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Arithmetic Progression formula (i.e. AP) is not a part of

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

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.

[------------------------------------------------------------------------------------------------------------------------]

problem 1

How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3?

A. 14
B. 15
C. 16
D. 17
E. 18

[------------------------------------------------------------------------------------------------------------------------]

problem 2

If a sequence of 8 consecutive odd integers with increasing values has 9 as its 7th term, what is the sum of the terms of the sequence?

A. 22
B. 32
C. 36
D. 40
E. 44

[------------------------------------------------------------------------------------------------------------------------]

problem 3

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

[------------------------------------------------------------------------------------------------------------------------]

solution 1

Using AP:

First find the Progression

(1, 4, 7… 49) = so p=3

Apply AP formula

49 = 1+ (n-1)*3

n = 17

The answer is (D)

[------------------------------------------------------------------------------------------------------------------------]

solution 2

x, x+2, x+4…x+14,x+16

Finding for x+14 = 9 (the 7th term)

Progression = 2

9 = x + (7-1)*2 =

x = -3

Sum (-3,-1, 1, 3, 5, 7, 9, 11) = 32

The answer is (B)


[------------------------------------------------------------------------------------------------------------------------]

solution 3

7*n+1 = x

8,15,22,29....

3*n+2 = x

5,8,11,14,17,20,23,26,29....

Using the AP formula:

Combined the lists:

8, 29, 50 … 491 to find p

Progression = 29-8 = 21

491 = 8+ (n-1)*21 = 21*n-13

n = 24

the answer is (D)

:-D
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CIO
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 [#permalink] New post 29 Jun 2007, 07:59
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.
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Re: Using the Arithmetic Progression formula - 101 [#permalink] New post 29 Jun 2007, 08:10
nice one!
thanks killer.
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 [#permalink] New post 29 Jun 2007, 08:45
ian7777 wrote:
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:

8,29,50, ... 491

21*n-8 = ?

21*22-8 = 454
21*23-8 = 475
21*24-8 = 496
21*25-8 = 517 (too big)

n = 24

:-D

Last edited by KillerSquirrel on 29 Jun 2007, 08:54, edited 2 times in total.
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 [#permalink] New post 29 Jun 2007, 08:49
KillerSquirrel wrote:
ian7777 wrote:
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:

491 - 8 = 483
483/21 = 23
23+1 = 24

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 [#permalink] New post 29 Jun 2007, 10:03
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
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 [#permalink] New post 31 Jul 2007, 19:34
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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


can someone please explain this to me?
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 [#permalink] New post 31 Jul 2007, 20:31
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?
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 [#permalink] New post 14 Aug 2007, 08:26
awsome discussion...

now lets have one on geometric progression shall we???
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 [#permalink] New post 14 Aug 2007, 08:31
beckee: here is what kevin is saying

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?
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Re: Using the Arithmetic Progression formula - 101 [#permalink] New post 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 :wink: ).

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.

[------------------------------------------------------------------------------------------------------------------------]

problem 1

How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3?

A. 14
B. 15
C. 16
D. 17
E. 18

[------------------------------------------------------------------------------------------------------------------------]

problem 2

If a sequence of 8 consecutive odd integers with increasing values has 9 as its 7th term, what is the sum of the terms of the sequence?

A. 22
B. 32
C. 36
D. 40
E. 44

[------------------------------------------------------------------------------------------------------------------------]

problem 3

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

[------------------------------------------------------------------------------------------------------------------------]

solution 1

Using AP:

First find the Progression

(1, 4, 7… 49) = so p=3

Apply AP formula

49 = 1+ (n-1)*3

n = 17

The answer is (D)

[------------------------------------------------------------------------------------------------------------------------]

solution 2

x, x+2, x+4…x+14,x+16

Finding for x+14 = 9 (the 7th term)

Progression = 2

9 = x + (7-1)*2 =

x = -3

Sum (-3,-1, 1, 3, 5, 7, 9, 11) = 32

The answer is (B)


[------------------------------------------------------------------------------------------------------------------------]

solution 3

7*n+1 = x

8,15,22,29....

3*n+2 = x

5,8,11,14,17,20,23,26,29....

Using the AP formula:

Combined the lists:

8, 29, 50 … 491 to find p

Progression = 29-8 = 21

491 = 8+ (n-1)*21 = 21*n-13

n = 24

the answer is (D)

:-D


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?
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Re: Using the Arithmetic Progression formula - 101 [#permalink] New post 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

:-D
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Re: Using the Arithmetic Progression formula - 101 [#permalink] New post 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

:-D


22/39
part/whole

22 is the remainder
39 parts comprise the total
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Re: Arithmetic Progression formula (i.e. AP) is not a part of [#permalink] New post 16 May 2014, 05:20
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Re: Arithmetic Progression formula (i.e. AP) is not a part of   [#permalink] 16 May 2014, 05:20
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