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Interesting problem. Tests a good concept......... How many

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Senior Manager
Joined: 29 Mar 2008
Posts: 339
Interesting problem. Tests a good concept......... How many  [#permalink]

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07 Oct 2008, 23:19
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Interesting problem. Tests a good concept.........

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

A. 15
B. 16
C. 17
D. 18
E. 19

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Senior Manager
Joined: 26 Jan 2008
Posts: 260
Re: PS: Number of Integers- Good one  [#permalink]

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07 Oct 2008, 23:34
leonidas wrote:
Interesting problem. Tests a good concept.........

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

A. 15
B. 16
C. 17
D. 18
E. 19

[(49 - 1) / 3] + 1 = 17

First number (larger than 0, inclusive) = 1
Last number (smaller than 50, inclusive) = 49

Take difference, divide by 3, and add 1 to include the number 1
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Re: PS: Number of Integers- Good one  [#permalink]

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07 Oct 2008, 23:44
incognito1 wrote:
leonidas wrote:
Interesting problem. Tests a good concept.........

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

A. 15
B. 16
C. 17
D. 18
E. 19

[(49 - 1) / 3] + 1 = 17

First number (larger than 0, inclusive) = 1
Last number (smaller than 50, inclusive) = 49

Take difference, divide by 3, and add 1 to include the number 1

I did it this way.....
The numbers when divided by 3 that give a remainder 1 are : 1, 4, 7, 9,....49
This adds up to 17 [Using, 1+(n-1)3=49)]

I did not pay attention to 1 (1/3 gives a remainder of 1; 2/3 gives a remainder of 2)....something I always forget to consider.

OA is (C) i.e 17.
_________________

To find what you seek in the road of life, the best proverb of all is that which says:
"Leave no stone unturned."
-Edward Bulwer Lytton

Senior Manager
Joined: 26 Jan 2008
Posts: 260
Re: PS: Number of Integers- Good one  [#permalink]

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07 Oct 2008, 23:53
leonidas wrote:
incognito1 wrote:
leonidas wrote:
Interesting problem. Tests a good concept.........

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

A. 15
B. 16
C. 17
D. 18
E. 19

[(49 - 1) / 3] + 1 = 17

First number (larger than 0, inclusive) = 1
Last number (smaller than 50, inclusive) = 49

Take difference, divide by 3, and add 1 to include the number 1

I did it this way.....
The numbers when divided by 3 that give a remainder 1 are : 1, 4, 7, 9,....49
This adds up to 17 [Using, 1+(n-1)3=49)]

I did not pay attention to 1 (1/3 gives a remainder of 1; 2/3 gives a remainder of 2)....something I always forget to consider.

OA is (C) i.e 17.

No doubt 1 was a gotcha. When I see the word remainder, I know I'm supposed to be working with whole numbers 0 through n

Btw, your method (of using the Arithmetic series) is a fail-proof way of solving something like this, so I'd definitely use that.
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Joined: 17 Jun 2008
Posts: 1474
Re: PS: Number of Integers- Good one  [#permalink]

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08 Oct 2008, 01:13
Nice approach from leondas.

I used the following approach.

Since, there are 17 integers between 0 and 50 that are divisible by 3 (including 0) and since the first such integer is 0 and the last is 48....hence there should be 17 such integers that have remainder as 1 when divided by 3 as 0+1 = and 48 + 1 = 49 and both these numbers are within the limits of 0 and 50.
Director
Joined: 20 Sep 2006
Posts: 630
Re: PS: Number of Integers- Good one  [#permalink]

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11 Oct 2008, 07:54

How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3 and have a remainder of 3 when divided by 5 ?

Any idea Gurus?
Senior Manager
Joined: 29 Mar 2008
Posts: 339
Re: PS: Number of Integers- Good one  [#permalink]

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11 Oct 2008, 08:44
1
rao_1857 wrote:

How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3 and have a remainder of 3 when divided by 5 ?

Any idea Gurus?

remainder of 1 when divided by 3: 3k+1 = 1, 4, 7, 10, (13), 16, 19, 22, 25, (28), 31, 34, 37, 40, (43), 46, 49...
remainder of 3 when divided by 5: 5k+3= 3, 8, (13), 18, 23, 28, 33, 38, (43), 48...

Here, 13 and 43 are common in the two series. Other numbers will be 58, 73, so on and so forth.
The general equation is: (28-13)k+13= 15k+ 13 (where, k=0,1,2,...)
Hence, if we want to find the number of integers even upto 1000, we can use the above formula.

Going back to the problem, I see 3 numbers that satisfies the above criteria i.e 13, 28 and 43. (The next one will be 58 which is greater than 50).
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To find what you seek in the road of life, the best proverb of all is that which says:
"Leave no stone unturned."
-Edward Bulwer Lytton

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Joined: 27 Sep 2008
Posts: 76
Re: PS: Number of Integers- Good one  [#permalink]

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11 Oct 2008, 10:24
useing the A.P formula a+(n-1)*d = an {a,a1,a2...an)

a=1 d=3 an=49 n=?

49 = a+(n-1)*d = 1+(n-1)*3 = 1+3n-3

51 = 3n

n = 17

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re: PS: Number of Integers- Good one &nbs [#permalink] 11 Oct 2008, 10:24
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