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Arithmetic Progression formula (i.e. AP) is not a part of [#permalink]
29 Jun 2007, 01:57

2

This post received KUDOS

Arithmetic Progression formula (i.e. AP) is not a part of the OG prep material. I have never seen it on GmatPrep but maybe that’s because I never get questions that are above the 600 level (adaptive test, remember ).

I attached some questions that can be solved by using the AP formula but can be solved just as well by using common sense or other methods of solving.

If anyone encountered AP questions on the real GMAT or on GmatPrep, please let me know.

X is positive integer less than 500.When x is divided by 7 the remainder is 1; when x is divided by 3, the remainder is 2. How many such numbers are possible?

I like this topic. In my mind, the fact that you can do all of these without the AP formula is important. I always just use what I know about consecutive numbers to solve. But I guess it doesn't really matter - the first part of each solution is the same - still need to know about remainders, about number properties, and a few other things. Once you get the list of numbers written out, it's up to you how you want to solve it.

I like this topic. In my mind, the fact that you can do all of these without the AP formula is important. I always just use what I know about consecutive numbers to solve. But I guess it doesn't really matter - the first part of each solution is the same - still need to know about remainders, about number properties, and a few other things. Once you get the list of numbers written out, it's up to you how you want to solve it.

I agree ! Using fancy formulas is worthless unless you understand the logic behind them. Some of the above questions can be solved without AP and only by applying common sense. Question number three, can be solved by using simple trial and error:

I like this topic. In my mind, the fact that you can do all of these without the AP formula is important. I always just use what I know about consecutive numbers to solve. But I guess it doesn't really matter - the first part of each solution is the same - still need to know about remainders, about number properties, and a few other things. Once you get the list of numbers written out, it's up to you how you want to solve it.

I agree ! Using fancy formulas is worthless unless you understand the logic behind them. Some of the questions above can be solved without AP and only by apllying common sense. Question number three, can be solved by using simple trial and error:

8,29,50, ... n

21*22-8 = 454 21*23-8 = 475 21*24-8 = 496

n = 24

Once i figured out the list (8, 29, 50...) i realized that we want all multiples of 21 with remainder 8 below 500. So I divided 500 by 21, and got 23 remainder 17. So 23*21 is 483, and I added 8 to get to 491, which is the last one.

So then it was just a matter of counting number of numbers between 8 and 491 when the difference is 21. My way:

X is positive integer less than 500.When x is divided by 7 the remainder is 1; when x is divided by 3, the remainder is 2. How many such numbers are possible?

A. 18
B. 20
C. 22
D. 24
E. 25

x = 7k+1 =3m +2 for some k,m nonnegative integers

7k = 3m + 1

k = (3m + 1)/7
Since k must be an integer,
m = 7r + 2 where r is a nonnegative integer

so x = 21mr + 8

0 < 21mr + 8 < 500
-8 < 21mr < 492
mr is between 0 and 23 inclusive
24 possibilities

X is positive integer less than 500.When x is divided by 7 the remainder is 1; when x is divided by 3, the remainder is 2. How many such numbers are possible?

A. 18 B. 20 C. 22 D. 24 E. 25

x = 7k+1 =3m +2 for some k,m nonnegative integers

7k = 3m + 1

k = (3m + 1)/7 Since k must be an integer, m = 7r + 2 where r is a nonnegative integer

so x = 21mr + 8

0 < 21mr + 8 < 500 -8 < 21mr < 492 mr is between 0 and 23 inclusive 24 possibilities

can someone please explain this to me?
_________________

Success is my only option, failure is not -- Eminem

X is positive integer less than 500.When x is divided by 7 the remainder is 1; when x is divided by 3, the remainder is 2. How many such numbers are possible?

A. 18 B. 20 C. 22 D. 24 E. 25

x = 7k+1 =3m +2 for some k,m nonnegative integers

7k = 3m + 1

k = (3m + 1)/7 Since k must be an integer, m = 7r + 2 where r is a nonnegative integer

so x = 21mr + 8

0 < 21mr + 8 < 500 -8 < 21mr < 492 mr is between 0 and 23 inclusive 24 possibilities

I got lost after that first part. Can somebody explain the part in red?

we know that x is an integer ... then 7K=3M+1 has to b an integer

which mean K is an integer..and

in order for k to be an integer (3m+1)/7 has to be an integer correct??

OK...

the only way 3m+1 can be an integer is m*3 is a multiple of 7 and 1 somehow becomes a multiple of 7

m=(7r+2)*3 ; suppose r=1

then 3(7 +2) +1= will always result in a multiple of 7!

hope this helps!!

Brilliant piece of work by Kevin..

beckee529 wrote:

kevincan wrote:

X is positive integer less than 500.When x is divided by 7 the remainder is 1; when x is divided by 3, the remainder is 2. How many such numbers are possible?

A. 18 B. 20 C. 22 D. 24 E. 25

x = 7k+1 =3m +2 for some k,m nonnegative integers

7k = 3m + 1

k = (3m + 1)/7 Since k must be an integer, m = 7r + 2 where r is a nonnegative integer

so x = 21mr + 8

0 < 21mr + 8 < 500 -8 < 21mr < 492 mr is between 0 and 23 inclusive 24 possibilities

I got lost after that first part. Can somebody explain the part in red?

Re: Using the Arithmetic Progression formula - 101 [#permalink]
14 Aug 2007, 09:34

KillerSquirrel wrote:

Arithmetic Progression formula (i.e. AP) is not a part of the OG prep material. I have never seen it on GmatPrep but maybe that’s because I never get questions that are above the 600 level (adaptive test, remember ).

I attached some questions that can be solved by using the AP formula but can be solved just as well by using common sense or other methods of solving.

If anyone encountered AP questions on the real GMAT or on GmatPrep, please let me know.

X is positive integer less than 500.When x is divided by 7 the remainder is 1; when x is divided by 3, the remainder is 2. How many such numbers are possible?

For question#1, do you consider 1 as an option?
I imagine you should start from 4 since by definition, 1/3 is really no remainder. Shouldn't the answer be 16?

Re: Using the Arithmetic Progression formula - 101 [#permalink]
15 Aug 2007, 00:41

bkk145 wrote:

For question#1, do you consider 1 as an option? I imagine you should start from 4 since by definition, 1/3 is really no remainder. Shouldn't the answer be 16?

No - to me 1/3 is zero (0) with remainder 1, since "3 goes into 1 zero (0) times" leaving a remainder of 1, and 3*0+1 = 1

Re: Using the Arithmetic Progression formula - 101 [#permalink]
30 Jan 2008, 01:16

KillerSquirrel wrote:

bkk145 wrote:

For question#1, do you consider 1 as an option? I imagine you should start from 4 since by definition, 1/3 is really no remainder. Shouldn't the answer be 16?

No - to me 1/3 is zero (0) with remainder 1, since "3 goes into 1 zero (0) times" leaving a remainder of 1, and 3*0+1 = 1

22/39 part/whole

22 is the remainder 39 parts comprise the total
_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

gmatclubot

Re: Using the Arithmetic Progression formula - 101
[#permalink]
30 Jan 2008, 01:16