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Arithmetic Progression formula (i.e. AP) is not a part of [#permalink]
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29 Jun 2007, 02:57
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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 ).
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+ (n1)*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 + (71)*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 = 298 = 21
491 = 8+ (n1)*21 = 21*n13
n = 24
the answer is (D)



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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]
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29 Jun 2007, 09:10
nice one!
thanks killer.



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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*n8 = ?
21*228 = 454
21*238 = 475
21*248 = 496
21*258 = 517 (too big)
n = 24
Last edited by KillerSquirrel on 29 Jun 2007, 09:54, edited 2 times in total.



CIO
Joined: 09 Mar 2003
Posts: 463

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*228 = 454 21*238 = 475 21*248 = 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
Ian



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



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awsome discussion...
now lets have one on geometric progression shall we???



Current Student
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Posts: 3357
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Schools: Wharton'11 HBS'12

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]
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14 Aug 2007, 10:34
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. [] problem 1How 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 2If 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 3X 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 1Using AP: First find the Progression (1, 4, 7… 49) = so p=3 Apply AP formula 49 = 1+ (n1)*3 n = 17 The answer is (D) [] solution 2x, x+2, x+4…x+14,x+16 Finding for x+14 = 9 (the 7th term) Progression = 2 9 = x + (71)*2 = x = 3 Sum (3,1, 1, 3, 5, 7, 9, 11) = 32 The answer is (B) [] solution 37*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 = 298 = 21 491 = 8+ (n1)*21 = 21*n13 n = 24 the answer is (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]
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15 Aug 2007, 01:41
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



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Re: Using the Arithmetic Progression formula  101 [#permalink]
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30 Jan 2008, 02:16
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
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Re: Arithmetic Progression formula (i.e. AP) is not a part of [#permalink]
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