Mar 02, 2012 1:47 pm
[quote="toddg"]
krusta80 wrote:
toddg wrote:
OK, so maybe I'm missing a good shortcut with this problem, but hopefully my explanations help build a deeper understanding of these sorts of problems. I welcome anyone else to offer a simpler method of solution.
Trust me, they are very helpful. Your explanations have been helping me grasp some of the more abstract number property questions.

Also, any shortcuts to see if a number is divisible by 7? I have all of the other divisibility rules memorized.
To be honest, I've never known of one, but a quick Google came up with this: http://www.johndcook.com/blog/2010/10/27/divisibility-by-7/

Honestly, it's probably easier just to actually divide by 7. lol

toddg Just gettin' started!
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Mon Mar 05, 2012 7:40 pm
Thanks again for all of your help. I had my GMAT today and while I didn't get my target score of 700, I came reasonably close. Your tips helped me on multiple quant questions significantly. I will more than likely retake it soon, so I'll be sure to get back in touch with you on some of these number properties.

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